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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.

[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.

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?

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.

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.

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.

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?

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.