Stochastic Calculus Solution Manual

Stochastic Calculus for Finance, Volume I and II by Yan Zeng Last updated: August 20, 2007 This is a solution manual for the two-volume textbook Stochastic calculus for ? nance, by Steven Shreve. If you have any comments or ? nd any typos/errors, please email me at [email  protected] edu. The current version omits the following problems. Volume I: 1. 5, 3. 3, 3. 4, 5. 7; Volume II: 3. 9, 7. 1, 7. 2, 7. 5–7. 9, 10. 8, 10. 9, 10. 10. Acknowledgment I thank Hua Li (a graduate student at Brown University) for reading through this solution manual and communicating to me several mistakes/typos. 1. 1. Stochastic Calculus for Finance I: The Binomial Asset Pricing Model 1. The Binomial No-Arbitrage Pricing Model Proof. If we get the up sate, then X1 = X1 (H) = ? 0 uS0 + (1 + r)(X0 ? ?0 S0 ); if we get the down state, then X1 = X1 (T ) = ? 0 dS0 + (1 + r)(X0 ? ?0 S0 ). If X1 has a positive probability of being strictly positive, then we must either have X1 (H) > 0 or X1 (T ) > 0. (i) If X1 (H) > 0, then ? 0 uS0 + (1 + r)(X0 ? ?0 S0 ) > 0. Plug in X0 = 0, we get u? 0 > (1 + r)? 0 . By condition d < 1 + r < u, we conclude ? 0 > 0.In this case, X1 (T ) = ? 0 dS0 + (1 + r)(X0 ? ?0 S0 ) = ? 0 S0 [d ? (1 + r)] < 0. (ii) If X1 (T ) > 0, then we can similarly deduce ? 0 < 0 and hence X1 (H) < 0. So we cannot have X1 strictly positive with positive probability unless X1 is strictly negative with positive probability as well, regardless the choice of the number ? 0 . Remark: Here the condition X0 = 0 is not essential, as far as a property de? nition of arbitrage for arbitrary X0 can be given. Indeed, for the one-period binomial model, we can de? ne arbitrage as a trading strategy such that P (X1 ?X0 (1 + r)) = 1 and P (X1 > X0 (1 + r)) > 0. First, this is a generalization of the case X0 = 0; second, it is â€Å"proper† because it is comparing the result of an arbitrary investment involving money and stock markets with that of a safe investment involving only money market. This can also be seen by regarding X0 as borrowed from money market account. Then at time 1, we have to pay back X0 (1 + r) to the money market account. In summary, arbitrage is a trading strategy that beats â€Å"safe† investment. Accordingly, we revise the proof of Exercise 1. 1. as follows.If X1 has a positive probability of being strictly larger than X0 (1 + r), the either X1 (H) > X0 (1 + r) or X1 (T ) > X0 (1 + r). The ? rst case yields ? 0 S0 (u ? 1 ? r) > 0, i. e. ?0 > 0. So X1 (T ) = (1 + r)X0 + ? 0 S0 (d ? 1 ? r) < (1 + r)X0 . The second case can be similarly analyzed. Hence we cannot have X1 strictly greater than X0 (1 + r) with positive probability unless X1 is strictly smaller than X0 (1 + r) with positive probability as well. Finally, we comment that the above formulation of arbitrage is equivalent to the one in the textbook. For details, see Shreve [7], Exercise 5. . 1. 2. 1 5 Proof. X1 (u) = ? 0 ? 8 + ? 0 ? 3 ? 5 (4? 0 + 1. 20? 0 ) = 3? 0 + 1. 5? 0 , a nd X1 (d) = ? 0 ? 2 ? 4 (4? 0 + 1. 20? 0 ) = 4 ? 3? 0 ? 1. 5? 0 . That is, X1 (u) = ? X1 (d). So if there is a positive probability that X1 is positive, then there is a positive probability that X1 is negative. Remark: Note the above relation X1 (u) = ? X1 (d) is not a coincidence. In general, let V1 denote the ? ? payo? of the derivative security at time 1. Suppose X0 and ? 0 are chosen in such a way that V1 can be ? 0 ? ?0 S0 ) + ? 0 S1 = V1 . Using the notation of the problem, suppose an agent begins ? replicated: (1 + r)(X with 0 wealth and at time zero buys ? 0 shares of stock and ? 0 options. He then puts his cash position ? 0 S0 ? ?0 X0 in a money market account. At time one, the value of the agent’s portfolio of stock, option and money market assets is ? X1 = ? 0 S1 + ? 0 V1 ? (1 + r)(? 0 S0 + ? 0 X0 ). Plug in the expression of V1 and sort out terms, we have ? X1 = S0 (? 0 + ? 0 ? 0 )( S1 ? (1 + r)). S0 ? Since d < (1 + r) < u, X1 (u) and X1 (d) have opposite signs. So if the price of the option at time zero is X0 , then there will no arbitrage. 1. 3. S0 1 Proof. V0 = 1+r 1+r? d S1 (H) + u? ? r S1 (T ) = 1+r 1+r? d u + u? 1? r d = S0 . This is not surprising, since u? d u? d u? d u? d this is exactly the cost of replicating S1 . Remark: This illustrates an important point. The â€Å"fair price† of a stock cannot be determined by the risk-neutral pricing, as seen below. Suppose S1 (H) and S1 (T ) are given, we could have two current prices, S0 and S0 . Correspondingly, we can get u, d and u , d . Because they are determined by S0 and S0 , respectively, it’s not surprising that risk-neutral pricing formula always holds, in both cases. That is, 1+r? d u? d S1 (H) S0 = + u? 1? r u? d S1 (T ) 1+r S0 = 1+r? d u ? d S1 (H) + u ? 1? r u ? d S1 (T ) 1+r . Essentially, this is because risk-neutral pricing relies on fair price=replication cost. Stock as a replicating component cannot determine its own â€Å"fair† price via the risk-n eutral pricing formula. 1. 4. Proof. Xn+1 (T ) = = ? n dSn + (1 + r)(Xn ? ?n Sn ) ?n Sn (d ? 1 ? r) + (1 + r)Vn pVn+1 (H) + q Vn+1 (T ) ? ? Vn+1 (H) ? Vn+1 (T ) (d ? 1 ? r) + (1 + r) = u? d 1+r = p(Vn+1 (T ) ? Vn+1 (H)) + pVn+1 (H) + q Vn+1 (T ) ? ? ? = pVn+1 (T ) + q Vn+1 (T ) ? ? = Vn+1 (T ). 1. 6. 2 Proof. The bank’s trader should set up a replicating portfolio whose payo? s the opposite of the option’s payo?. More precisely, we solve the equation (1 + r)(X0 ? ?0 S0 ) + ? 0 S1 = ? (S1 ? K)+ . 1 Then X0 = ? 1. 20 and ? 0 = ? 2 . This means the trader should sell short 0. 5 share of stock, put the income 2 into a money market account, and then transfer 1. 20 into a separate money market account. At time one, the portfolio consisting of a short position in stock and 0. 8(1 + r) in money market account will cancel out with the option’s payo?. Therefore we end up with 1. 20(1 + r) in the separate money market account. Remark: This problem illustrates why we are in terested in hedging a long position.In case the stock price goes down at time one, the option will expire without any payo?. The initial money 1. 20 we paid at time zero will be wasted. By hedging, we convert the option back into liquid assets (cash and stock) which guarantees a sure payo? at time one. Also, cf. page 7, paragraph 2. As to why we hedge a short position (as a writer), see Wilmott [8], page 11-13. 1. 7. Proof. The idea is the same as Problem 1. 6. The bank’s trader only needs to set up the reverse of the replicating trading strategy described in Example 1. 2. 4. More precisely, he should short sell 0. 1733 share of stock, invest the income 0. 933 into money market account, and transfer 1. 376 into a separate money market account. The portfolio consisting a short position in stock and 0. 6933-1. 376 in money market account will replicate the opposite of the option’s payo?. After they cancel out, we end up with 1. 376(1 + r)3 in the separate money market ac count. 1. 8. (i) 2 s s Proof. vn (s, y) = 5 (vn+1 (2s, y + 2s) + vn+1 ( 2 , y + 2 )). (ii) Proof. 1. 696. (iii) Proof. ?n (s, y) = vn+1 (us, y + us) ? vn+1 (ds, y + ds) . (u ? d)s 1. 9. (i) Proof. Similar to Theorem 1. 2. 2, but replace r, u and d everywhere with rn , un and dn .More precisely, set pn = 1+rn ? dn and qn = 1 ? pn . Then un ? dn Vn = pn Vn+1 (H) + qn Vn+1 (T ) . 1 + rn (ii) Proof. ?n = (iii) 3 Vn+1 (H)? Vn+1 (T ) Sn+1 (H)? Sn+1 (T ) = Vn+1 (H)? Vn+1 (T ) . (un ? dn )Sn 10 10 Proof. un = Sn+1 (H) = Sn +10 = 1+ Sn and dn = Sn+1 (T ) = Sn ? 10 = 1? Sn . So the risk-neutral probabilities Sn Sn Sn Sn at time n are pn = u1? dnn = 1 and qn = 1 . Risk-neutral pricing implies the price of this call at time zero is ? ? 2 2 n ? d 9. 375. 2. Probability Theory on Coin Toss Space 2. 1. (i) Proof. P (Ac ) + P (A) = (ii) Proof. By induction, it su? ces to work on the case N = 2.When A1 and A2 are disjoint, P (A1 ? A2 ) = A1 ? A2 P (? ) = A1 P (? ) + A2 P (? ) = P (A1 ) + P (A2 ). When A1 and A2 are arbitrary, using the result when they are disjoint, we have P (A1 ? A2 ) = P ((A1 ? A2 ) ? A2 ) = P (A1 ? A2 ) + P (A2 ) ? P (A1 ) + P (A2 ). 2. 2. (i) 1 3 1 Proof. P (S3 = 32) = p3 = 8 , P (S3 = 8) = 3p2 q = 3 , P (S3 = 2) = 3pq 2 = 8 , and P (S3 = 0. 5) = q 3 = 8 . 8 Ac P (? ) + A P (? ) = P (? ) = 1. (ii) Proof. E[S1 ] = 8P (S1 = 8) + 2P (S1 = 2) = 8p + 2q = 5, E[S2 ] = 16p2 + 4  · 2pq + 1  · q 2 = 6. 25, and 3 1 E[S3 ] = 32  · 1 + 8  · 8 + 2  · 3 + 0.  · 8 = 7. 8125. So the average rates of growth of the stock price under P 8 8 5 are, respectively: r0 = 4 ? 1 = 0. 25, r1 = 6. 25 ? 1 = 0. 25 and r2 = 7. 8125 ? 1 = 0. 25. 5 6. 25 (iii) 8 1 Proof. P (S3 = 32) = ( 2 )3 = 27 , P (S3 = 8) = 3  · ( 2 )2  · 1 = 4 , P (S3 = 2) = 2  · 1 = 2 , and P (S3 = 0. 5) = 27 . 3 3 3 9 9 9 Accordingly, E[S1 ] = 6, E[S2 ] = 9 and E[S3 ] = 13. 5. So the average rates of growth of the stock price 9 6 under P are, respectively: r0 = 4 ? 1 = 0. 5, r1 = 6 ? 1 = 0. 5, and r2 = 13. 5 ? 1 = 0. 5. 9 2. 3. Proof. Apply conditional Jensen’s inequality. 2. 4. (i) Proof.En [Mn+1 ] = Mn + En [Xn+1 ] = Mn + E[Xn+1 ] = Mn . (ii) 2 n+1 Proof. En [ SSn ] = En [e? Xn+1 e? +e ] = 2 ? Xn+1 ] e? +e E[e = 1. 2. 5. (i) 2 2 Proof. 2In = 2 j=0 Mj (Mj+1 ? Mj ) = 2 j=0 Mj Mj+1 ? j=1 Mj ? j=1 Mj = 2 j=0 Mj Mj+1 + n? 1 n? 1 n? 1 n? 1 2 2 2 2 2 2 2 2 Mn ? j=0 Mj+1 ? j=0 Mj = Mn ? j=0 (Mj+1 ? Mj ) = Mn ? j=0 Xj+1 = Mn ? n. n? 1 n? 1 n? 1 n? 1 n? 1 (ii) Proof. En [f (In+1 )] = En [f (In + Mn (Mn+1 ? Mn ))] = En [f (In + Mn Xn+1 )] = 1 [f (In + Mn ) + f (In ? Mn )] = 2 v v v g(In ), where g(x) = 1 [f (x + 2x + n) + f (x ? 2x + n)], since 2In + n = |Mn |. 2 2. 6. 4 Proof. En [In+1 ?In ] = En [? n (Mn+1 ? Mn )] = ? n En [Mn+1 ? Mn ] = 0. 2. 7. Proof. We denote by Xn the result of n-th coin toss, where Head is represented by X = 1 and Tail is 1 represented by X = ? 1. We also suppose P (X = 1) = P (X = ? 1) = 2 . De? ne S1 = X1 and Sn+1 = n Sn +bn (X1 ,  ·  ·  · , Xn )Xn+1 , where bn ( ·) is a bounded function on {? 1, 1} , to be determined later on. Clearly (Sn )n? 1 is an adapted stochastic process, and we can show it is a martingale. Indeed, En [Sn+1 ? Sn ] = bn (X1 ,  ·  ·  · , Xn )En [Xn+1 ] = 0. For any arbitrary function f , En [f (Sn+1 )] = 1 [f (Sn + bn (X1 ,  ·  ·  · , Xn )) + f (Sn ? n (X1 ,  ·  ·  · , Xn ))]. Then 2 intuitively, En [f (Sn+1 ] cannot be solely dependent upon Sn when bn ’s are properly chosen. Therefore in general, (Sn )n? 1 cannot be a Markov process. Remark: If Xn is regarded as the gain/loss of n-th bet in a gambling game, then Sn would be the wealth at time n. bn is therefore the wager for the (n+1)-th bet and is devised according to past gambling results. 2. 8. (i) Proof. Note Mn = En [MN ] and Mn = En [MN ]. (ii) Proof. In the proof of Theorem 1. 2. 2, we proved by induction that Xn = Vn where Xn is de? ned by (1. 2. 14) of Chapter 1. In other words, the sequence (Vn )0? n?N can be realized as the value process of a portfolio, Xn which consists of stock and money market accounts. Since ( (1+r)n )0? n? N is a martingale under P (Theorem Vn 2. 