Assume V(S, t) is a smooth function that satisfies the partial differential equation 1 + +r...
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Assume V(S, t) is a smooth function that satisfies the partial differential equation 1 + +r S as? -rV = 0, as subject to V(S,T) = max(S – K,0), V(0, t) = 0, V(S, t) - S as S → 0. Here r, o 2 0, K >0 are constants. The purpose of the exercise is to reduce equation (1), by a suitable change of variables, to the heat equation and then solve for V(S, t). (a) Use the following change of variables t = T - S = K e", V = K v(x, 7) to show that (1) becomes dv dv - kv, + (k – 1) dx where k = r/}o? and v(x, 0) = max(e" – 1,0). (b) Now let v(x, 7) = ear+3T u(x, 7), for so constants a and 8 and show that v(x, T) = e (k-1) *- (k+1)²7 u(x, 7), where du for – 0 < a <0, T>0, with u(x.0) = max (et(*+1)= - et (k-1)=,0) and = - (k – 1), B = (c) Show that u(x, 7) = ez (k+1) r+ (k+1)? r (d1)- ež (k-1)r+ (k-1)?r (d). where (y) is the normal cumulative density function and log(S(t)/K) + (r -²) (T – t) o VT - t d2 = di = d2 +o VT– t. and (d) Finally, show that V(S, t) = S -E=r(T-t) K Þ(d2). Assume V(S, t) is a smooth function that satisfies the partial differential equation 1 + +r S as? -rV = 0, as subject to V(S,T) = max(S – K,0), V(0, t) = 0, V(S, t) - S as S → 0. Here r, o 2 0, K >0 are constants. The purpose of the exercise is to reduce equation (1), by a suitable change of variables, to the heat equation and then solve for V(S, t). (a) Use the following change of variables t = T - S = K e", V = K v(x, 7) to show that (1) becomes dv dv - kv, + (k – 1) dx where k = r/}o? and v(x, 0) = max(e" – 1,0). (b) Now let v(x, 7) = ear+3T u(x, 7), for so constants a and 8 and show that v(x, T) = e (k-1) *- (k+1)²7 u(x, 7), where du for – 0 < a <0, T>0, with u(x.0) = max (et(*+1)= - et (k-1)=,0) and = - (k – 1), B = (c) Show that u(x, 7) = ez (k+1) r+ (k+1)? r (d1)- ež (k-1)r+ (k-1)?r (d). where (y) is the normal cumulative density function and log(S(t)/K) + (r -²) (T – t) o VT - t d2 = di = d2 +o VT– t. and (d) Finally, show that V(S, t) = S -E=r(T-t) K Þ(d2).
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