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Let $B$ be a continuous Brownian motion, for every $t \geqslant 0$. I've to calculate :

$E\left[B_t | B_t^2 \right]$

They told me that the result is 0 by symmetry ... can anyone explain it to me please.

Could the resolution be like $$E[E\left[B_t | B_t^2 \right] | I_{(-a,a)}(B_t)]=E[B_t | I_{(-a,a)}(B_t)]=E[B_tI_{(-a,a)}(B_t)] = \int^a_{-a} {\frac{x}{\sqrt {2 \pi t}}e^{\frac{-x^2}{2t}}} =0$$

because the sigma field $\sigma\left({B^2_t}\right)$ contains the one of the indicator function of $B_t$.

Then $a$ it's arbitrary so $E\left[B_t | B_t^2 \right]=0$ $\mathbb P$-a.s.

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  • $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. $\endgroup$ Dec 23, 2017 at 10:26
  • $\begingroup$ Thanks for the suggestion $\endgroup$
    – Virginie
    Dec 23, 2017 at 12:05

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Let us prove a more general result by simpler arguments, namely, let us show that $$E(X\mid X^2)=0$$ for every integrable random variable $X$ with symmetric distribution, that is, such that $$X\stackrel{d}{=}-X$$

First recall that, by definition of conditional expectation, $E(X\mid X^2)=u(X^2)$, for some measurable function $u$ which depends only on the joint distribution of $(X,X^2)$.

Now, let $Y=-X$, then $X\stackrel{d}{=}Y$ hence $(X,X^2)\stackrel{d}{=}(Y,Y^2)$. The property that conditional expectations depend only on joint distributions recalled above implies that $E(Y\mid Y^2)=u(Y^2)$ for the same function $u$.

To summarize, $$E(X\mid X^2)=-E(Y\mid Y^2)=-u(Y^2)=-u(X^2)=-E(X\mid X^2)$$ hence $$E(X\mid X^2)=0$$

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Heuristically you can think of it like this: If you know the value of $B_t^2$, then $B_t$ will either be $+\sqrt{B_t^2}$ or $-\sqrt{B_t^2}$, each with probability one half. So the expectation should be 0. With this educated guess you can now proceed by just checking if $0$ fulfills the requirements of being the conditional expectation:

Measurability: Since $0$ is constant, it is measurable w.r.t. to any $\sigma$-algebra, especially $\sigma(B_t^2)$

Integrals Let $A \in \sigma(B_t^2)$. Then $A = (B_t^2)^{-1}(C)$ for some $C \in \mathcal{B}(\mathbb{R})$. $$\int_A B_t dP = \int_{(B_t^2)^{-1}(C) \cap B_t > 0} B_t dP + \int_{(B_t^2)^{-1}(C) \cap B_t < 0} B_t dP = 0 = \int_A 0 dP$$

so $E[B_t | B_t^2] = 0$ a.s.. The last step holds because the mentioned symmetry. To make it more precise, you can try to write $(B_t^2)^{-1}(C)$ in terms of $B_t^{-1}(D)$ for some set $D$.

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