4. 5), ( (1+r)n )0? n? N is a martingale under P . (iii) Proof. (iv) Proof. Combine (ii) and (iii), then use (i). 2. 9. (i) (H) S1 (H) 1 = 2, d0 = S1S0 = 2 , S0 (T and d1 (T ) = S21 (TT)) = 1. S 1 1 0 ? d So p0 = 1+r? d0 0 = 2 , q0 = 2 , p1 (H) u0 5 q1 (T ) = 6 . Therefore P (HH) = p0 p1 (H) = 1 , 4 5 q0 q1 (T ) = 12 . Vn (1+r)n = En VN (1+r)N , so V0 , V1 1+r ,  ·Ã‚ ·Ã‚ ·, VN ? 1 , VN (1+r)N ? 1 (1+r)N is a martingale under P . Proof. u0 = u1 (H) = =S2 (HH) S1 (H) = 1. 5, d1 (H) = S2 (HT ) S1 (H) = 1, u1 (T ) = S2 (T H) S1 (T ) =4 1+r1 (H)? d1 (H) u1 (H)? d1 (H) 1 = 1 , q1 (H) = 2 , p1 (T ) = 2 1 4, 1+r1 (T )? d1 (T ) u1 (T )? d1 (T ) 1 12 1 = 6 , and P (HT ) = p0 q1 (H) = P (T H) = q0 p1 (T ) = and P (T T ) = The proofs of Theorem 2. 4. 4, Theorem 2. 4. 5 and Theorem 2. 4. 7 still work for the random interest rate m odel, with proper modi? cations (i. e. P would be constructed according to conditional probabilities P (? n+1 = H|? 1 ,  ·  ·  · , ? n ) := pn and P (? n+1 = T |? 1 ,  ·  ·  · , ? n ) := qn . Cf. notes on page 39. ). So the time-zero value of an option that pays o?V2 at time two is given by the risk-neutral pricing formula V0 = E (1+r0V2 1 ) . )(1+r (ii) Proof. V2 (HH) = 5, V2 (HT ) = 1, V2 (T H) = 1 and V2 (T T ) = 0. So V1 (H) = 2. 4, V1 (T ) = p1 (T )V2 (T H)+q1 (T )V2 (T T ) 1+r1 (T ) p1 (H)V2 (HH)+q1 (H)V2 (HT ) 1+r1 (H) = = 1 9, and V0 = p0 V1 (H)+q0 V1 (T ) 1+r0 ? 1. 5 (iii) Proof. ?0 = (iv) Proof. ?1 (H) = 2. 10. (i) Xn+1 Proof. En [ (1+r)n+1 ] = En [ ? n Yn+1 Sn + (1+r)n+1 (1+r)(Xn n Sn ) ] (1+r)n+1 Xn (1+r)n . V2 (HH)? V2 (HT ) S2 (HH)? S2 (HT ) V1 (H)? V1 (T ) S1 (H)? S1 (T ) = 1 2. 4? 9 8? 2 = 0. 4 ? 1 54 ? 0. 3815. = 5? 1 12? 8 = 1. = ?n Sn (1+r)n+1 En [Yn+1 ] + Xn Sn (1+r)n = ?n Sn (1+r)n+1 (up + dq) + Xn n Sn (1+r)n = ?n Sn +Xn n Sn (1+r)n = (ii) Proof . From (2. 8. 2), we have ? n uSn + (1 + r)(Xn ? ?n Sn ) = Xn+1 (H) ? n dSn + (1 + r)(Xn ? ?n Sn ) = Xn+1 (T ). So ? n = Xn+1 (H)? Xn+1 (T ) uSn ? dSn and Xn = En [ Xn+1 ]. To make the portfolio replicate the payo? at time N , we 1+r VN X must have XN = VN . So Xn = En [ (1+r)N ? n ] = En [ (1+r)N ? n ]. Since (Xn )0? n? N is the value process of the N unique replicating portfolio (uniqueness is guaranteed by the uniqueness of the solution to the above linear VN equations), the no-arbitrage price of VN at time n is Vn = Xn = En [ (1+r)N ? ]. (iii) Proof. En [ Sn+1 ] (1 + r)n+1 = = < = 1 En [(1 ? An+1 )Yn+1 Sn ] (1 + r)n+1 Sn [p(1 ? An+1 (H))u + q(1 ? An+1 (T ))d] (1 + r)n+1 Sn [pu + qd] (1 + r)n+1 Sn . (1 + r)n Sn (1+r)n+1 (1? a)(pu+qd) Sn+1 If An+1 is a constant a, then En [ (1+r)n+1 ] = Sn (1+r)n (1? a)n . = Sn (1+r)n (1? a). Sn+1 So En [ (1+r)n+1 (1? a)n+1 ] = 2. 11. (i) Proof. FN + PN = SN ? K + (K ? SN )+ = (SN ? K)+ = CN . (ii) CN FN PN Proof. Cn = En [ (1+r)N ? n ] = En [ (1+ r)N ? n ] + En [ (1+r)N ? n ] = Fn + Pn . (iii) FN Proof. F0 = E[ (1+r)N ] = 1 (1+r)N E[SN ? K] = S0 ? K (1+r)N . (iv) 6 Proof.At time zero, the trader has F0 = S0 in money market account and one share of stock. At time N , the trader has a wealth of (F0 ? S0 )(1 + r)N + SN = ? K + SN = FN . (v) Proof. By (ii), C0 = F0 + P0 . Since F0 = S0 ? (vi) SN ? K Proof. By (ii), Cn = Pn if and only if Fn = 0. Note Fn = En [ (1+r)N ?n ] = Sn ? So Fn is not necessarily zero and Cn = Pn is not necessarily true for n ? 1. (1+r)N S0 (1+r)N ? n (1+r)N S0 (1+r)N = 0, C0 = P0 . = Sn ? S0 (1 + r)n . 2. 12. Proof. First, the no-arbitrage price of the chooser option at time m must be max(C, P ), where C=E (SN ? K)+ (K ? SN )+ , and P = E . (1 + r)N ? m (1 + r)N ? That is, C is the no-arbitrage price of a call option at time m and P is the no-arbitrage price of a put option at time m. Both of them have maturity date N and strike price K. Suppose the market is liquid, then the chooser option is equivalent to receiving a payo? of max(C, P ) at time m. Therefore, its current no-arbitrage price should be E[ max(C,P ) ]. (1+r)m K K By the put-call parity, C = Sm ? (1+r)N ? m + P . So max(C, P ) = P + (Sm ? (1+r)N ? m )+ . Therefore, the time-zero price of a chooser option is E K (Sm ? (1+r)N ? m )+ P +E (1 + r)m (1 + r)m =E K (Sm ? (1+r)N ? m )+ (K ? SN )+ . +E (1 + r)N (1 + r)mThe ? rst term stands for the time-zero price of a put, expiring at time N and having strike price K, and the K second term stands for the time-zero price of a call, expiring at time m and having strike price (1+r)N ? m . If we feel unconvinced by the above argument that the chooser option’s no-arbitrage price is E[ max(C,P ) ], (1+r)m due to the economical argument involved (like â€Å"the chooser option is equivalent to receiving a payo? of max(C, P ) at time m†), then we have the following mathematically rigorous argument. First, we can construct a portfolio ? 0 ,  ·  ·  · , ? m? 1 , whose payo? at time m is max(C, P ).Fix ? , if C(? ) > P (? ), we can construct a portfolio ? m ,  ·  ·  · , ? N ? 1 whose payo? at time N is (SN ? K)+ ; if C(? ) < P (? ), we can construct a portfolio ? m ,  ·  ·  · , ? N ? 1 whose payo? at time N is (K ? SN )+ . By de? ning (m ? k ? N ? 1) ? k (? ) = ? k (? ) ? k (? ) if C(? ) > P (? ) if C(? ) < P (? ), we get a portfolio (? n )0? n? N ? 1 whose payo? is the same as that of the chooser option. So the no-arbitrage price process of the chooser option must be equal to the value process of the replicating portfolio. In Xm particular, V0 = X0 = E[ (1+r)m ] = E[ max(C,P ) ]. (1+r)m 2. 13. (i) Proof.Note under both actual probability P and risk-neutral probability P , coin tosses ? n ’s are i. i. d.. So n+1 without loss of generality, we work on P . For any function g, En [g(Sn+1 , Yn+1 )] = En [g( SSn Sn , Yn + = pg(uSn , Yn + uSn ) + qg(dSn , Yn + dSn ), which is a function of (Sn , Yn ). So (Sn , Yn )0? n? N is Markov un der P . (ii) 7 Sn+1 Sn Sn )] Proof. Set vN (s, y) = f ( Ny ). Then vN (SN , YN ) = f ( +1 Vn = where En [ Vn+1 ] 1+r = n+1 En [ vn+1 (S1+r ,Yn+1 ) ] N n=0 Sn N +1 ) = VN . Suppose vn+1 is given, then = 1 1+r [pvn+1 (uSn , Yn + uSn ) + qvn+1 (dSn , Yn + dSn )] = vn (Sn , Yn ), vn (s, y) = n+1 (us, y + us) + vn+1 (ds, y + ds) . 1+r 2. 14. (i) Proof. For n ? M , (Sn , Yn ) = (Sn , 0). Since coin tosses ? n ’s are i. i. d. under P , (Sn , Yn )0? n? M is Markov under P . More precisely, for any function h, En [h(Sn+1 )] = ph(uSn ) + h(dSn ), for n = 0, 1,  ·  ·  · , M ? 1. For any function g of two variables, we have EM [g(SM +1 , YM +1 )] = EM [g(SM +1 , SM +1 )] = pg(uSM , uSM )+ n+1 n+1 qg(dSM , dSM ). And for n ? M +1, En [g(Sn+1 , Yn+1 )] = En [g( SSn Sn , Yn + SSn Sn )] = pg(uSn , Yn +uSn )+ qg(dSn , Yn + dSn ), so (Sn , Yn )0? n? N is Markov under P . (ii) y Proof. Set vN (s, y) = f ( N ? M ).Then vN (SN , YN ) = f ( N K=M +1 Sk N ? M ) = VN . Suppose vn+1 is already given. a) If n > M , then En [vn+1 (Sn+1 , Yn+1 )] = pvn+1 (uSn , Yn + uSn ) + qvn+1 (dSn , Yn + dSn ). So vn (s, y) = pvn+1 (us, y + us) + qvn+1 (ds, y + ds). b) If n = M , then EM [vM +1 (SM +1 , YM +1 )] = pvM +1 (uSM , uSM ) + vn+1 (dSM , dSM ). So vM (s) = pvM +1 (us, us) + qvM +1 (ds, ds). c) If n < M , then En [vn+1 (Sn+1 )] = pvn+1 (uSn ) + qvn+1 (dSn ). So vn (s) = pvn+1 (us) + qvn+1 (ds). 3. State Prices 3. 1. Proof. Note Z(? ) := P (? ) P (? ) = 1 Z(? ) . Apply Theorem 3. 1. 1 with P , P , Z replaced by P , P , Z, we get the nalogous of properties (i)-(iii) of Theorem 3. 1. 1. 3. 2. (i) Proof. P (? ) = (ii) Proof. E[Y ] = (iii) ? Proof. P (A) = (iv) Proof. If P (A) = A Z(? )P (? ) = 0, by P (Z > 0) = 1, we conclude P (? ) = 0 for any ? ? A. So P (A) = A P (? ) = 0. (v) Proof. P (A) = 1 P (Ac ) = 0 P (Ac ) = 0 P (A) = 1. (vi) A P (? ) = Z(? )P (? ) = E[Z] = 1. Y (? )P (? ) = Y (? )Z(? )P (? ) = E[Y Z]. Z(? )P (? ). Since P (A) = 0, P (? ) = 0 for any ? ? A. So P (A) = 0. 8 Proof. Pick ? 0 such that P (? 0 ) > 0, de? ne Z(? ) = 1 P (? 0 ) 0, 1 P (? 0 ) , if ? = ? 0 Then P (Z ? 0) = 1 and E[Z] = if ? = ? 0 .  · P (? 0 ) = 1. =? 0 Clearly P (? {? 0 }) = E[Z1? {? 0 } ] = Z(? )P (? ) = 0. But P (? {? 0 }) = 1 ? P (? 0 ) > 0 if P (? 0 ) < 1. Hence in the case 0 < P (? 0 ) < 1, P and P are not equivalent. If P (? 0 ) = 1, then E[Z] = 1 if and only if Z(? 0 ) = 1. In this case P (? 0 ) = Z(? 0 )P (? 0 ) = 1. And P and P have to be equivalent. In summary, if we can ? nd ? 0 such that 0 < P (? 0 ) < 1, then Z as constructed above would induce a probability P that is not equivalent to P . 3. 5. (i) Proof. Z(HH) = (ii) Proof. Z1 (H) = E1 [Z2 ](H) = Z2 (HH)P (? 2 = H|? 1 = H) + Z2 (HT )P (? 2 = T |? 1 = H) = 3 E1 [Z2 ](T ) = Z2 (T H)P (? 2 = H|? = T ) + Z2 (T T )P (? 2 = T |? 1 = T ) = 2 . (iii) Proof. V1 (H) = [Z2 (HH)V2 (HH)P (? 2 = H|? 1 = H) + Z2 (HT )V2 (HT )P (? 2 = T |? 1 = T )] = 2. 4, Z1 (H)(1 + r1 (H)) [Z2 (T H)V2 (T H)P (? 2 = H|? 1 = T ) + Z2 (T T )V2 (T T )P (? 2 = T |? 1 = T )] 1 = , Z1 (T )(1 + r1 (T )) 9 3 4. 9 16 , Z(HT ) = 9 , Z(T H) = 8 3 8 and Z(T T ) = 15 4 . Z1 (T ) = V1 (T ) = and V0 = Z2 (HH)V2 (HH) Z2 (HT )V2 (HT ) Z2 (T H)V2 (T H) P (HH) + P (T H) + 0 ? 1. 1 1 1 1 P (HT ) + 1 (1 + 4 )(1 + 4 ) (1 + 4 )(1 + 4 ) (1 + 4 )(1 + 1 ) 2 3. 6. Proof. U (x) = have XN = 1 x, (1+r)N ? Z so I(x) = = 1 Z] 1 x. Z (3. 3. 26) gives E[ (1+r)N 1 X0 (1 + r)n Zn En [Z  ·X0 N Z (1 + r) . 0 = Xn , where ? Hence Xn = (1+r)N ? Z X En [ (1+r)N ? n ] N ] = X0 . So ? = = En [ X0 (1+r) Z n 1 X0 . By (3. 3. 25), we 1 ] = X0 (1 + r)n En [ Z ] = the second to last â€Å"=† comes from Lemma 3. 2. 6. 3. 7. Z ? Z Proof. U (x) = xp? 1 and so I(x) = x p? 1 . By (3. 3. 26), we have E[ (1+r)N ( (1+r)N ) p? 1 ] = X0 . Solve it for ? , we get ? ?p? 1 1 1 ? ? =? ? X0 p E 1 Z p? 1 Np ? ? ? = p? 1 X0 (1 + r)N p (E[Z p? 1 ])p? 1 1 p . (1+r) p? 1 ? Z So by (3. 3. 25), XN = ( (1+r)N ) p? 1 = 1 1 Np ? p? 1 Z p? 1 N (1+r) p? 1 = X0 (1+r) p? 1 E[Z p p? 1 Z p? 1 N (1+r) p? 1 = (1+r)N X0 Z p? 1 E[Z p p? 1 1 . ] ] 3. 8. (i) 9 d d Proof. x (U (x) ? yx) = U (x) ? y. So x = I(y) is an extreme point of U (x) ? yx. Because dx2 (U (x) ? yx) = U (x) ? 0 (U is concave), x = I(y) is a maximum point. Therefore U (x) ? y(x) ? U (I(y)) ? yI(y) for every x. 2 (ii) Proof. Following the hint of the problem, we have E[U (XN )] ? E[XN ? Z ? Z ? Z ? Z ] ? E[U (I( ))] ? E[ I( )], N N N (1 + r) (1 + r) (1 + r) (1 + r)N ? ? ? ? ? i. e. E[U (XN )] ? ?X0 ? E[U (XN )] ? E[ (1+r)N XN ] = E[U (XN )] ? ?X0 . So E[U (XN )] ? E[U (XN )]. 3. 9. (i) X Proof. Xn = En [ (1+r)N ? n ]. So if XN ? 0, then Xn ? 0 for all n. N (ii) 1 Proof. a) If 0 ? x < ? and 0 < y ? ? , then U (x) ? yx = ? yx ? and U (I(y)) ? yI(y) = U (? ) ? y? = 1 ? y? ? 0. So U (x) ? yx ? U (I(y)) ? yI(y). 1 b) If 0 ? x < ? and y > ? , then U (x) ? yx = ? yx ? 0 and U (I(y)) ? yI(y) = U (0) ? y  · 0 = 0. So U (x) ? yx ? U (I(y)) ? yI(y). 1 c) If x ? ? and 0 < y ? ? , then U (x) ? yx = 1 ? yx and U (I(y)) ? yI(y) = U (? ) ? y? = 1 ? y? ? 1 ? yx. So U (x) ? yx ? U (I(y)) ? yI(y). 