calculate a conditional expectation

Question

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?

Answer 1

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.) $$ \mathbb E(\zeta;\zeta\gt-z)=\int_{-z}^{+\infty}t\varphi(t)\mathrm dt=\left[-\varphi(t)\right]^{+\infty}_{-z}=\varphi(-z)=\varphi(z). $$