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Doing some exercises from a mathematical finance book, I got stuck at the following point. It is a purely probability question. Let $S_t^1 = \sigma W_t$, where $W_t$ is a brownian motion and $\sigma>0$ a parameter. Furthermore let $K>0$ also be a positive constant. I want to compute the price of a call option under $Q$, i.e.

$$E_Q[(S_T^1-K)^+|\mathcal{F}_t]$$

So far I was able to do this: Let $A:=\{S^1_T>K\}$

$$E_Q[(S_T^1-K)^+|\mathcal{F}_t]=E_Q[S_T^1\mathbf1_A|\mathcal{F}_t]-KE_Q[\mathbf1_A|\mathcal{F}_t]$$

Writing $S^1_T=S_t^1+\sigma(W_T-W_t)$ leads to

$$\sigma E_Q[(W_T-W_t)\mathbf1_A|\mathcal{F}_t]+(S^1_t-K)E_Q[\mathbf1_A|\mathcal{F}_t]$$

Now here is the point, where I got stuck. I know $(W_T-W_t)$ is independent of $\mathcal{F}_t$ but I do not see if $A\in \mathcal{F}_t$. Or how else should I simplify this?

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What is $T$ here? You also have to use $B:=\{\sigma W_t>K\}$ (then $(S_t^1-K)^+$ will appear). –  Davide Giraudo Oct 13 '12 at 15:14
    
@DavideGiraudo $T$ should be a fixed time horizon, $T\ge t$. –  user20869 Oct 14 '12 at 12:02
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1 Answer

up vote 2 down vote accepted

Here is a general result.

Let $\xi$ and $\eta$ denote two random variables on a given probability space $(\Omega,\mathcal F,\mathbb P)$, $\mathcal G\subseteq\mathcal F$ any sigma-algebra and $u$ any nonnegative function. Assume that $\xi$ is $\mathcal G$-measurable and that $\eta$ is independent of $\mathcal G$. Then, $\mathbb E(u(\xi,\eta)\mid \mathcal G)=v(\xi)$, where $v:x\mapsto\mathbb E(u(x,\eta))$.

Applying this result to $\xi=W_t$, $\eta=W_T-W_t$, $\mathcal G=\mathcal F_t$ and $u(x,y)=(\sigma(x+y)-K)^+$ yields the formula $\mathbb E((\sigma W_T-K)^+\mid \mathcal F_t)=v(W_t)$ with $v:x\mapsto\mathbb E((\sigma\sqrt{T-t}\cdot\zeta+\sigma x-K)^+)$, where $\zeta$ is a standard normal random variable. Thus, $$ v(x)=\sigma\sqrt{T-t}\cdot g\left(\frac{\sigma x-K}{\sigma\sqrt{T-t}}\right), $$ where, for every $z$, $$ g(z)=\mathbb E((\zeta+z)^+)=z\Phi(z)+\varphi(z),\qquad\varphi(z)=\frac{\mathrm e^{-z^2/2}}{\sqrt{2\pi}},\quad\Phi(z)=\int_{-\infty}^z\varphi(t)\,\mathrm dt. $$ Edit: (This is to answer a question asked in the comments.) $$ \mathbb E(\zeta;\zeta\gt-z)=\int_{-z}^{+\infty}t\varphi(t)\mathrm dt=\left[-\varphi(t)\right]^{+\infty}_{-z}=\varphi(-z)=\varphi(z). $$

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Thanks for your answer. I know this result and I also wanted to use it! That's the reason why I asked if $A\in\mathcal{F}_t$. But I defined the wrong variables. However there is one point which I do not get: $E[(\zeta+z)^+]=z\Phi(z)+\phi(z)$. I did the calculation and the term $z\Phi(z)$ is coming from $zP[\zeta> -z]$. But why is $E[\zeta\mathbf1_{\{\zeta > -z\}}]=\phi(z)$? –  user20869 Oct 14 '12 at 11:59
    
Because $t\varphi(t)=-\varphi'(t)$. (Of course, $A$ is not in $\mathcal F_t$ as soon as $t\lt T$, one cannot predict with certainty whether $W_T\geqslant K/\sigma$ or not, using only the path $(W_s)_{s\leqslant t}$.) –  Did Oct 14 '12 at 13:14
    
Sorry for bothering you, but how does $t\varphi (t)=-\varphi^`(t)$ imply $E[\zeta\mathbf1_{\{\zeta>-z\}}]=\varphi(z)$ ? –  user20869 Oct 14 '12 at 13:36
    
See Edit. $ $ $ $ –  Did Oct 14 '12 at 13:45
    
thanks so much! I did very bad mistake in reasoning. –  user20869 Oct 14 '12 at 16:08
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