1 d) If x ? ? and y > ? , then U (x) ? yx = 1 ? yx < 0 and U (I(y)) ? yI(y) = U (0) ? y  · 0 = 0. So U (x) ? yx ? U (I(y)) ? yI(y). (iii) XN ? Z Proof. Using (ii) and set x = XN , y = (1+r)N , where XN is a random variable satisfying E[ (1+r)N ] = X0 , we have ?Z ? Z ? E[U (XN )] ? E[ XN ] ? E[U (XN )] ? E[ X ? ]. (1 + r)N (1 + r)N N ? ? That is, E[U (XN )] ? ?X0 ? E[U (XN )] ? ?X0 . So E[U (XN )] ? E[U (XN )]. (iv) Proof. Plug pm and ? m into (3. 6. 4), we have 2N 2N X0 = m=1 pm ? m I( m ) = m=1 1 pm ? m ? 1{ m ? ? } . So X0 ? X0 ? {m : = we are looking for positive solution ? > 0). Conversely, suppose there exists some K so that ? K < ? K+1 and K X0 1 m=1 ? m pm = ? . Then we can ? nd ? > 0, such that ? K < < ? K+1 . For such ? , we have Z ? Z 1 E[ I( )] = pm ? m 1{ m ? ? } ? = pm ? m ? = X0 . N (1 + r) (1 + r)N m=1 m=1 Hence (3. 6. 4) has a solution. 0 2N K 2N X0 1 m=1 pm ? m 1{ m ? ? } . Suppose there is a solution ? to (3. 6. 4), note ? > 0, we then can conclude 1 1 1 m ? ? } = ?. Let K = max{m : m ? ? }, then K ? ? < K+1 . So ? K < ? K+1 and K N m=1 pm ? m (Note, however, that K could be 2 . In this case, ? K+1 is interpreted as ?. Also, note = (v) ? 1 Proof. XN (? m ) = I( m ) = ? 1{ m ? ? } = ?, if m ? K . 0, if m ? K + 1 4. American Derivative Securities Before proceeding to the exercise problems, we ? rst give a brief summary of pricing American derivative securities as presented in the textbook. We shall use the notation of the book.From the buyer’s perspective: At time n, if the derivative security has not been exercised, then the buyer can choose a policy ? with ? ? Sn . The valuation formula for cash ? ow (Theorem 2. 4. 8) gives a fair price for the derivative security exercised according to ? : N Vn (? ) = k=n En 1{? =k} 1 1 Gk = En 1{? ?N } G? . (1 + r)k? n (1 + r)? ?n The buyer wants to consider all the possible ? ’s, so that he c an ? nd the least upper bound of security value, which will be the maximum price of the derivative security acceptable to him. This is the price given by 1 De? nition 4. 4. 1: Vn = max? ?Sn En [1{? ?N } (1+r)? n G? ]. From the seller’s perspective: A price process (Vn )0? n? N is acceptable to him if and only if at time n, he can construct a portfolio at cost Vn so that (i) Vn ? Gn and (ii) he needs no further investing into the portfolio as time goes by. Formally, the seller can ? nd (? n )0? n? N and (Cn )0? n? N so that Cn ? 0 and Sn Vn+1 = ? n Sn+1 + (1 + r)(Vn ? Cn ? ?n Sn ). Since ( (1+r)n )0? n? N is a martingale under the risk-neutral measure P , we conclude En Cn Vn+1 Vn =? ? 0, ? n+1 n (1 + r) (1 + r) (1 + r)n Vn i. e. ( (1+r)n )0? n? N is a supermartingale. This inspired us to check if the converse is also true.This is exactly the content of Theorem 4. 4. 4. So (Vn )0? n? N is the value process of a portfolio that needs no further investing if and only if Vn (1+r)n Vn (1+r)n is a supermartingale under P (note this is independent of the requirement 0? n? N Vn ? Gn ). In summary, a price process (Vn )0? n? N is acceptable to the seller if and only if (i) Vn ? Gn ; (ii) is a supermartingale under P . 0? n? N Theorem 4. 4. 2 shows the buyer’s upper bound is the seller’s lower bound. So it gives the price acceptable to both. Theorem 4. 4. 3 gives a speci? c algorithm for calculating the price, Theorem 4. 4. establishes the one-to-one correspondence between super-replication and supermartingale property, and ? nally, Theorem 4. 4. 5 shows how to decide on the optimal exercise policy. 4. 1. (i) Proof. V2P (HH) = 0, V2P (HT ) = V2P (T H) = 0. 8, V2P (T T ) = 3, V1P (H) = 0. 32, V1P (T ) = 2, V0P = 9. 28. (ii) Proof. V0C = 5. (iii) Proof. gS (s) = |4 ? s|. We apply Theorem 4. 4. 3 and have V2S (HH) = 12. 8, V2S (HT ) = V2S (T H) = 2. 4, V2S (T T ) = 3, V1S (H) = 6. 08, V1S (T ) = 2. 16 and V0S = 3. 296. (iv) 11 Proof. First, we note the simple inequality max(a1 , b1 ) + max(a2 , b2 ) ? max(a1 + a2 , b1 + b2 ). >† holds if and only if b1 > a1 , b2 < a2 or b1 < a1 , b2 > a2 . By induction, we can show S Vn = max gS (Sn ), S S pVn+1 + Vn+1 1+r C P P pV C + Vn+1 pVn+1 + Vn+1 + n+1 1+r 1+r C C pVn+1 + Vn+1 1+r ? max gP (Sn ) + gC (Sn ), ? max gP (Sn ), P C = Vn + Vn . P P pVn+1 + Vn+1 1+r + max gC (Sn ), S P C As to when â€Å" C C pVn+1 +qVn+1 1+r or gP (Sn ) > P P pVn+1 +qVn+1 1+r and gC (Sn ) < C C pVn+1 +qVn+1 }. 1+r 4. 2. Proof. For this problem, we need Figure 4. 2. 1, Figure 4. 4. 1 and Figure 4. 4. 2. Then ? 1 (H) = and ? 0 = V2 (HH) ? V2 (HT ) 1 V2 (T H) ? V2 (T T ) = ? , ? 1 (T ) = = ? 1, S2 (HH) ? S2 (HT ) 12 S2 (T H) ?S2 (T T ) V1 (H) ? V1 (T ) ? ?0. 433. S1 (H) ? S1 (T ) The optimal exercise time is ? = inf{n : Vn = Gn }. So ? (HH) = ? , ? (HT ) = 2, ? (T H) = ? (T T ) = 1. Therefore, the agent borrows 1. 36 at time zero and buys the put. At the same time, to hedge the long position, he needs to borr ow again and buy 0. 433 shares of stock at time zero. At time one, if the result of coin toss is tail and the stock price goes down to 2, the value of the portfolio 1 is X1 (T ) = (1 + r)(? 1. 36 ? 0. 433S0 ) + 0. 433S1 (T ) = (1 + 4 )(? 1. 36 ? 0. 433 ? 4) + 0. 433 ? 2 = ? 3. The agent should exercise the put at time one and get 3 to pay o? is debt. At time one, if the result of coin toss is head and the stock price goes up to 8, the value of the portfolio 1 is X1 (H) = (1 + r)(? 1. 36 ? 0. 433S0 ) + 0. 433S1 (H) = ? 0. 4. The agent should borrow to buy 12 shares of stock. At time two, if the result of coin toss is head and the stock price goes up to 16, the value of the 1 1 portfolio is X2 (HH) = (1 + r)(X1 (H) ? 12 S1 (H)) + 12 S2 (HH) = 0, and the agent should let the put expire. If at time two, the result of coin toss is tail and the stock price goes down to 4, the value of the portfolio is 1 1 X2 (HT ) = (1 + r)(X1 (H) ? 12 S1 (H)) + 12 S2 (HT ) = ? 1.The agent should exercise the put to get 1. This will pay o? his debt. 4. 3. Proof. We need Figure 1. 2. 2 for this problem, and calculate the intrinsic value process and price process of the put as follows. 2 For the intrinsic value process, G0 = 0, G1 (T ) = 1, G2 (T H) = 3 , G2 (T T ) = 5 , G3 (T HT ) = 1, 3 G3 (T T H) = 1. 75, G3 (T T T ) = 2. 125. All the other outcomes of G is negative. 12 2 5 For the price process, V0 = 0. 4, V1 (T ) = 1, V1 (T H) = 3 , V1 (T T ) = 3 , V3 (T HT ) = 1, V3 (T T H) = 1. 75, V3 (T T T ) = 2. 125. All the other outcomes of V is zero. Therefore the time-zero price of the derivative security is 0. and the optimal exercise time satis? es ? (? ) = ? if ? 1 = H, 1 if ? 1 = T . 4. 4. Proof. 1. 36 is the cost of super-replicating the American derivative security. It enables us to construct a portfolio su? cient to pay o? the derivative security, no matter when the derivative security is exercised. So to hedge our short position after selling the put, there is no need to charge t he insider more than 1. 36. 4. 5. Proof. The stopping times in S0 are (1) ? ? 0; (2) ? ? 1; (3) ? (HT ) = ? (HH) = 1, ? (T H), ? (T T ) ? {2, ? } (4 di? erent ones); (4) ? (HT ), ? (HH) ? {2, ? }, ? (T H) = ? (T T ) = 1 (4 di? rent ones); (5) ? (HT ), ? (HH), ? (T H), ? (T T ) ? {2, ? } (16 di? erent ones). When the option is out of money, the following stopping times do not exercise (i) ? ? 0; (ii) ? (HT ) ? {2, ? }, ? (HH) = ? , ? (T H), ? (T T ) ? {2, ? } (8 di? erent ones); (iii) ? (HT ) ? {2, ? }, ? (HH) = ? , ? (T H) = ? (T T ) = 1 (2 di? erent ones). ? 4 For (i), E[1{? ?2} ( 4 )? G? ] = G0 = 1. For (ii), E[1{? ?2} ( 5 )? G? ] ? E[1{? ? ? 2} ( 4 )? G? ? ], where ? ? (HT ) = 5 5 1 4 4 ? 2, ? ? (HH) = ? , ? ? (T H) = ? ? (T T ) = 2. So E[1{? ? ? 2} ( 5 )? G? ? ] = 4 [( 4 )2  · 1 + ( 5 )2 (1 + 4)] = 0. 96. For 5 (iii), E[1{? ?2} ( 4 )? G? has the biggest value when ? satis? es ? (HT ) = 2, ? (HH) = ? , ? (T H) = ? (T T ) = 1. 5 This value is 1. 36. 4. 6. (i) Proof. The value of the put at time N , if it is not exercised at previous times, is K ? SN . Hence VN ? 1 = VN K max{K ? SN ? 1 , EN ? 1 [ 1+r ]} = max{K ? SN ? 1 , 1+r ? SN ? 1 } = K ? SN ? 1 . The second equality comes from the fact that discounted stock price process is a martingale under risk-neutral probability. By induction, we can show Vn = K ? Sn (0 ? n ? N ). So by Theorem 4. 4. 5, the optimal exercise policy is to sell the stock at time zero and the value of this derivative security is K ?S0 . Remark: We cheated a little bit by using American algorithm and Theorem 4. 4. 5, since they are developed for the case where ? is allowed to be ?. But intuitively, results in this chapter should still hold for the case ? ? N , provided we replace â€Å"max{Gn , 0}† with â€Å"Gn †. (ii) Proof. This is because at time N , if we have to exercise the put and K ? SN < 0, we can exercise the European call to set o? the negative payo?. In e? ect, throughout the portfolio’s lifetime, the portfolio has intrinsic values greater than that of an American put stuck at K with expiration time N . So, we must have V0AP ? V0 + V0EC ? K ?S0 + V0EC . (iii) 13 Proof. Let V0EP denote the time-zero value of a European put with strike K and expiration time N . Then V0AP ? V0EP = V0EC ? E[ K SN ? K ] = V0EC ? S0 + . (1 + r)N (1 + r)N 4. 7. VN K K Proof. VN = SN ? K, VN ? 1 = max{SN ? 1 ? K, EN ? 1 [ 1+r ]} = max{SN ? 1 ? K, SN ? 1 ? 1+r } = SN ? 1 ? 1+r . K By induction, we can prove Vn = Sn ? (1+r)N ? n (0 ? n ? N ) and Vn > Gn for 0 ? n ? N ? 1. So the K time-zero value is S0 ? (1+r)N and the optimal exercise time is N . 5. Random Walk 5. 1. (i) Proof. E[ 2 ] = E[? (? 2 1 )+? 1 ] = E[? (? 2 1 ) ]E[ 1 ] = E[ 1 ]2 . (ii) Proof. If we de? ne Mn = Mn+? ? M? m (m = 1, 2,  ·  ·  · ), then (M · )m as random functions are i. i. d. with (m) distributions the same as that of M . So ? m+1 ? ?m = inf{n : Mn = 1} are i. i. d. with distributions the same as that of ? 1 . Therefore E [ m ] = E[? (? m m? 1 )+(? m? 1 m? 2 )+ ·Ã‚ ·Ã‚ ·+? 1 ] = E[ 1 ]m . (m) (m) (iii) Proof. Yes, since the argument of (ii) still works for asymmetric random walk. 5. 2. (i) Proof. f (? ) = pe? ? qe , so f (? ) > 0 if and only if ? > f (? ) > f (0) = 1 for all ? > 0. (ii) 1 1 1 n+1 Proof. En [ SSn ] = En [e? Xn+1 f (? ) ] = pe? f (? ) + qe f (? ) = 1. 1 2 (ln q ? ln p). Since 1 2 (ln q ln p) < 0, (iii) 1 Proof. By optional stopping theorem, E[Sn 1 ] = E[S0 ] = 1. Note Sn 1 = e? Mn 1 ( f (? ) )n 1 ? e?  ·1 , by bounded convergence theorem, E[1{? 1 1 for all ? > ? 0 . v (ii) 1 1 Proof. As in Exercise 5. 2, Sn = e? Mn ( f (? ) )n is a martingale, and 1 = E[S0 ] = E[Sn 1 ] = E[e? Mn 1 ( f (? ) )? 1 ? n ]. Suppose ? > ? 0 , then by bounded convergence theorem, 1 = E[ lim e? Mn 1 ( n>? 1 n 1 1 ? 1 ) ] = E[1{? 1 K} ] = P (ST > K). Moreover, by Girsanov’s Theorem, Wt = Wt + in Theorem 5. 4. 1. ) (iii) Proof. ST = xe? WT +(r? 2 ? 1 2 1 2 t ( )du 0 = Wt ? ?t is a P -Brownian motion (set ? )T = xe? WT +(r+ 2 ? 1 2 1 2 )T . So WT v > ? d+ (T, x) T = N (d+ (T, x)). P (ST > K) = P (xe? WT +(r+ 2 ? )T > K) = P 46 5. 4. First, a few typos. In the SDE for S, â€Å"? (t)dW (t)† > â€Å"? (t)S(t)dW (t)†. In the ? rst equation for c(0, S(0)), E > E. In the second equation for c(0, S(0)), the variable for BSM should be ? ? 