# Stochastic Calculus Solution Manual

Topics: Trigraph, Gh, Martingale Pages: 85 (19710 words) Published: November 18, 2012
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 ﬁnance, by Steven Shreve. If you have any comments or ﬁnd any typos/errors, please email me at yz44@cornell.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.

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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 deﬁnition of arbitrage for arbitrary X0 can be given. Indeed, for the one-period binomial model, we can deﬁne 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 “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 “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 ﬁrst 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 5.7. 1.2.

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5 Proof. X1 (u) = ∆0 × 8 + Γ0 × 3 − 5 (4∆0 + 1.20Γ0 ) = 3∆0 + 1.5Γ0 , and X1 (d) = ∆0 × 2 − 4 (4∆0 + 1.20Γ0 ) = 4 −3∆0 − 1.5Γ0 . 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 ¯ ¯ payoﬀ 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 Γ0 options. He then puts his cash position ¯ −∆0 S0 − Γ0 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 + Γ0 V1 − (1 + r)(∆0 S0 + Γ0 X0 ). Plug in the expression of V1 and sort out terms, we have ¯ X1 = S0 (∆0 + ∆0 Γ0 )( 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...