By Daniel A. Murray

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**Additional resources for A first course in infinitesimal calculus**

**Example text**

Ii) Proof. We prove the following more general lemma. Lemma 1. e. ) defined on the probability space (Ω, G, P ). s. Proof. s. Let A = {Y = 0}, we shall show P (A) = 0. In∞ 1 deed, note A ∈ F, 0 = E[Y IA ] = E[E[X|F]IA ] = E[XIA ] = E[X1A∩{X≥1} ] + n=1 E[X1A∩{ n1 >X≥ n+1 }] ≥ ∞ 1 1 1 P (A∩{X ≥ 1})+ n=1 n+1 P (A∩{ n1 > X ≥ n+1 }). So P (A∩{X ≥ 1}) = 0 and P (A∩{ n1 > 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. s..

By Itˆ o’s formula, d(Bt Wt2 ) = Bt dWt2 + Wt2 dBt + dBt dWt2 = Bt (2Wt dWt + dt) + Wt2 sign(Wt )dWt + sign(Wt )dWt (2Wt dWt + dt) = 2Bt Wt dWt + Bt dt + sign(Wt )Wt2 dWt + 2sign(Wt )Wt dt. So t E[Bt Wt2 ] = E[ t Bs ds] + 2E[ sign(Ws )Ws ds] 0 0 t = t E[Bs ]ds + 2 0 E[sign(Ws )Ws ]ds 0 t = (E[Ws 1{Ws ≥0} ] − E[Ws 1{Ws <0} ])ds 2 0 ∞ t = 4 0 0 x2 e− 2s dxds x√ 2πs t = = s ds 2π 0 0 = E[Bt ] · E[Wt2 ]. 4 Since E[Bt Wt2 ] = E[Bt ] · E[Wt2 ], Bt and Wt are not independent. 20. (i) Proof. f (x) = if x > K 1, x − K, if x ≥ K So f (x) = undefined, if x = K and f (x) = 0, if x < K.

2 So we must have ft + (r + 21 v)fx + (a − bv + ρσv)fv + 12 fxx v + 12 fvv σ 2 v + σvρfxv = 0. 32). (iv) Proof. 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.

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