1 T 2 1 T r(t)dt, ? (t)dt? . BSM ? T, S(0); K, T 0 T 0 (i) Proof. d ln St = X = ? is a Gaussian with X ? N ( (ii) Proof. For the standard BSM model with constant volatility ? and interest rate R, under the risk-neutral measure, we have ST = S0 eY , where Y = (R? 1 ? 2 )T +? WT ? N ((R? 1 ? )T, ? 2 T ), and E[(S0 eY ? K)+ ] = 2 2 eRT BSM (T, S0 ; K, R, ? ). Note R = 1 T (rt 0 T T dSt 1 2 1 1 2 2 St ? 2St d S t = rt dt + ? t dWt ? 2 ? t dt. So ST = S0 exp{ 0 (rt ? 2 ? t )dt + 0 T 1 2 2 ? t )dt + 0 ? t dWt . The ? rst term in the expression of X is a number and the T 2 random variable N (0, 0 ? t dt), since both r and ? ar deterministic. Th erefore, T T 2 2 (rt ? 1 ? t )dt, 0 ? t dt),. 2 0 ?t dWt }. Let second term ST = S0 eX , 1 T (E[Y ] + 1 V ar(Y )) and ? = 2 T, S0 ; K, 1 T 1 T V ar(Y ), we can get 1 V ar(Y ) . T E[(S0 eY ? K)+ ] = eE[Y ]+ 2 V ar(Y ) BSM So for the model in this problem, c(0, S0 ) = = e? ? T 0 1 E[Y ] + V ar(Y ) , 2 rt dt E[(S0 eX ? K)+ ] e BSM T, S0 ; K, 1 T T 0 T 0 1 rt dt E[X]+ 2 V ar(X) 1 T ? 1 E[X] + V ar(X) , 2 1 V ar(X) T ? = 1 BSM ? T, S0 ; K, T 0 T rt dt, 2 ? t dt? . 5. 5. (i) 1 1 Proof. Let f (x) = x , then f (x) = ? x2 and f (x) = 2 x3 . Note dZt = ? Zt ? t dWt , so d 1 Zt 1 1 1 2 2 2 ? t ? 2 t = f (Zt )dZt + f (Zt )dZt dZt = ? 2 (? Zt )? t dWt + 3 Zt ? t dt = Z dWt + Z dt. 2 Zt 2 Zt t t (ii) Proof. By Lemma 5. 2. 2. , for s, t ? 0 with s < t, Ms = E[Mt |Fs ] = E Zs Ms . So M = Z M is a P -martingale. (iii) Zt Mt Zs |Fs . That is, E[Zt Mt |Fs ] = 47 Proof. dMt = d Mt  · 1 Zt = 1 1 1 ? M t ? t M t ? 2 ? t ? t t dMt + Mt d + dMt d = dWt + dWt + dt + dt. Zt Zt Zt Zt Zt Zt Zt (iv) Proof. In part (iii), we have dMt = Let ? t = 5. 6. Proof. By Theorem 4. 6. 5, it su? ces to show Wi (t) is an Ft -martingale under P and [Wi , Wj ](t) = t? ij (i, j = 1, 2). Indeed, for i = 1, 2, Wi (t) is an Ft -martingale under P if and only if Wi (t)Zt is an Ft -martingale under P , since Wi (t)Zt E[Wi (t)|Fs ] = E |Fs . Zs By It? ’s product formula, we have o d(Wi (t)Zt ) = Wi (t)dZt + Zt dWi (t) + dZt dWi (t) = Wi (t)(? Zt )? (t)  · dWt + Zt (dWi (t) + ? i (t)dt) + (? Zt ? t  · dWt )(dWi (t) + ? i (t)dt) d t M t ? t M t ? 2 ? t ? t ? t M t ? t t dWt + dWt + dt + dt = (dWt + ? t dt) + (dWt + ? t dt). Zt Zt Zt Zt Zt Zt then dMt = ? t dWt . This proves Corollary 5. 3. 2. ?t +Mt ? t , Zt = Wi (t)(? Zt ) j=1 d ?j (t)dWj (t) + Zt (dWi (t) + ? i (t)dt) ? Zt ? i (t)dt = Wi (t)(? Zt ) j=1 ?j (t)dWj (t) + Zt dWi (t) This shows Wi (t)Zt is an Ft -martingale under P . So Wi (t) is an Ft -martingale under P . Moreover,  ·  · [Wi , Wj ](t) = Wi + 0 ?i (s)ds, Wj + 0 ?j (s)ds (t) = [Wi , Wj ](t) = t? ij . Combined, this proves the two-dimensional Girsanov’s Theorem. 5. 7. (i) Proof. Let a be any strictly positive number. We de? e X2 (t) = (a + X1 (t))D(t)? 1 . Then P X2 (T ) ? X2 (0) D(T ) = P (a + X1 (T ) ? a) = P (X1 (T ) ? 0) = 1, and P X2 (T ) > X2 (0) = P (X1 (T ) > 0) > 0, since a is arbitrary, we have proved the claim of this problem. D(T ) Remark: The intuition is that we invest the positive starting fund a into the money market account, and construct portfolio X1 from zero cost. Their sum should be able to beat the return of money market account. (ii) 48 Proof. We de? ne X1 (t) = X2 (t)D(t) ? X2 (0). Then X1 (0) = 0, P (X1 (T ) ? 0) = P X2 (T ) ? X2 (0) D(T ) = 1, P (X1 (T ) > 0) = P X2 (T ) > X2 (0) D(T ) > 0. 5. 8.The basic idea is that for any positive P -martingale M , dMt = Mt  · sentation Theorem, dMt = ? t dWt for some adapted process ? t . So martingale must be the exponential of an integral w. r. t. Brownian motion. Taking into account d iscounting factor and apply It? ’s product rule, we can show every strictly positive asset is a generalized geometric o Brownian motion. (i) Proof. Vt Dt = E[e? 0 Ru du VT |Ft ] = E[DT VT |Ft ]. So (Dt Vt )t? 0 is a P -martingale. By Martingale Represent tation Theorem, there exists an adapted process ? t , 0 ? t ? T , such that Dt Vt = 0 ? s dWs , or equivalently, ? 1 t ? 1 t ? 1 Vt = Dt 0 ? dWs . Di? erentiate both sides of the equation, we get dVt = Rt Dt 0 ? s dWs dt + Dt ? t dWt , i. e. dVt = Rt Vt dt + (ii) Proof. We prove the following more general lemma. Lemma 1. Let X be an almost surely positive random variable (i. e. X > 0 a. s. ) de? ned on the probability space (? , G, P ). Let F be a sub ? -algebra of G, then Y = E[X|F] > 0 a. s. Proof. By the property of conditional expectation Yt ? 0 a. s. Let A = {Y = 0}, we shall show P (A) = 0. In? 1 1 deed, note A ? F, 0 = E[Y IA ] = E[E[X|F]IA ] = E[XIA ] = E[X1A? {X? 1} ] + n=1 E[X1A? { n >X? n+1 } ] ? 1 1 1 1 1 P (A? {X ? 1})+ n=1 n+1 P (A? n > X ? n+1 }). So P (A? {X ? 1}) = 0 and P (A? { n > X ? n+1 }) = 0, ? 1 1 ? n ? 1. This in turn implies P (A) = P (A ? {X > 0}) = P (A ? {X ? 1}) + n=1 P (A ? { n > X ? n+1 }) = 0. ? ? t Dt dWt . T 1 Mt dMt . By Martingale Repre? dMt = Mt ( Mtt )dWt , i. e. any positive By the above lemma, it is clear that for each t ? [0, T ], Vt = E[e? t Ru du VT |Ft ] > 0 a. s.. Moreover, by a classical result of martingale theory (Revuz and Yor [4], Chapter II, Proposition (3. 4)), we have the following stronger result: for a. s. ?, Vt (? ) > 0 for any t ? [0, T ]. (iii) 1 1 Proof. By (ii), V > 0 a. s. so dVt = Vt Vt dVt = Vt Vt Rt Vt dt + ? t Dt dWt ? t = Vt Rt dt + Vt Vt Dt dWt = Rt Vt dt + T ?t Vt dWt , where ? t = 5. 9. ?t Vt Dt . This shows V follows a generalized geometric Brownian motion. Proof. c(0, T, x, K) = xN (d+ ) ? Ke? rT N (d? ) with d ± = then f (y) = ? yf (y), cK (0, T, x, K) = xf (d+ ) 1 v ? T x (ln K + (r  ± 1 ? 2 )T ). Let f (y) = 2 y v1 e? 2 2? 2 , ?d+ ? d? ? e? rT N (d? ) ? Ke? rT f (d? ) ? y ? y ? 1 1 = xf (d+ ) v ? e? rT N (d? ) + e? rT f (d? ) v , ? TK ? T 49 and cKK (0, T, x, K) x ? d? e? rT 1 ? d+ d? ? v ? e? rT f (d? ) + v (? d? )f (d? ) xf (d+ ) v f (d+ )(? d+ ) 2 ? y ? y ? y ? TK ? TK ?T x xd+ ? 1 ? 1 e? rT d? ?1 v v ? e? rT f (d? ) v ? v f (d? ) v f (d+ ) + v f (d+ ) ? T K2 ? TK K? T K? T ? T K? T x d+ e? rT f (d? ) d? v [1 ? v ] + v f (d+ ) [1 + v ] 2? T K ? T K? T ? T e? rT x f (d? )d+ ? 2 2 f (d+ )d? . K? 2 T K ? T = = = = 5. 10. (i) Proof. At time t0 , the value of the chooser option is V (t0 ) = max{C(t0 ), P (t0 )} = max{C(t0 ), C(t0 ) ? F (t0 )} = C(t0 ) + max{0, ? F (t0 )} = C(t0 ) + (e? r(T ? t0 ) K ? S(t0 ))+ . (ii) Proof. By the risk-neutral pricing formula, V (0) = E[e? rt0 V (t0 )] = E[e? rt0 C(t0 )+(e? rT K ? e? rt0 S(t0 )+ ] = C(0) + E[e? rt0 (e? r(T ? t0 ) K ? S(t0 ))+ ]. The ? st term is the value of a call expiring at time T with strike price K and the second term is the value of a put expiring at time t0 with strike price e? r(T ? t0 ) K. 5. 11. Proof. We ? rst make an analysis which leads to the hint, then we give a formal proof. (Analysis) If we want to construct a portfolio X that exactly replicates the cash ? ow, we must ? nd a solution to the backward SDE dXt = ? t dSt + Rt (Xt ? ?t St )dt ? Ct dt XT = 0. Multiply Dt on both sides of the ? rst equation and apply It? ’s product rule, we get d(Dt Xt ) = ? t d(Dt St ) ? o T T Ct Dt dt. Integrate from 0 to T , we have DT XT ? D0 X0 = 0 ? d(Dt St ) ? 0 Ct Dt dt. By the terminal T T ? 1 condition, we get X0 = D0 ( 0 Ct Dt dt ? 0 ? t d(Dt St )). X0 is the theoretical, no-arbitrage price of the cash ? ow, provided we can ? nd a trading strategy ? that solves the BSDE. Note the SDE for S ? R gives d(Dt St ) = (Dt St )? t (? t dt + dWt ), where ? t = ? t? t t . Take the proper change of measure so that Wt = t ? ds 0 s + Wt is a Brownian motion under the new measure P , we get T T T Ct Dt dt = D0 X0 + 0 T 0 ?t d(Dt St ) = D0 X0 + 0 ?t (Dt St )? t dWt . T This says the random variable 0 Ct Dt dt has a stochastic integral representation D0 X0 + 0 ? t Dt St ? dWt . T This inspires us to consider the martingale generated by 0 Ct Dt dt, so that we can apply Martingale Representation Theorem and get a formula for ? by comparison of the integrands. 50 (Formal proof) Let MT = Xt = ?1 Dt (D0 X0 T 0 Ct Dt dt, and Mt = E[MT |Ft ]. Then by Martingale Representation Theot 0 rem, we can ? nd an adapted process ? t , so that Mt = M0 + + t 0 ?t dWt . If we set ? t = T 0 ?u d(Du Su ) ? t 0 ?t Dt St ? t , we can check Cu Du du), with X0 = M0 = E[ Ct Dt dt] solves the SDE dXt = ? t dSt + Rt (Xt ? ?t St )dt ? Ct dt XT = 0. Indeed, it is easy to see that X satis? es the ? rst equation.To check the terminal condition, we note T T T XT DT = D0 X0 + 0 ? t Dt St ? t dWt ? 0 Ct Dt dt = M0 + 0 ? t dWt ? MT = 0. So XT = 0. Thus, we have found a trading strategy ? , so that the corresponding portfolio X replicates the cash ? ow and has zero T terminal value. So X0 = E[ 0 Ct Dt dt] is the no-arbitrage price of the cash ? ow at time zero. Remark: As shown in the analysis, d(Dt Xt ) = ? t d(Dt St ) ? Ct Dt dt. Integrate from t to T , we get T T 0 ? Dt Xt = t ? u d(Du Su ) ? t Cu Du du. Take conditional expectation w. r. t. Ft on both sides, we get T T ? 1 ? Dt Xt = ? E[ t Cu Du du|Ft ]. So Xt = Dt E[ t Cu Du du|Ft ].This is the no-arbitrage price of the cash ? ow at time t, and we have justi? ed formula (5. 6. 10) in the textbook. 5. 12. (i) Proof. dBi (t) = dBi (t) + ? i (t)dt = martingale. Since dBi (t)dBi (t) = P. (ii) Proof. dSi (t) = = = R(t)Si (t)dt + ? i (t)Si (t)dBi (t) + (? i (t) ? R(t))Si (t)dt ? ?i (t)Si (t)? i (t)dt d d ? ij (t) ? ij (t) d d j=1 ? i (t) ? j (t)dt = j=1 ? i (t) dWj (t) + ? ij (t)2 d e j=1 ? i (t)2 dt = dt, by L? vy’s Theorem, Bi ? ij (t) d j=1 ? i (t) dWj (t). So Bi is a is a Brownian motion under R(t)Si (t)dt + ? i (t)Si (t)dBi (t) + j=1 ?ij (t)? j (t)Si (t)dt ? Si (t) j=1 ?ij (t)? j (t)dt R(t)Si (t)dt + ? (t)Si (t)dBi (t). (iii) Proof. dBi (t)dBk (t) = (dBi (t) + ? i (t)dt)(dBj (t) + ? j (t)dt) = dBi (t)dBj (t) = ? ik (t)dt. (iv) Proof. By It? ’s product rule and martingale property, o t t t E[Bi (t)Bk (t)] = E[ 0 t Bi (s)dBk (s)] + E[ 0 t Bk (s)dBi (s)] + E[ 0 dBi (s)dBk (s)] = E[ 0 ?ik (s)ds] = 0 ?ik (s)ds. t 0 Similarly, by part (iii), we can show E[Bi (t)Bk (t)] = (v) ?ik (s)ds. 51 Proof. By It? ’s product formula, o t t E[B1 (t)B2 (t)] = E[ 0 sign(W1 (u))du] = 0 [P (W1 (u) ? 0) ? P (W1 (u) < 0)]du = 0. Meanwhile, t E[B1 (t)B2 (t)] = E[ 0 t sign(W1 (u))du [P (W1 (u) ? 0) ? P (W1 (u) < 0)]du = 0 t = 0 t [P (W1 (u) ? ) ? P (W1 (u) < u)]du 2 0 = < 0, 1 ? P (W1 (u) < u) du 2 for any t > 0. So E[B1 (t)B2 (t)] = E[B1 (t)B2 (t)] for all t > 0. 5. 13. (i) Proof. E[W1 (t)] = E[W1 (t)] = 0 and E[W2 (t)] = E[W2 (t) ? (ii) Proof. Cov[W1 (T ), W2 (T )] = E[W1 (T )W2 (T )] T T t 0 W1 (u)du] = 0, for all t ? [0, T ]. = E 0 T W1 (t)dW2 (t) + 0 W2 ( t)dW1 (t) T = E 0 W1 (t)(dW2 (t) ? W1 (t)dt) + E 0 T W2 (t)dW1 (t) = ? E 0 T W1 (t)2 dt tdt = ? 0 1 = ? T 2. 2 5. 14. Equation (5. 9. 