Brownian motion conditional expectation

Question

I need to solve for the following in my model: $E[X_t^i|X_s < K_1, X_t > K_2]$ where $X$ is Brownian motion and $i$ is a real number. any suggestion? I already know about the simpler case: $E[X_t^i|X_s < K_1]$

Answer 1

A partial result:

Let $W_t$ be the Brownian motion, and assume $s<t$: $$ \mathbb{E}(W_t^i | W_s<K_1\cup W_t>K_2)=\mathbb{E}(\mathbb{E}(W_t^i | W_s<K_1\cup W_t>K_2)|W_s)= $$ Use the independence of $\sigma(W_v)_{v=0}^s$ and $W_t-W_s$, so we can write this as: $$ \int_{-\infty}^{K_1}\frac{1}{\sqrt{2\pi s}}\exp\left(\frac{-x^2}{2s}\right)\int_{K_2-x}^{\infty}y^{i}\frac{1}{\sqrt{2\pi(t-s)}}\exp\left(\frac{-(y-x)^2}{2(t-s)}\right)\mathrm{d}y \,\mathrm{d}x $$ (Interpretation of the PDFs: the BM goes under $k_1$ at time $s$, then goes above the level $k_2$ at $t$)