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[https://www.math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Hansen.pdf][page 12] and [peter morters][page 47] Let $B$ be a stanrd Brownian Motion and $R$ a function defined on $\mathbb{R}^2$ such that $$ \forall x \in \mathbb{R}^2 \text{: } R(x)=\mathcal{L}_{2}(B[0,1]\cap (x+B(t+2)-B(2)+B(1))$$ $$Y=B(2)-B(1)$$ $$\forall A \subset \mathbb{R}^2 \text{ and } x \in \mathbb{R}^2 A+x:=\{a+x/a\in A\}$$ my question is why $$E(R(Y))=1/2pi\int_{\mathbb{R}^2}e^{-x^2/2}E(R(x))dx?$$

$Y$ is a gaussian random variable , why do we still have $E$ in the second part of the equality? how to proove the mesurability of $R$?

Currently I'm reading from Peter Morter's book to understand the Lebesgue measure and Hausdorff dimension of Brownian motion's path... if you possess any well detailed proofs or resources on this topic, I'd greatly appreciate your sharing.

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    $\begingroup$ did you have any specific questions about it? The "potential theory" and BM chapter in Peres-Morters goes into detail. There are many notes on this topic math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Hansen.pdf $\endgroup$
    – Thomas Kojar
    Commented Aug 12, 2023 at 19:00
  • $\begingroup$ Let $B$ be a standart brownian motion , and $R$ a function defined on $\mathbb{R}^2$ and take $x$ to the Lebesgue measure of $B[0,1]\cap (x+B(t+2)-B(2)+B(1))$ \\ $Y=B(2)-B(1)$ \\ why $E(R(Y))=1/2pi\int_{\mathbb{R}^2}e^{-|x|^2}E(R(x))dx$ $\endgroup$
    – sara
    Commented Aug 12, 2023 at 20:00
  • $\begingroup$ @sara ok I edit that in your question above. $\endgroup$
    – Thomas Kojar
    Commented Aug 12, 2023 at 20:14
  • $\begingroup$ Re, what does "take $x$ to be the Lebesgue measure of [a set involving $x$]" mean? $\endgroup$
    – LSpice
    Commented Aug 12, 2023 at 20:21
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    $\begingroup$ @sara you can edit your above post and add those details so that a person can answer it below. Specify what R is. and what you mean by x... do you mean any set A that contains x? $\endgroup$
    – Thomas Kojar
    Commented Aug 12, 2023 at 20:33

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So according to the notes "BROWNIAN MOTION AND HAUSDORFF DIMENSION" by Peter Hansen, he lets

$$R(x):=m_{2}(B_{1}[0,1]\cap (x+B_{2}([2,3])))$$

where $B_{1}(t):=B_{t}$ and $B_{2}(t):=B(t+2)-B_{2}+B_{1}$. The variable $Y:=B_{2}-B_{1}$ is independent of both BMs $B_{1},B_{2}$ by Markov property.

Then the claim is that

$$0=m_{2}(B_{1}[0,1]\cap (x+B_{2}([2,3])))=E[R(Y)]=(2\pi)^{-1}\int_{\mathbb{R}}e^{-x^{2}/2}E[R(x)].$$

The first equality he proves it before. The second one is just definition. And the third equality follows because $Y\sim N(0,1)$ and it is independent of $X=B_{1}[0,1],B_{2}([2,3])$, and so by conditioning on them

$$E[R(Y)]=E[E[R(Y)|\sigma(X)]]=E[(2\pi)^{-1}\int_{\mathbb{R}}e^{-x^{2}/2}R(x)dx]=(2\pi)^{-1}\int_{\mathbb{R}}e^{-x^{2}/2}E[R(x)].$$

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  • $\begingroup$ i got it , thank you so much $\endgroup$
    – sara
    Commented Aug 12, 2023 at 22:35

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