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We are given two random variables defined on $[0,1]$:

$$X(\omega) = 2 \omega -1 + |2 \omega -1|$$ $$Y(\omega) = 1-|2 \omega^2 -1|$$

I am supposed to find $\mathbb{E}(X|Y)$ which by definition is a $\sigma (Y)-$measurable random variable and $\forall B \in \sigma (Y): \ \int_B XdP = \int_B \mathbb{E}(X|Y)dP$. Here $P$ denotes Lebesgue measure on $[0,1]$.

I think that the sigma algebra generated by $Y$ is $$\sigma(Y) = \{B \in \mathcal{B}[0,1] : \ \forall x \in B: \sqrt{1-x^2} \in B\}$$

Then $B = [a,b] \cup [\sqrt{1-b^2}, \sqrt{1-a^2}], \ a \le b \le \frac{\sqrt{2}}{2}$, would be a generator of $\sigma(Y) $.

So I computed $$\int_B XdP = \int_{1/2}^b (4x-2)dx + \int_{\sqrt{1-b^2}}^{\sqrt{1-a^2}} (4x-2)dx = 4b^2 - 2a^2 + 2 (\sqrt{1-b^2} - \sqrt{1-a^2}) -2b + \frac{1}{2}$$

My problem is that $X=0$ on $[0, \frac{1}{2}]$ and none of the sets contained in $[0, \frac{1}{2}]$ is $\sigma(Y)-$measurable . I don't know how to construct $\mathbb{E}(X|Y)$ in order for the integrals to be equal. I suppose we have to take $\sqrt{1- \frac{Y}{2}}$ for $\omega \in [\frac{\sqrt{2}}{2},1]$ and $\sqrt{\frac{Y}{2}}$ on $[0,\frac{\sqrt{2}}{2}]$ but it's not obvious to me how.

Could you help me with that?

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    $\begingroup$ What is the source of this exercise? $\endgroup$ – Did May 16 '15 at 8:44
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    $\begingroup$ $$E(X\mid Y)=\frac{(\sqrt{2Y}-1)\,\sqrt{2-Y}\,\mathbf 1_{2Y>1}+(\sqrt{2-Y}-1)\,\sqrt{2Y}}{2(\sqrt{2Y}+\sqrt{2-Y})}$$ $\endgroup$ – Did May 16 '15 at 9:05
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    $\begingroup$ @Did I agree with your reasoning, if I understood it correctly, that the 2 $X$ values are weighted by their corresponding $|d\omega/dy|$. (I think I was wrong in my prev. comment that the 2 $\omega$ values are equally likely.) However, I get: $$E[X|Y]=\dfrac{\left|\frac{d\omega_1}{dy}\right| X_1+\left|\frac{d\omega_2}{dy}\right| X_2}{\left|\frac{d\omega_1}{dy}\right|+\left|\frac{d\omega_2}{dy}\right|}= \cdots$$ [Continued in next comment.] $\endgroup$ – Mick A May 16 '15 at 12:44
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    $\begingroup$ @Did [Continued] $$E[X|Y]=\dfrac{2\sqrt{2-Y}\left(\sqrt{2Y}-1 \right)I_{2Y\gt 1}+2\sqrt{Y}\left(\sqrt{4-2Y}-1\right)}{\sqrt{Y}+\sqrt{2-Y}}.$$ As a check: I think we should have $E[X|Y=1]=X(1/\sqrt{2})=2(\sqrt{2}-1).$ $\endgroup$ – Mick A May 16 '15 at 12:50
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    $\begingroup$ @Hagrid The canonical approach is to find some measurable function $h$ such that, for every test function $g$, $$E(Xg(Y))=E(h(Y)g(Y)),$$ then one knows that $E(X\mid Y)=h(Y)$ almost surely. In the present case, $X(\omega)=0$ when $\omega<1/2$ hence one asks that $$\int_{1/2}^12(2u-1)g(1-|2u^2-1|)du=\int_0^1h(1-|2u^2-1|)g(1-|2u^2-1|)du,$$ and the rest is a matter of careful changes of variable in integrals. $\endgroup$ – Did May 16 '15 at 13:26

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