s****y
发帖数: 390 | 1 dS = mu*dt+sigma*dW
V(S(T),T)=1/S(T)
ask for V(0)
How can I do this??? >>>_<<< | m******e
发帖数: 45 | 2 dS = mu*dt + sigma*dW or dS/S = mu*dt + sigma*dW, huge difference btw these
two. | b******w
发帖数: 52 | 3
There might be smart solution, hopefully somebody can point that out!
In the risk-neutral world, dS = rS dt + \sigma dw',
it can shown that S_t = S_0 e^{rt} + \int_{0}^t e{r(t-s)} \sigma dW_s,
which is normally distributed.
Then you might compute the following expectation, but I did not work that
out.
E[e^{-rT} 1/S(T)].
【在 s****y 的大作中提到】
: dS = mu*dt+sigma*dW
: V(S(T),T)=1/S(T)
: ask for V(0)
: How can I do this??? >>>_<<<
| w******w
发帖数: 92 | 4 Suppose it is ds/s=mu*dt+sigma*dw.
In risk neutral it is ds/s=r*dt+sigma*dW.
Denote Y=1/s
From Ito's lemma, you can get
d(ln Y)=( -r + sigma^2/2) *dt-sigma*dW
So Y is lognormal. Then it is easy to obtain that
E(e^{-rT}/s(T))=(1/s_0)*e^{ -2rT + sigma^2*T }
【在 s****y 的大作中提到】
: dS = mu*dt+sigma*dW
: V(S(T),T)=1/S(T)
: ask for V(0)
: How can I do this??? >>>_<<<
| q*****r
发帖数: 43 | 5 我的结果是E(Y(T))=(1/s_0)*e^(-rT+0.5*sigma^2*T)
【在 w******w 的大作中提到】
: Suppose it is ds/s=mu*dt+sigma*dw.
: In risk neutral it is ds/s=r*dt+sigma*dW.
: Denote Y=1/s
: From Ito's lemma, you can get
: d(ln Y)=( -r + sigma^2/2) *dt-sigma*dW
: So Y is lognormal. Then it is easy to obtain that
: E(e^{-rT}/s(T))=(1/s_0)*e^{ -2rT + sigma^2*T }
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