Expectation as an integral I wish to express as a Lebesgue integral the following expectation,
$E[\varphi(B_t)\varphi(B_s)]=\int ?$
for $0\leq s\leq t$, where $B_t$ is a Brownian motion with law $N(0,\sigma^2 t)$. Any ideas? I guess the point is to use independent increments since I would like to avoid the joint distribution.
Thank you very much! :)
 A: Let $T\colon (x,y)\mapsto (x-y,y)$. Then 
$$E[g(B_t,B_s)]=E[h(B_t-B_s,B_s)],$$
where $h(T(x,y))=h(x-y,y)=g(x,y)$. This gives, using independence of the increments of Brownian motion, and a density of $(B_t-B_s,B_s)$, and a substitution,
\begin{align}
E[g(B_t,B_s)]&=\int_{\Bbb R^2}h(u,v)\frac 1{\sqrt{2\pi(t-s)}\sigma}\frac 1{\sqrt{2\pi s}\sigma}\exp\left(-\frac 1{2\sqrt{\sigma}}\left(\frac{u^2}{\sqrt{t-s}}+\frac{v^2}{\sqrt s}\right)\right)dudv\\
&=\frac 1{2\pi\sigma^2}\int_{\Bbb R^2}g(x_1,x_2)\exp\left(-\frac 1{2\sqrt{\sigma}}\left(\frac{(x_1-x_2)^2}{\sqrt{t-s}}+\frac{x_2^2}{\sqrt s}\right)\right)dx_1dx_2.
\end{align} 
We can generalize this: if $t_1<\dots<t_n$, let $T$ the map given by 
$$T(x_n,\dots,x_1)=(x_n-x_{n-1},x_{n-1}-x_{n-2},\dots,x_2-x_1,x_1).$$
Let $h$ such that $h(T(x_n,\dots,x_1))=g(x_n,\dots,x_1)$. Then, writing $t_0=0=x_0$, 
\begin{align}
E[g(B_{t_n},\dots,B_{t_1})]&=\int_{\Bbb R^n}h(T(x_n,\dots,x_1))\prod_{j=1}^n\frac 1{\sqrt{2\pi(t_j-t_{j-1})}\sigma}\exp\left(-\frac{s_j^2}{2\sigma\sqrt{t_j-t_{j-1}}}\right)ds\\
&=\int_{\Bbb R^n}g(x_n,\dots,x_1)\prod_{j=1}^n\frac 1{\sqrt{2\pi(t_j-t_{j-1})}\sigma}\exp\left(-\frac{(x_j-x_{j-1})^2}{2\sigma\sqrt{t_j-t_{j-1}}}\right)dx.
\end{align}
