Let $S_t$ be a geometric Brownian motion defined as:

$dS_t = \mu S_t dt +\sigma S_t dW_t $ Where $W_t$ is a Wiener Process or Brownian Motion, $\mu$ is the drift term and $\sigma$ is the volatility and both are constants.

Does a closed-form solution exists for the Expectation of this geometric brownian motion. In other words, what is $\mathbb{E}[S_r | (S_r > a)\cap(S_t > b)]$ for $t \geq r$?

  • 1
    $\begingroup$ With respect to which sigma field are you taking the expectation? Because if it is $\mathcal{F}_{t}$ of the natural filtration, then $S_{r}$ is $\mathcal{F}_{t}$-measurable $\endgroup$ Sep 26, 2020 at 7:16

1 Answer 1


Let's assume you are using the natural filtration.

For $t=r$ we have $$\mathbb{E}[S_r | (S_r > a)\cap(S_t > b)] = E[S_r | S_r > \max\{a,b\}] = \frac{E[S_r1_{S_r > \max\{a,b\}}]}{P(S_r > \max\{a,b\})}$$

For $t>r$ it holds that $S_r$ and $\frac{S_t}{S_r}$ are independent and so:

\begin{align*}E[S_r1_{S_r > a}1_{S_t>b}] &= E\left[S_r1_{S_r > a}1_{\frac{S_t}{S_r}S_r>b}\right] \\ \\&= \int_a^\infty \int_{\frac{b}{x}}^\infty xf(x,y) \,dy\,dx\end{align*}

With $$f(x,y) = f_{S_t}(x)f_\frac{S_t}{S_r}(y)$$ where $f_X$ denotes the density function of $X$ so in our cases the density of the related lognormal distribution.

For the same reasons we get: $$P((S_r > a)\cap(S_t > b)) = \int_a^\infty \int_{\frac{b}{x}}^\infty f(x,y) \,dy\,dx = \int_a^\infty f_{S_t}(x) \int_{\frac{b}{x}}^\infty f_\frac{S_t}{S_r}(y) \,dy\,dx$$

And we conclude for $t>r$: $$\mathbb{E}[S_r | (S_r > a)\cap(S_t > b)] = \frac{E[S_r1_{S_r > a}1_{S_t>b}]}{P((S_r > a) \cap (S_t>b))} = \frac{\int\limits_a^\infty \int\limits_{\frac{b}{x}}^\infty xf(x,y) \,dy\,dx}{\int\limits_a^\infty \int\limits_{\frac{b}{x}}^\infty f(x,y) \,dy\,dx}$$


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