Let $B$ be one-dimensional brownian motion, and let $s<t$ be fixed. I want to find the joint probability distribution of $B_s$ and $B_t$.
My work:
We know that $B_s \sim N(0,s)$ and $B_t|B_s = y \sim N(y, t-s)$. Therefore the joint PDF of $B_s$ and $B_t$ is $$f_{B_t, B_s}(x,y) = f_{B_t | B_s =y}(x) f_{B_s}(y)$$ $$ = \frac{1}{\sqrt{2\pi}(t-s)}\exp\left[-\frac{x - y}{2(t-s)} \right] \frac{1}{\sqrt{2\pi}s}\exp\left[-\frac{y}{2s} \right] $$ $$ = \frac{1}{2\pi s(t-s)}\exp\left[-\frac{x - y}{2(t-s)} \right] $$ $$ = \frac{1}{2\pi s(t-s)}\exp\left[-\frac{s(x-2y) +ty}{2s(t-s)} \right] $$
Is this correct? I thought it might be simpler. Thanks!