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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! :)

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What is $\varphi$? – Davide Giraudo Oct 21 '12 at 18:53
A function/transformation. Like when you do: $E[\varphi(X)]=\int_{\mathbb{R}} \varphi(x)f(x)dx$ where $f$ is the density of $X$. – Daniel Oct 21 '12 at 19:17
But, I meant, are there conditions on this function or not? I guess at least for example measurable bounded, or something like that. We can try to compute a density of $(B_t,B_s)$ as we know those of $(B_t-B_s,B_s)$. – Davide Giraudo Oct 21 '12 at 19:18
Yes of course, put whatever conditions you need on $\varphi$ but yes, must be at least measurable so that the expectation makes sense. – Daniel Oct 21 '12 at 19:24
@Daniel: your accounts have been merged. In the future please register your account to avoid accidentally creating duplicates. – Qiaochu Yuan Oct 23 '12 at 3:34

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}

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Thanks a lot! That was really useful :) I was wondering if one could iterate the same idea for $E g(B_t,B_s,B_r)$ ? How would you use the independent increments here? Seems a bit tricky :P :D – Daniel Oct 22 '12 at 15:17
@Daniel: I've added it. – Davide Giraudo Oct 22 '12 at 19:34

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