6) can be transformed into d(e? rt Xt ) = ? t [d(e? rt St ) ? ae? rt dt] = ? t e? rt [dSt ? rSt dt ? adt]. So, to make the discounted portfolio value e? t Xt a martingale, we are motivated to change the measure t in such a way that St ? r 0 Su du? at is a martingale under the new measure. To do this, we note the SDE for S is dSt = ? t St dt+? St dWt . Hence dSt ? rSt dt? adt = [(? t ? r)St ? a]dt+? St dWt = ? St Set ? t = (? t ? r)St ? a ? St (? t ? r)St ? a dt ? St + dWt . and Wt = t ? ds 0 s + Wt , we can ? nd an equivalent probability measure P , under which S satis? es the SDE dSt = rSt dt + ? St dWt + adt and Wt is a BM. This is the rational for formula (5. 9. 7). This is a good place to pause and think about the meaning of â€Å"martingale measure. † What is to be a martingale?The new measure P should be such that the discounted value pro cess of the replicating 52 portfolio is a martingale, not the discounted price process of the underlying. First, we want Dt Xt to be a martingale under P because we suppose that X is able to replicate the derivative payo? at terminal time, XT = VT . In order to avoid arbitrage, we must have Xt = Vt for any t ? [0, T ]. The di? culty is how to calculate Xt and the magic is brought by the martingale measure in the following line of reasoning: ? 1 ? 1 Vt = Xt = Dt E[DT XT |Ft ] = Dt E[DT VT |Ft ]. You can think of martingale measure as a calculational convenience.That is all about martingale measure! Risk neutral is a just perception, referring to the actual e? ect of constructing a hedging portfolio! Second, we note when the portfolio is self-? nancing, the discounted price process of the underlying is a martingale under P , as in the classical Black-Scholes-Merton model without dividends or cost of carry. This is not a coincidence. Indeed, we have in this case the relation d(Dt Xt ) = ? t d(Dt St ). So Dt Xt being a martingale under P is more or less equivalent to Dt St being a martingale under P . However, when the underlying pays dividends, or there is cost of carry, d(Dt Xt ) = ? d(Dt St ) no longer holds, as shown in formula (5. 9. 6). The portfolio is no longer self-? nancing, but self-? nancing with consumption. What we still want to retain is the martingale property of Dt Xt , not that of Dt St . This is how we choose martingale measure in the above paragraph. Let VT be a payo? at time T , then for the martingale Mt = E[e? rT VT |Ft ], by Martingale Representation rt t Theorem, we can ? nd an adapted process ? t , so that Mt = M0 + 0 ? s dWs . If we let ? t = ? t e t , then the ? S value of the corresponding portfolio X satis? es d(e? rt Xt ) = ? t dWt . So by setting X0 = M0 = E[e? T VT ], we must have e? rt Xt = Mt , for all t ? [0, T ]. In particular, XT = VT . Thus the portfolio perfectly hedges VT . This justi? es the risk-neutral pricing of Europea n-type contingent claims in the model where cost of carry exists. Also note the risk-neutral measure is di? erent from the one in case of no cost of carry. Another perspective for perfect replication is the following. We need to solve the backward SDE dXt = ? t dSt ? a? t dt + r(Xt ? ?t St )dt XT = VT for two unknowns, X and ?. To do so, we ? nd a probability measure P , under which e? rt Xt is a martingale, t then e? rt Xt = E[e? T VT |Ft ] := Mt . Martingale Representation Theorem gives Mt = M0 + 0 ? u dWu for some adapted process ?. This would give us a theoretical representation of ? by comparison of integrands, hence a perfect replication of VT . (i) Proof. As indicated in the above analysis, if we have (5. 9. 7) under P , then d(e? rt Xt ) = ? t [d(e? rt St ) ? ae? rt dt] = ? t e? rt ? St dWt . So (e? rt Xt )t? 0 , where X is given by (5. 9. 6), is a P -martingale. (ii) 1 1 Proof. By It? ’s formula, dYt = Yt [? dWt + (r ? 2 ? 2 )dt] + 2 Yt ? 2 dt = Yt (? dWt + rdt). So d(e? rt Yt ) = o t a ? e? rt Yt dWt and e? rt Yt is a P -martingale.Moreover, if St = S0 Yt + Yt 0 Ys ds, then t dSt = S0 dYt + 0 a dsdYt + adt = Ys t S0 + 0 a ds Yt (? dWt + rdt) + adt = St (? dWt + rdt) + adt. Ys This shows S satis? es (5. 9. 7). Remark: To obtain this formula for S, we ? rst set Ut = e? rt St to remove the rSt dt term. The SDE for U is dUt = ? Ut dWt + ae? rt dt. Just like solving linear ODE, to remove U in the dWt term, we consider Vt = Ut e Wt . It? ’s product formula yields o dVt = = e Wt dUt + Ut e Wt 1 ( )dWt + ? 2 dt + dUt  · e Wt 2 1 ( )dWt + ? 2 dt 2 1 e Wt ae? rt dt ? ? 2 Vt dt. 2 53 Note V appears only in the dt term, so multiply the integration factor e 2 ? e get 1 2 1 2 d(e 2 ? t Vt ) = ae? rt Wt + 2 ? t dt. Set Yt = e? Wt +(r? 2 ? (iii) Proof. t 1 2 1 2 t on both sides of the equation, )t , we have d(St /Yt ) = adt/Yt . So St = Yt (S0 + t ads ). 0 Ys E[ST |Ft ] = S0 E[YT |Ft ] + E YT 0 t a ds + YT Ys T t T a ds|Ft Ys E YT |Ft ds Ys E[YT ? s ]ds t = S0 E[YT |Ft ] + 0 a dsE[YT |Ft ] + a Ys t t T = S0 Yt E[YT ? t ] + 0 t a dsYt E[YT ? t ] + a Ys T t = = S0 + 0 t a ds Yt er(T ? t) + a Ys ads Ys er(T ? s) ds S0 + 0 a Yt er(T ? t) ? (1 ? er(T ? t) ). r In particular, E[ST ] = S0 erT ? a (1 ? erT ). r (iv) Proof. t dE[ST |Ft ] = aer(T ? t) dt + S0 + 0 t ads Ys a (er(T ? ) dYt ? rYt er(T ? t) dt) + er(T ? t) (? r)dt r = S0 + 0 ads Ys er(T ? t) ? Yt dWt . So E[ST |Ft ] is a P -martingale. As we have argued at the beginning of the solution, risk-neutral pricing is valid even in the presence of cost of carry. So by an argument similar to that of  §5. 6. 2, the process E[ST |Ft ] is the futures price process for the commodity. (v) Proof. We solve the equation E[e? r(T ? t) (ST ? K)|Ft ] = 0 for K, and get K = E[ST |Ft ]. So F orS (t, T ) = F utS (t, T ). (vi) Proof. We follow the hint. First, we solve the SDE dXt = dSt ? adt + r(Xt ? St )dt X0 = 0. By our analysis in part (i), d(e? t Xt ) = d(e? rt St ) ? ae? rt dt. Integrate fr om 0 to t on both sides, we get Xt = St ? S0 ert + a (1 ? ert ) = St ? S0 ert ? a (ert ? 1). In particular, XT = ST ? S0 erT ? a (erT ? 1). r r r Meanwhile, F orS (t, T ) = F uts (t, T ) = E[ST |Ft ] = S0 + t ads 0 Ys Yt er(T ? t) ? a (1? er(T ? t) ). So F orS (0, T ) = r S0 erT ? a (1 ? erT ) and hence XT = ST ? F orS (0, T ). After the agent delivers the commodity, whose value r is ST , and receives the forward price F orS (0, T ), the portfolio has exactly zero value. 54 6. Connections with Partial Di? erential Equations 6. 1. (i) Proof. Zt = 1 is obvious.Note the form of Z is similar to that of a geometric Brownian motion. So by It? ’s o formula, it is easy to obtain dZu = bu Zu du + ? u Zu dWu , u ? t. (ii) Proof. If Xu = Yu Zu (u ? t), then Xt = Yt Zt = x  · 1 = x and dXu = = = = Yu dZu + Zu dYu + dYu Zu au ? ?u ? u ? u du + dWu Zu Zu [Yu bu Zu + (au ? ?u ? u ) + ? u ? u ]du + (? u Zu Yu + ? u )dWu Yu (bu Zu du + ? u Zu dWu ) + Zu (bu Xu + au )du + (? u Xu + ? u )dWu . + ? u Z u ? u du Zu Remark: To see how to ? nd the above solution, we manipulate the equation (6. 2. 4) as follows. First, to u remove the term bu Xu du, we multiply on both sides of (6. 2. 4) the integrating factor e? bv dv . Then d(Xu e? ? Let Xu = e? u t u t bv dv ) = e? u t bv dv (au du + (? u + ? u Xu )dWu ). u t bv dv Xu , au = e? ? u t bv dv au and ? u = e? ? bv dv ? ? u , then X satis? es the SDE ? ? ? dXu = au du + (? u + ? u Xu )dWu = (? u du + ? u dWu ) + ? u Xu dWu . ? ? a ? ? ? ? To deal with the term ? u Xu dWu , we consider Xu = Xu e? ? dXu = e? u t u t ?v dWv . Then ?v dWv ?v dWv ? ? [(? u du + ? u dWu ) + ? u Xu dWu ] + Xu e? a ? u t u t 1 ( u )dWu + e? 2 u t ?v dWv 2 ? u du ? +(? u + ? u Xu )( u )e? ? ?v dWv du 1 ? 2 ? ? ? = au du + ? u dWu + ? u Xu dWu ? ?u Xu dWu + Xu ? u du ? ?u (? u + ? u Xu )du ? ? ? 1 ? 2 = (? u ? ?u ? u ? Xu ? u )du + ? u dWu , a ? ? 2 where au = au e? ? ? ? 1 d Xu e 2 u t ?v dWv 2 ? v dv and ? u = ? u e? ? ? = e2 1 u t 2 ? v dv u t ?v dWv . Finally, use the integrating factor e u t 2 ? v dv u 1 2 ? dv t 2 v , we have u t 1 ? ? 1 2 (dXu + Xu  · ? u du) = e 2 2 [(? u ? ?u ? u )du + ? u dWu ]. a ? ? Write everything back into the original X, a and ? , we get d Xu e? i. e. d u t bv dv? u t 1 ? v dWv + 2 u t 2 ? v dv = e2 1 u t 2 ? v dv? u t ?v dWv ? u t bv dv [(au ? ?u ? u )du + ? u dWu ], Xu Zu = 1 [(au ? ?u ? u )du + ? u dWu ] = dYu . Zu This inspired us to try Xu = Yu Zu . 6. 2. (i) 55 Proof.The portfolio is self-? nancing, so for any t ? T1 , we have dXt = ? 1 (t)df (t, Rt , T1 ) + ? 2 (t)df (t, Rt , T2 ) + Rt (Xt ? ?1 (t)f (t, Rt , T1 ) ? ?2 (t)f (t, Rt , T2 ))dt, and d(Dt Xt ) = ? Rt Dt Xt dt + Dt dXt = Dt [? 1 (t)df (t, Rt , T1 ) + ? 2 (t)df (t, Rt , T2 ) ? Rt (? 1 (t)f (t, Rt , T1 ) + ? 2 (t)f (t, Rt , T2 ))dt] 1 = Dt [? 1 (t) ft (t, Rt , T1 )dt + fr (t, Rt , T1 )dRt + frr (t, Rt , T1 )? 2 (t, Rt )dt 2 1 +? 2 (t) ft (t, Rt , T2 )dt + fr (t, Rt , T2 )dRt + frr (t, Rt , T2 )? 2 (t, Rt )dt 2 ? Rt (? 1 (t)f (t, Rt , T1 ) + ? 2 (t)f (t, Rt , T2 ))dt] 1 = ? 1 (t)Dt [? Rt f (t, Rt , T1 ) + ft (t, Rt , T1 ) + ? t, Rt )fr (t, Rt , T1 ) + ? 2 (t, Rt )frr (t, Rt , T1 )]dt 2 1 +? 2 (t)Dt [? Rt f (t, Rt , T2 ) + ft (t, Rt , T2 ) + ? (t, Rt )fr (t, Rt , T2 ) + ? 2 (t, Rt )frr (t, Rt , T2 )]dt 2 +Dt ? (t, Rt )[Dt ? (t, Rt )[? 1 (t)fr (t, Rt , T1 ) + ? 2 (t)fr (t, Rt , T2 )]]dWt = ? 1 (t)Dt [? (t, Rt ) ? ?(t, Rt , T1 )]fr (t, Rt , T1 )dt + ? 2 (t)Dt [? (t, Rt ) ? ?(t, Rt , T2 )]fr (t, Rt , T2 )dt +Dt ? (t, Rt )[? 1 (t)fr (t, Rt , T1 ) + ? 2 (t)fr (t, Rt , T2 )]dWt . (ii) Proof. Let ? 1 (t) = St fr (t, Rt , T2 ) and ? 2 (t) = ? St fr (t, Rt , T1 ), then d(Dt Xt ) = Dt St [? (t, Rt , T2 ) ? ?(t, Rt , T1 )]fr (t, Rt , T1 )fr (t, Rt , T2 )dt = Dt |[? t, Rt , T1 ) ? ?(t, Rt , T2 )]fr (t, Rt , T1 )fr (t, Rt , T2 )|dt. Integrate from 0 to T on both sides of the above equation, we get T DT XT ? D0 X0 = 0 Dt |[? (t, Rt , T1 ) ? ?(t, Rt , T2 )]fr (t, Rt , T1 )fr (t, Rt , T2 )|dt. If ? (t, Rt , T1 ) = ? (t, Rt , T 2 ) for some t ? [0, T ], under the assumption that fr (t, r, T ) = 0 for all values of r and 0 ? t ? T , DT XT ? D0 X0 > 0. To avoid arbitrage (see, for example, Exercise 5. 7), we must have for a. s. ?, ? (t, Rt , T1 ) = ? (t, Rt , T2 ), ? t ? [0, T ]. This implies ? (t, r, T ) does not depend on T . (iii) Proof. In (6. 9. 4), let ? 1 (t) = ? (t), T1 = T and ? (t) = 0, we get d(Dt Xt ) = 1 ? (t)Dt ? Rt f (t, Rt , T ) + ft (t, Rt , T ) + ? (t, Rt )fr (t, Rt , T ) + ? 2 (t, Rt )frr (t, Rt , T ) dt 2 +Dt ? (t, Rt )? (t)fr (t, Rt , T )dWt . This is formula (6. 9. 5). 1 If fr (t, r, T ) = 0, then d(Dt Xt ) = ? (t)Dt ? Rt f (t, Rt , T ) + ft (t, Rt , T ) + 2 ? 2 (t, Rt )frr (t, Rt , T ) dt. We 1 2 choose ? (t) = sign ? Rt f (t, Rt , T ) + ft (t, Rt , T ) + 2 ? (t, Rt )frr (t, Rt , T ) . To avoid arbitrage in this case, we must have ft (t, Rt , T ) + 1 ? 2 (t, Rt )frr (t, Rt , T ) = Rt f (t, Rt , T ), or equivalently, for any r in the 2 range of Rt , ft (t, r, T ) + 1 ? (t, r)frr (t, r, T ) = rf (t, r, T ). 2 56 6. 3. Proof. We note d ? e ds s 0 bv dv C(s, T ) = e? s 0 bv dv [C(s, T )(? bs ) + bs C(s, T ) ? 1] = ? e? s 0 bv dv . So integrate on both sides of the equation from t to T, we obtain e? T 0 bv dv C(T, T ) ? e? t 0 t 0 T bv dv C(t, T ) = ? t s 0 e? T t s 0 bv dv ds. Since C(T, T ) = 0, we have C(t, T ) = e 1 ? a(s)C(s, T ) + 2 ? 2 (s)C 2 (s, T ), we get A(T, T ) ? A(t, T ) = ? bv dv T t e? bv dv ds = T e t s bv dv ds. Finally, by A (s, T ) = T a(s)C(s, T )ds + t 1 2 ? 2 (s)C 2 (s, T )ds. t

