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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 $\varphi$?2012-10-21
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    A function/transformation. Like when you do: $E[\varphi(X)]=\int_{\mathbb{R}} \varphi(x)f(x)dx$ where $f$ is the density of $X$.2012-10-21
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    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 $(B_t-B_s,B_s)$.2012-10-21
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    Yes of course, put whatever conditions you need on $\varphi$ but yes, must be at least measurable so that the expectation makes sense.2012-10-21
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    @Daniel: your accounts have been merged. In the future please register your account to avoid accidentally creating duplicates.2012-10-23

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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

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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 :D2012-10-22
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    @Daniel: I've added it.2012-10-22