# Brownian Motion - Closed Form Solution

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 $$t \geq r$$?

• 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 Sep 26, 2020 at 7:16

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