Analyse how the development of relationship between characters helped you to understand the characters in the written texts. Lord of the Flies Essay

Published in 1954, William Golding’s novel the ‘Lord of the Flies’ developed many relationships to shed light on the characters. Perhaps the most important of these would be between the two main characters Jack and Ralph, however relationships between other characters including Ralph and Piggy’s relationship and Jack and Rodgers relationship are equally as important and help provide evidence of the behaviours of the children. The relationship between the two main characters Ralph and Jack at the beginning of the novel is an indication for how different the boys truly are. When the boys crash and arrive onto the deserted island they attempt friendship despite their differing personalities, â€Å"Jack and Ralph smile at each other with shy liking† just as any school boy would do. After being elected as chief of the tribe, Ralph makes his first move of power in electing Jack as chief of the hunting group to ease his embarrassment of wanting to be over all chief but not receiving this role highlighting Ralphs caring side to his personality. The boy’s continue to work together as a team and acknowledge one another’s leadership differences but manage to work together still in creating the best island possibly for the tribe. As the tension between them builds arguments begin to break out as their priorities begin to drift â€Å"they look at one another, baffled by love and hate†. With Ralph being a democratic leader, he takes in the opinions of the other tribe members in order to create the best society possible. It becomes an obvious barrier between the two boys drawing them back from being able to create the perfect society as they simply cannot respect one another’s way of leading. Jacks desire towards hunting and disobeying Ralph’s rules situation is similar to the biblical reference of Jack likening to Satan and Ralph Likening to God. People choose to follow Satan because he persuades and tricks people into believing his way is the more enjoyable way as it is a lot easier to go with Satan’s way then to follow Gods way, just as the tribe begins to fall under Jacks excitement towards hunting and ignoring the need to get rescued with a fire this being Ralphs rule. However this is all expected as for a child the more exciting and enjoyable approach to life is the path an average child would choose to follow proving the actions of the tribe to be considerably normal for them to want to begin following Jacks way of life. At the end of the novel the relationship between the characters Jack and Ralph proves the boys personality differences and helps the reader understand them. Throughout the whole novel it was evident of their unspoken hate towards one another and this tension is finally broken when the remainder of the tribe turns on Ralph as their previous chief and choose to follow Jack and his way of life. â€Å"And you shut up! Who are you anyway? Sitting there telling people what to do. You can’t hunt you can’t sing.† Jacks idea of a perfect leader is someone who had the previous roles of leadership in other scenarios, such as Jacks role of being leader of a school boys’ choir, giving Jack the impression that he is indeed perfect for the position of tribe leader. Jacks displays of power and leadership is Golding’s representation of an autocratic government where the power was taken from the tribe and not voted upon whereas Ralphs leadership and power was voted on just like a democratic government where he takes to mind the opinions of the other children. Understanding this, the reader can then begin to understand how dramatically different the characters each are and Golding uses the idea of ‘good versus evil’ in the characters Ralph and Jack by showing them opposites in their positions on social responsibility, personality, and appearance. The only thing Ralph and Jacks have in common is their utter hatred towards each other. Jacks arrogant and self-centred mind set causes him to believe that once hunted and killed, the pigs he catches make him worthy of the role of leader. However, in a tribe today any person who can bring back meat to feed the rest of the tribe is generally the best leader of the group, but due to the boys situation, it is only a bonus that the boys can have meat and the real focus should be getting rescued, this being Ralphs focus. With the tension being released within the last few chapters, Ralph is hunted due to the tribe’s sudden hatred towards him and their complete savagery takes over despite Ralphs caring nature and civilised personality. Throughout the novel Piggy and Ralphs relationship has been an eye opener to expose the boy’s nature and helps the reader to understand each character independently. When Ralph is introduced to Piggy instantly the differences between the two is shown by the way they speak: â€Å"sucks to your ass-mar!† said Ralph as he mocked Piggy’s unusual accent; from Ralph’s mockery, the reader can see that the two upbringings were very dissimilar. Piggy is in a lower socio-economic class and the two boys being forced into co-habitation together highlights the contrast between the two defined classes of society. Being a typical child Ralph chooses to name Piggy as Piggy just by looking  at his physical appearance once again showing Ralphs back round of being from a more upper-class region in England highlighting his ‘bullying the poorer child’ mind set. Understanding this, it is not expected that the boys would have a good relationship due to Ralphs immaturity but as Ralph soon realises the capability of Piggy’s knowledge and how that would help his leadership skills they soon become friends amongst the chaos. Piggy’s knowledge represents the law and order of the adult world. Throughout the novel, Piggy attempts to condition the society the boys had made to mirror the society they all lived in in England. Piggy’s continual references to his auntie demonstrate this philosophy. He tries to pull Ralph towards the reason-oriented side of human nature. The overall characteristics of each of these two important characters proves that amongst the islands nightmare state it had spiralled into there was still a sense of hope for humanity amongst it due to the maturity of Piggy who then taught Ralph his ways making Ralph an overall stronger leader, but despite all this the rest of the tribe still fell short and turned on both Ralph and Piggy making them â€Å"outsiders†. Ralph’s character can be once again shown through his relationship with Simon where Simon’s characteristics help shape Ralph into the boy he becomes. Simon can be seen as pure goodness, and he is something of a Christ like figure. Simon is fearless as he is the only one who goes to confront ‘The Beast’. The pureness in Simon, like Jesus Christ, prevents him from submitting to the beast-like nature of mankind just as Christ did not succumb into the terrible ways of man at his time. Simon also is the only character who stays true to his personal morals and beliefs although Ralph did try to stay true to his beliefs he did fall into the temptations of the tribe by joining them in a tribal dance around a pig they had just caught. Simon gives Ralph encouragement. Whenever Ralph becomes despondent or loses hope that they will be found, Simon lifts his spirits: â€Å"I just think you’ll get back all right.† Golding portrays Simon, the first boy to be killed (followed by the death of Piggy) as a symbol of a Christ-figure, of grace and light. Ralph’s characteristics of goodness, on the other hand is in contrast to Jack’s evil. Ralph’s goodness is based on order, logic, rationality, and reason. But due to Ralphs beliefs it ends up turning on him as he believed everyone has good in them, but through the novel the reader is introduced to the idea that not all people have good in them, and this is  highlighted through the imagery of school children turning into a â€Å"pack of painted niggers†. Throughout the novel â€Å"The Lord of the Flies† William Golding uses the relationships to ultimately shed light on the characters. The relationship between Ralph and Jack lets the reader begin to understand the true vision of good versus evil which is portrayed throughout the entire novel by Jack being the evil character and Ralph as being the good. Ralphs relationship with Piggy and Simon accompanies Golding’s idea of the good characters as Piggy and Simon pay important roles in aiding Ralph to make him the character he is at the end of the novel.

