By Chuang C.T. (ed.)

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2 ϕ(t) γ cosh(γ(T − t)) + 12 b sinh(γ(T − t)) (vi) Proof. 8), A (t, T ) = 2aϕ (t) σ 2 ϕ(t) . Hence T A(T, T ) − A(t, T ) = t 2aϕ (s) 2a ϕ(T ) ds = 2 ln , 2 σ ϕ(s) σ ϕ(t) and 1 A(t, T ) = − 2a ϕ(T ) 2a γe 2 b(T −t) ln = − ln . 5. (i) Proof. Since g(t, X1 (t), X2 (t)) = E[h(X1 (T ), X2 (T ))|Ft ] and e−rt f (t, X1 (t), X2 (t)) = E[e−rT h(X1 (T ), X2 (T ))|Ft ], iterated conditioning argument shows g(t, X1 (t), X2 (t)) and e−rt f (t, X1 (t), X2 (t)) ar both martingales. (ii) and (iii) Proof. We note dg(t, X1 (t), X2 (t)) 1 1 = gt dt + gx1 dX1 (t) + gx2 dX2 (t) + gx1 x2 dX1 (t)dX1 (t) + gx2 x2 dX2 (t)dX2 (t) + gx1 x2 dX1 (t)dX2 (t) 2 2 1 2 2 = gt + gx1 β1 + gx2 β2 + gx1 x1 (γ11 + γ12 + 2ργ11 γ12 ) + gx1 x2 (γ11 γ21 + ργ11 γ22 + ργ12 γ21 + γ12 γ22 ) 2 1 2 2 + gx2 x2 (γ21 + γ22 + 2ργ21 γ22 ) dt + martingale part.

S. by the continuity of Y . This implies m j=1 (Ytj − Ytj−1 )(Stj − Stj−1 ) → 0. 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∗ ].

D(Dt St ) = St dDt + Dt dSt + dDt dSt = −St Rt Dt dt + Dt αt St dt + Dt σt St dWt = Dt St (αt − Rt )dt + Dt St σt dWt . 20). 2. DT VT ZT Zt Proof. , E[DT VT |Ft ] = E |Ft . 30) is equivalent to Dt Vt Zt = E[DT VT ZT |Ft ]. 3. (i) Proof. cx (0, x) = = = = 1 2 d E[e−rT (xeσWT +(r− 2 σ )T − K)+ ] dx 1 2 d E e−rT h(xeσWT +(r− 2 σ )T ) dx 1 E e−rT eσWT +(r− 2 σ − 21 σ 2 T e E e σ WT 2 )T 1 = e √ σ T E e WT √ T 1 1 e− 2 σ 2 ∞ −∞ = >K} 2 σ 1 √ T (ln K x √ −(r− 12 σ 2 )T )−σ T } √ z2 1 √ e− 2 eσ T z 1{z−σ√T >−d+ (T,x)} dz 2π T ∞ = x √ W { √T −σ T > T = 2 )T 1{WT > 1 (ln K −(r− 1 σ2 )T )} σ − 21 σ 2 T 1 {xeσWT +(r− 2 σ √ (z−σ 1 2 √ e− 2π −∞ N (d+ (T, x)).