Answers for Stochastic Calculus for Finance I

Answers for Stochastic Calculus for Finance I; Steven Shreve VJul 15 2009
Marco Cabral [e-mail] Department of Mathematics Federal University of Rio de Janeiro July 25, 2009

Chapter 1
−→ 1.1: Since S1 (H) = uS0 , S1 (T ) = dS0 , X1 (H) = ∆0 S0 (u − (1 + r)) and X1 (T ) = ∆0 S0 (d − (1 + r)). Therefore, X1 (H) positive implies X1 (T ) negative and vice-versa. −→ 1.2: Check that X1 (H) = 3∆0 − 3/2Γ0 = −X1 (T ). Therefore if X1 (H) is positive, X1 (T ) is negative and vice-versa. −→ 1.3: By (1.1.6), V0 = S0 . −→ 1.4: mutatis-mutandis the proof in Theorem 1.2.2 replacing u by d and H by T . −→ 1.5: many computations. −→ 1.6: We have 1.5 −V1 = ∆0 S1 +(−∆0 S0 )(1+r). We determine ∆0 = −1/2. So we should sell short 1/2 stocks. −→ 1.7: see last exercise. −→ 1.8: (i) vn (s, y) = 2/5[vn+1 (2s, y + 2s) + vn+1 (s/2, y + s/2)] (ii) v0 (4, 4) = 2.375. One can check that v2 (16, 28) = 8, v2 (4, 16) = 1.25, v2 (4, 10) = 0.25 and v2 (1, 7) = 0. (iii) δ(s, y) = [vn+1 (2s, y + 2s) − vn+1 (s/2, y + s/2)]/[2s − s/2]. −→ 1.9: (i) Vn (w) = 1/(1+rn (w))[pn (w)Vn+1 (wH)+qn (w)Vn+1 (wT )] where pn (w) = (1 + rn (w) − dn (w))/(un (w) − dn (w)) and qn (w) = 1 − pn (w). (ii) ∆n (w) = [Vn+1 (wH) − Vn+1 (wT )]/[Sn+1 (wH) − Sn+1 (wT )] (iii) p = q = 1/2. V0 = 9.375. One can check that V2 (HH) = 21.25, V2 (HT ) = V2 (T H) = 7.5 and V2 (T T ) = 1.25.

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Chapter 2
−→ 2.1: (i) A and A are disjoint and their union is the whole space. Therefore P (A ∪ A [= whole space]) = 1 = P (A) + P (A ) [they are disjoint]. (ii) it is enough to show it for two events A and B. By induction it follows for any finite set of events. Since A ∪ B = A ∪ (B/A), and this is a disjoint union, P (A ∪ B) = P (A) + P (B/A). Since B/A is a subset of B, P (B/A) ≤ P (B) by definition (2.1.5). −→ 2.2: (i) P (S3 = 32) = P (S3 = 0.5) = (1/2)3 = 1/8; P (S3 = 2) = P (S3 = 8) = 3(1/2)3 = 3/8. (ii) ES1 = (1 + r)S0 = 4(1 +