Some Theories of Personality Essay

Personality Introduction   Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚   There are a number of personalities that people display in life in accordance with different life situations that they encounter or are living in. The most common personality that a typical person is likely to display today that has been described by Sigmund Freud is that of ego. This is so because in the most part of our lives, we rely on unconscious part of mind (George, 2010). This part is the source reality of what can be afforded, how and when. It controls our desires on the basis of what we really should have at a particular time and if we cannot afford it at that time, the ego informs a person to postpone it for another day.   Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚   We are occasionally obliged to negate or repel becoming cognizant of some drives and thus, they in most cases manifest in a hidden form but the ego makes these drives real(George, 2010). Although human beings desire to have things that they think are of importance to them, this part of mind makes them aware of what they can afford or what they cannot afford. It is the ego that helps the typical person to relate things that he or she wants to the reality (Sow & Chan, 2010). It thus leads a person into looking for objects that can satisfy what he or she is in need of but because of the inability to acquire them, they compensate with some other source of satisfaction.   Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚   Relating to the world today, people are spending millions of money in search for self-realization through programs that are offered by several firms. These programs help them become enlightened and they quickly turn their lives around. To this end, these people feel that if they realize themselves, they will be able to live a happy life. They thus are willing to spend whatever the money they deem necessary to enroll in the programs that will help them realize themselves. Most of these individuals are wealthy or somewhat able of sustaining themselves if not wealthy but, they do not realize happiness in their wealth or ability. The only way they can get this happiness is through spending their money not on buying expensive possessions but by spending that money to seek for self-satisfaction (George, 2010).   Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚   Similarly, depending on the level of needs, most people are driven by a desire to accumulate wealth in their life. As they perceive, being wealthy means being happy because, they can afford to buy all that they desire with their wealth. Therefore, they would do anything within their power to satisfy their desire. Sometimes, the justified means of acquiring wealth that have been dictated by the society may not favor such individuals. They are, therefore, driven by their desires to create other means to achieve their goals. These other means may be lawful for example, through investing or starting a business or may be unlawful for example, the selling of drugs, corruption, stealing, and forceful amassing among other illegal means (Sow& Chan, 2010).   Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚   Unfortunately, when one desire is achieved, the body keeps on yearning for more. Such individuals form a disease out of a habit in the process of satisfying their motivations (Weiten et al, 2011). They thus identify by the means in which they achieve their desires and this suppresses their perception on the existing societal laws. If the activity that they achieve their desires through is unlawful, they think of it as lawful and usual. Later on, after such individuals accumulate more wealth, they realize that the wealth is not enough to make them happy. At this point, they invest in programs that are geared towards achieving self-realization (Weiten et al, 2011).   Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚   Such a personality affects a typical person’s behavior in several ways. First, he or she learns to balance between the lusty desires and the actual world expectations (Sow and Chan, 2010). When one is dominated by lusty desires, he or she is able to rationalize those desires through the use of ego personality. If the person allows the lusty desires personality to dominate, he or she goes against the societal norms or the stipulated laws. To this end, such a person is said to be a criminal.   Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚   Secondly, when an individual is driven by the desire to become wealthy through amassing of wealth either through conformity to stipulated laws or negating the laws, they are turned into wealth making machines whose role is to make wealth. This has contributed to failure in social structures such as the family because of the little time these individuals have to concentrate on building social relationships. No wonder the high rate of broken marriages and families. This has left the little children without the most basic contributors to their personality at early age namely; father and mother. Deficiency in personality development on children keeps on adding to the miseries of the world today (George, 2010).   Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚   Through engagement in illegal means of satisfying an individual’s desire, the rate of crimes in the country keeps on rising. It becomes difficult for the individual to do what is right because the part of personality that is responsible for informing is numbed from recognizing the right from the wrong. This is dangerous because the individual would do anything to maintain his source of wealth. No wonder there are many assassinations in places subjugated by illegal business such as the drug vending cartels (Sow& Chan, 2010). References George,B. (2006). Personality Theories. Found online at: htttp://www.social-psychology.de/do/pt_freud.pdf Sow, Gaik and Chan, Wai. (2010). Personality Development. Found online at: http://elearning.ibc.ac.th/sites/default/files/personality%20development_0.pdfWeiten, W.,Dunn, D. & Hammer, E.Y. (2011). Psychology Applied to Modern Life; Adjustment in the 21st Century: Tenth Edition. Belmont, CA: Wadsworth-Cengage. Source document

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(For Engineering Applicants Only) If you are applying to the Pratt School of Engineering, please discuss why you want to study engineering and why you would like to study at Duke - Essay Example Besides, I chose Clark because it offers my desired 3/2 dual engineering program in conjunction with Columbia University which is another reputed university in U.S. It provides an opportunity to earn prestigious B.A. degree and B.S. degree in engineering from Columbia University (CU). I have the option to major in any of the fields of engineering available. The Engineering degree from CU would offer me a rewarding and lucrative career like job satisfaction, variety of career opportunities, working challenges including in scientific and technological areas , self professional development including creativity and critical thinking, financial security, status in society, beneficiary to the society etc. It has an academic program center in China. I would get an opportunity to experience all around American culture. In the changing world the Liberal Arts college provides career-based education than the broad-based, choosing of a right college for post-secondary education is therefore not only an individual choice but a big decision for a student. Lafayette College’s mission for commitment to intellectual integrity and achievement not only made my decision easy but is a right choice because I too follow the intellectual and achievement path for learning. I had been a student known for raising issues and discussing topics on different subjects related with my high school study in class as well as in school. This could become possible because of my intellectuality power like languages achievement (language fluency for both English and Chinese), visualizing, logical, problem creations and solving etc. Many honors and rewards conferred on me speak the truth for themselves. As a part of my ongoing intellectuality power I would of course try to participate in the distinctive McKelvy House Schol ars Program during my stay at Lafayette. I would utilize my intellectuality power to earn my Bachelors

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How Local Merchants Can Work Together to Slow the Money in Ashland - Essay Example This will circulate the money within the boundaries of Ashland and the profits generated by the local businesses would benefit the economy as a whole in terms of business expansions and growth. The merchants have started their businesses in Ashland for the convenience of residency they have in Ashland. They had put in their finances in different businesses like food, clothing etc. to make all necessary products available in the market for the people. For a success in business, Ashland’s entrepreneurs need to invest wisely in the business that has more demand and less supply. The reputation of businesses and awareness about ‘slow money’ ideas to develop the economy of Ashland is necessary to attract more local consumers. The local economy, with the help of proper marketing of the products, services and awareness about local circularization of money to improve the economy, would grow. The local merchants support each other as they have maintained good relationships between them. They may work together to improve the social and economic conditions of Ashland and reduce the crime rates, disputes and poverty that prevail

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People and organization development - Coursework Example An organisation is composed of different units with specific roles that contribute to the success of the operations and the achievement of the goal. One of the most important aspects on the establishment and operation of an organisation is the change that dictates organisational development. Organisational development (OD) is the force that moves the company forward in different aspects. It is described as by Huse in 1980 as ‘the deliberate, reasoned, introduction, establishment, reinforcement, and spread of change for†¦the improvement of the organization both in terms of effectiveness and health† (Nel, 2009, p.2). For that matter, the change brought about by the reaction of the organisation to different intrinsic and extrinsic factors can be considered included in the definition of organizational development. These changes affect the components of an organisation specifically the people. One aspect of organisational development is even targeted to improve the well b eing of the people such as the employees and the members of the organisation. Due to the importance of the development of the people, the research undertaken is focused on one of the important dynamics in an organisation that is related to the people, team and team building. Team and Team Building Teams are cooperative groups in an organisation established to achieve common specific goals. ... This is where the concept of team building comes in. Team building is defined as the planned activities with the primary objective of improving the group dynamics. The main roles of team building activities include â€Å"improvement of the accomplishment of tasks, interpersonal skills, problem solving skills and team performance.† Such activities apply for different classifications of groups such as work groups, temporary project teams and virtual teams. In addition, problems and challenges within an organisation or within the team itself can be resolved through team building activities namely lack of cooperation, loss of productivity, conflicts within the group and in the work environment, lack of innovation and initiation, and the failure to achieve goals and complete tasks leading to inefficient services. Team building activities cover the different aspect within the organisation such as employee involvement, work design, restructuring, and strategic change (Cummings and Wo rley, 2009, p. 2263). In general, the issues related to team and team building is clearly connected to the change that can challenge the company and can result to either excel or fail. For that matter, the methods and principles of the dynamics of the team and the methods and techniques in team building are considered to be included in the basic protocols in the establishment and operation of organisations. Thus, to be able to achieve an understanding of the team and team building concept, it is important to consider the different topics related to the subject matter understudy. Objectives of the Study The study is aimed to present an overview of the concepts of the team and team building and the role on people and organisational development. In general, the

