By Yan Zeng
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Additional resources for Stochastic Calculus for Finance, Vol. I and II, Solution
Example text
Similar to (iii). (v) Proof. c(T, s, v) = sf (T, log s, v) − e−r(T −t) Kg(T, log s, v) = s1{log s≥log K} − K1{log s≥log K} = 1{s≥K} (s − K) = (s − K)+ . 8. 61 Proof. We follow the hint. Suppose h is smooth and compactly supported, then it is legitimate to exchange integration and differentiation: ∂ ∂t gt (t, x) = ∞ ∞ h(y)p(t, T, x, y)dy = 0 ∞ gx (t, x) = h(y)pt (t, T, x, y)dy, 0 h(y)px (t, T, x, y)dy, 0 ∞ gxx (t, x) = h(y)pxx (t, T, x, y)dy. 45) implies 0 h(y) pt (t, T, x, y) + β(t, x)px (t, T, x, y) + 21 γ 2 (t, x)pxx (t, T, x, y) dy = 0.
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−rt 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 = αt St dt+σSt dWt . Hence dSt −rSt dt−adt = [(αt −r)St −a]dt+σSt dWt = σSt Set θt = (αt −r)St −a σSt and Wt = t θ ds 0 s (αt −r)St −a dt σSt + dWt . + Wt , we can find an equivalent probability measure P , under which S satisfies the SDE dSt = rSt dt + σSt dWt + adt and Wt is a BM.
8. 1. 2r x − σ2 −1 1 Proof. vL (L+) = (K − L)(− σ2r2 )( L ) L − σ2r 2 L (K − L) = −1. Solve for L, we get L = = − σ2r So vL (L+) = vL (L−) if and only if 2 L (K − L). x=L 2rK 2r+σ 2 . 2. Proof. 3, we can see v2 (x) ≥ (K2 − x)+ ≥ (K1 − x)+ , rv2 (x) − rxv2 (x) − 1 2 2 2 σ x v2 (x) ≥ 0 for all x ≥ 0, and for 0 ≤ x < L1∗ < L2∗ , 1 rv2 (x) − rxv2 (x) − σ 2 x2 v2 (x) = rK2 > rK1 > 0. 2 So the linear complementarity conditions for v2 imply v2 (x) = (K2 − x)+ = K2 − x > K1 − x = (K1 − x)+ on [0, L1∗ ].