# Conditional expectation of a stochastic exponential

Suppose we have a process $(X_t)$ such that under a measure $Q$ we know that

$$X_t=\mathcal{E}\left(\int_0^t\lambda(s) \, dW(s)\right)$$

for a deterministic function $\lambda(s)$ and a Brownian Motion $W$ under $Q$. Hence $X_t$ is lognormal distributed with mean $\mu:=-\frac{1}{2}\int_0^t |\lambda(s)|^2ds$ and variance $\sigma^2:=\int_0^t |\lambda(s)|^2ds$. Now, the process $(X_t)$ is adapted w.r.t to a filtration $(\mathcal{F}_t)$. Let $K$ be a positive constant, $t>s$ how do I calculate the following:

$$E_Q[\mathbf1_{\{X_t>K\}}|\mathcal{F}_s]$$

If we need, we can assume that $\mathcal{F}$ is generated by the Brownian motion $W$.

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Call $X_t^\lambda$ the exponential martingale defined in the post. Note that $X^\lambda_t=X^\lambda_sY$ where $Y$ is independent of $\mathcal F_s$ and distributed like $X^{\lambda_s}_{t-s}$ where $\lambda_s(\cdot)=\lambda(s+\cdot)$. Thus, $$Q(X^\lambda_t\gt K\mid\mathcal F_s)=G(K(X^\lambda_s)^{-1}),\qquad G:x\mapsto Q(X^{\lambda_s}_{t-s}\gt x).$$ Note that $X^{\lambda_s}_{t-s}=\exp(\mu+\sigma Z)$ where $\mu=\mu_t-\mu_s$, $\sigma^2=\sigma^2_t-\sigma^2_s$ and $Z$ is standard normal. Thus, for every $x$, $G(x)=Q(\sigma Z\gt\log x-\mu)$, that is, $G(x)=\Phi((\mu-\log x)/\sigma)$. Finally, $$Q(X_t\gt K\mid\mathcal F_s)=\Phi\left(\frac{\mu_t-\mu_s-\log K+\log X_s}{\sqrt{\sigma^2_t-\sigma^2_s}}\right).$$
thank you for the answer. Just two things: Why can we find such a $Y_{t-s}$ and why is $X_t=X_sY_{t-x}$? If we look at the stochastic integral we have something like $\int_0^s\lambda(u)dW(u)+\int_0^{t-s}\lambda(u)dW(u)\not=\int_0^t\lambda(u)dW(u)‌​$ –  user20869 Feb 1 '13 at 16:05
Which is the reason why $X_t\ne X_sX_{t-s}$... But (1) $X_t/X_s$ is based on $\bar W_u=W_{s+u}-W_u$ like $X_{t-s}$ is based on $(W_u)_{u\leqslant s}$ and (2) $\bar W$ is a Brownian motion independent of $X_s$ (can you see why?). This is all one needs. –  Did Feb 1 '13 at 16:31
I am not sure what you mean by "based on". It seems to me, that you want to do the same "trick" as if $\lambda$ would be a constant: $\int_0^t\lambda W_s=\lambda W_t=\lambda ( W_t-W_s) +\lambda W_s$ and then use independent increments. But here we have integrals, so I do not know how to apply this. We have proved in class that $\bar{W}$ is independent of $\sigma(W_l;l\le s)$ hence also of $X_s$ –  user20869 Feb 1 '13 at 16:44
No. Call $G(\lambda,W)=\mathcal E\left(\int_0^{t-s}\lambda(s)dW(s)\right)$, then $X_t=X_sY$ where $Y=G(\bar\lambda,\bar W)$, $\bar W(u)=W(s+u)-W(s)$ and $\bar\lambda(u)=\lambda(s+u)$ for every $u$. Then $X_s$ is measurable with respect to $(W(u))_{u\leqslant s}$, $Y$ is measurable with respect to $(\bar W(u))_{u\leqslant t-s}$ and $(\bar W(u))_{u\leqslant t-s}$ is independent of $(W(u))_{u\leqslant s}$. Hence $Y$ is independent of $X_s$ and distributred like $X_{t-s}$ would be distributed if it used $\bar\lambda$ instead of $\lambda$. (I will modify this $\lambda\to\bar\lambda$ stuff.) –  Did Feb 1 '13 at 16:53
Sorry, but I really don't get it. I agree that $Y$ and $X_s$ are independent. The things I do not get is: $X_t=X_sY$ and about the distribution of $Y$. To the first one: Again for the stochastic integrals, we have: $\int_0^s \lambda(r)dW(r)+\int_0^{t-s}\lambda(s+r)d\bar{W}(r)=\int_0^s \lambda(r)dW(r)+\int_0^{t-s}\lambda(s+r)dW(s+r)-\int_0^{t-s}\lambda(s+r)dW(r)$. The latter should be equal to $\int_0^t\lambda(r) dW(r)$, which I really not see. –  user20869 Feb 1 '13 at 17:31