Reflection and Improvement Assignment Example | Topics and Well Written Essays - 1500 words

Reflection and Improvement - Assignment Example Leaders are seeking to transform their schools in order provide effective knowledge and education to the students (Joyner, Ben-Avie & Comer, 2004). The key conceptual framework would productively encompass the necessary cultural and structural change efforts. Moreover, effective conceptual framework can enhance significant pipeline development of leadership. However, major focus of this leadership process would be to improve the learning and teaching process. Learning Focused Leadership Learning focused leadership is another effective strategy that can enhance transformation of schools. Learning focused leadership comprises investment in instructional leadership, new working relations across and within levels, reinvention of leadership practice, persistent public focus on learning and evidence as a medium of leadership (Aguilera, 2008). The meaning of learning focused leadership is investing in individuals and positions across and within the schools. The primary objective of this lea rning focused leadership is instructional leadership. Conceptualizing leadership is more significant and effective among these two leadership styles. This leadership style generally focuses on effective learning process. Effective utilization of effective data and resources, and supreme engagement with the community has motivated me to adopt conceptualizing leadership strategy. Evaluation of Leadership Style There are several significances of conceptualizing leadership process. A teacher or a or a leader within the institution can enhance effective transformation of a school through the implementation of conceptualizing leadership process. Under this leadership style, the leader’s ability to offer a consistent and clear focus on learning for all the students would be considered as the central part of their job. Effective utilization of data and evidence can enhance instructional improvement. Moreover, this leadership style will help the leaders to align resources with the lea rning improvement goals. Under this process, the leaders generally try to reallocate their resources and develop incentives to meet specific instructional improvement goals. These construction roles help the leaders to focus on the improvement of learning process. The leaders seek to provide effective learning environment and seer knowledge to the students of the schools. At last, leaders focus on engaging community citizens, support providers and parents that can promote effective learning agenda. However, it is quite important to change the blueprint of education sector. The old traditional blue print is reducing the quality of education process and system. The conceptualization leadership process will help the leaders to transform the system of learning process of several schools and education institutions. Three ways to improve leadership style It is clear from above discussion that the education sector needs effective transformation process to meet social and cultural demand. T here are several ways to improve the leadership style. First of all, the educational leaders should become the transformation leaders to implement effective change process. Successful transformation leaders should avail three capabilities like think ahead, deliver within and lead across to adopt and practice an effective transformation process (Covey, 2012). Transformation leaders should think ahead in a school transformation pr

Political Science Essay Example | Topics and Well Written Essays - 500 words - 11

Political Science - Essay Example Frederick Douglass defined the celebration of the Fourth of July from the point of view of an African American in his 1852 speech at Rochester, New York. At that time, slavery of African Americans was at its peak and he was the first to address the feelings true of an American Slave toward the celebration of Fourth of July as he sees t as â€Å"mere bombast, fraud, deception, impiety, and hypocrisy [†¦covering] up crimes which would disgrace a nation of savages† (Douglass). The Declaration of Independence was not only about getting independence from the Kingdom of Great Britain but was also about freedom, equality, and liberty. The members of the Continental Congress firmly believed that â€Å"all men are created equal† and thus have certain â€Å"unalienable rights† including those of â€Å"life, liberty, and the pursuit of happiness† (Digital History). They made it clear in the declaration that it is the responsibility of the government to secure these rights of every American citizen and should a government fail to do so, or becomes a threat itself, the people have the right to â€Å"alter or to abolish it, and to institute new Government†(Digital History). These are the foundations that were laid by the forefathers of America and this is where the true meaning of Fourth of July lies. In a study conducted by the National Assessment of Educational Progress in 2011, it was found that only 25 per cent of fourth grade students were aware of the purpose of Declaration of Independence while 26 per cent of the total adults surveyed by a public opinion institute were not aware of the fact that America gained independence from the Kingdom of Great Britain (Shammas). It is clear from this survey that the true spirit of Fourth of July is dying. For me, Fourth of July is still about freedom, liberty, and pursuit of happiness not just for myself, but for all the Americans. Unfortunately, the speech made by Frederick Douglas is still true today as not all

Stereotyping In An Organization Essay Example for Free

Stereotyping In An Organization Essay The fact that people are always judged for their actions and behavior could be one of the biggest problems in an organization. Stereotyping is a fixed notion of people, coming up with their own assumption and judgment even before giving the respective a chance to explain the reason for both their actions and behavior. Stereotyping is a closure in which each individual blocks the opportunities of what people really are. People often take in all the information they feel most comfortable with and then close up. Thus, stereotypes are formed. There are many different kinds of stereotype that can go round in an organization. Such as gender, sex, status, etc. All these are formed because people are not open-minded enough to accept the fact that all human beings are different in their own way. There are still organizations that possess high masculinity in this generation where the men have more mobility compared to the women. In such an organization, they do not believe that women are capable of performing as well as men in the same task. Thus, women are not given the equal opportunity to excel. Yet, what these organizations have failed to understand is that, both the male and the female have their own good points in their work ethic. Research shows that men are more assertive, women are more cooperative; men are focused, specific and logical, women are holistic, organic and wide-angle. However before they come to a realization, the women would have already left the organization due to feeling taken advantage of and unfairne ss. Another very common stereotype, which is found not only in organizations but also in the society, would be sexual orientation. Up till today, gays are still constantly being judged in our society when they actually make up 11% of the world’s population and are still growing. The few reasons why people discriminate against gays is because they do not understand why they choose to behave this way and they feel ashamed. Despite that, neither do they give the gays a chance to explain their actions and behavior thus they just choose to make an assumption of their own and put them down because they are â€Å"different†. However, there are some countries that actually legalize gay marriages and are at the advantage because the gays would definitely move there as they have found a place that accepts them for who they really are without being judged. For organizations to be successful, they have to learn how to be more open-minded about such sensitive issues. They must be able to accept and deal with diversity. Respecting an individual for who they are is very important for relationship bonding as it promotes trust that will definitely result in better teamwork. Research Question: How much influence does stereotyping have on an organization? Research Objectives: * To find out the many different types of stereotypes that can be found in an organization. * To determine the cause of stereotyping and how they are linked to the success of an organization. * To identify ways to reduce stereotyping and how to motivate employees. Getting information from the people on the job has always been the best way for overcoming a problem as it all starts from them. Therefore getting feedback is important. Firstly, managers can occasionally conduct meetings with their employees and have them voice out their unhappiness. Secondly, managers must try to understand the reasons for their employees’ unhappiness so as to be able to work on them. Lastly, the organization must be able to present their employees with benefits that can motivate them to work harder. References: 1. Juliet Andrews, 2010, Stereotypes in the workplace and a test that shows you are not innocent, http://www.couriermail.com.au/business/business/stereotypes-in-the-workplace-and-a-test-that-shows-you-are-not-innocent/story-e6freqo6-1225861554141. [18 Feb. 2012] 2. Brian Amble, 2005, Gender stereotypes block womens advancement, http://www.management-issues.com/2006/8/24/research/gender-stereotypes-block-

Series Of Hong Kong Mainland Conflict Media Essay

Series Of Hong Kong Mainland Conflict Media Essay In 2012 to 2013, there are a series of Hong Kong-Mainland conflict. Hatred towards Chinese is ignited as quarrels between Hong Kong people and Chinese are seen easily in daily life. Some Hong Kong people claim Mainland Chinese as locust since they think Chinese sap the resources and welfare in Hong Kong. The uncivil behavior of Chinese such as shouting in public area can be easily seen and it deepens the Hong Kong-Mainland conflict. The discord even rises into the political level. Thousands of protesters waving British colonial flags showed the view that Hong Kong has their unique history is endemic among Hong Kong people. The upholders think that Hong Kong should not be governed by Mainland government and the One Country Two System should be preserved. While some of the Chinese think that Hong Kong is a part of China so China has the right to interfere the Hong Kong government. In facts, it is time for Hong Kong people think about the question about personal identity of themselves a nd the future of Hong Kong. Marxism and the National Question written by J.V. Stalin discuss about the definition of a nation and the minorities autonomy or self-determination which have reference value in today situation. Background information of this passage The passage was written in 1913. At that time, the spread of newspaper and of literature generally, a certain freedom of the press and cultural institution, and increase in the number of national theatres and so forth, all unquestionably helped to strengthen national sentiments  [1]  . Since there were lots of national movements the Russian Social-Democracy need to express their view towards this phenomenon. Is Hong Kong a nation? According to the passage, Hong Kong has the condition of becoming a nation. Stalin thinks that nation is not racial or tribal which means different race and tribe can combine with a same nation finally. A nation can only be defined as a nation when four conditions are present. The conditions are: formed on the basis of a common language, territory, economic life, and psychological make-up manifested in a common culture. The idea discussed is influential in China as it directed China government to recognized different nation. In fact, I agree the argument stated by Stalin about the definition of a nation. As Born and bred in Hong Kong, I can see a lot of differences between Chinese and Hong Kong people especially when more interactions happened. I think the differences explained why there are lots of conflicts in nowadays. This argument provides evidence to me to support Hong Kong as a nation. The first condition, common language, refers to the spoken language among the citizen rather than the official language. The spoken language in Hong Kong called Hong Kong Cantonese is a unified and special language which mixes up with words and phrases from English and Japanese and dialect from Min Nan etc. or develop some proverb according to the historical factor. Even through there are similarities between Hong Kong Cantonese and Putonghua, the main point is that there is no other spoken language can replace the position of Hong Kong Cantonese. Although both American and British speak English, they are different nation. The reason behind is the second condition, common territory. Only when people live together can become a nation. A part of Hong Kong people migrated from other place such as China and Southeast Asia. They settled down in Hong Kong generation after generation and in the new territory, they build up a new Hong Kong nation. The third factor is economic cohesion. Hong Kong experienced its own economic life which is different from Mainland China. In the early colonial period, Hong Kong is an entrepot trade center in Asia. Until Korea War, the , the entrepot trade in Hong Kong is affected as the United Nation implemented the embargo towards China. The economic transition started. Hong Kong proceeded industrialization and there are lots of light industry such as textile factories, electronic component factories and etc. The economic transition started again when China instituted the Open Door Policy. Hong Kong has become the international financial center until now even though there are barrier such as Asian financial crisis in 1997 and the SARS outbreak in 2003. Besides, the ideology in Hong Kong is different from the Mainland China. Hong Kong government believed in the free market policy in capitalism. Hong Kong was once described by Milton Friedman as the worlds greatest experiment in laissez-faire capit alism while China believed in their socialist market economy. Even though the Hong Kong government is driving economic integration between Hong Kong and China, there are still a lot of differences between two places such as business law, business culture and corporate social responsibility etc. The forth factor, a psychological makeup or national character which is intangible for the observer but cannot be ignored. National character is a not a fixed thing, it vary from different condition of life. The special national character of Hong Kong began to develop in the colonial period. The British culture was diffused and invading the position of traditional Chinese culture or combine with the traditional Chinese culture. Lion rock spirit descried as never give up, hard-working and endurance became a symbol of Hong Kong local and significant culture. After 1997, Hong Kong developed their local core value included democracy, rule of law. In recent year, there are a series of social movements aim at protecting these values, for example, some scholars are planning Occupying Central to strive for true universal suffrage, as Hong Kong government ignore Hong Kongs people feeling and driving integration between two places (some Hong Kong people claim that Hong Kong is turning red). What is Hong Kong future? The right of self-determination means only the nation may arrange its life in the way it wishes. It has the right to arrange its life on the basis of autonomy It has the right to enter into federal relations with other nations. It has the right to complete secession. Nations are sovereign, and all nations have equal rights. Stalin think that a nation has the right to secede especially there is policy of oppression such as restriction of freedom of movement, disfranchisement and repression of language etc. However, it doesnt mean all nations should secession and self-determination is beneficial to all nations. Stalin though that the solution of nation depends on the historical condition in the nation itself. Besides, Stalin connect the fate of national movement with bourgeois movement while he also state that a democratic country giving the opportunity to the nation to have free development can reduce the national struggle to minimum. Rather than secession, I believed the way to solve the national problems by a democratic government and it was a more suitable way for Hong Kong people in the past. As I thought a mature democratic government will not allow oppression policy towards the other nation or tyranny of the majority. Through democratic government, the minority can take advantage and get respect form other nation. Using the Switzerland which has stated by Stalin as an example, the direct democracy system and high level of autonomy in the twenty six states allow different nation enjoy an equal position. However, my view changed after the Catalan independence in last year. Even though there is democracy and autonomy in Catalan, once there are some policy which erode the advantage of the nation, national struggle will be ignited due to the oppressing policy in the past. It reminds me if the Beijing government keeps ignoring Hong Kong peoples demands, for example, the universal suffrage, or even instituting oppr essing policy, self-determination will be the only way for Hong Kong.