Consider the probability space $\Omega = (-1,1)$ with the Borel-sigma algebra $\mathcal{B}((-1,1))$ with the uniform distribution $\mathcal{U}((-1,1))$ as probability measure. Now let $X$ and $Y$ be random variables on this space with

\begin{equation} X(x) = \max[0,x], Y(x) = x^{2}, ~~x \in(-1,1). \end{equation} We have to compute $\mathbb{E}(X|Y)$ and $\mathbb{E}(Y|X)$. I don't really understand the concept of conditional expectation. How should I begin? Anyone can help?


Let $P$ denote the uniform measure. [$P$ is just Lebesgue measure divided by 2]. $E(X|Y)$ can be expressed as a measurable function of $Y$, say $f(Y)$. We have to determine $f$ from the relation $EXI_{\{Y\leq t\}} =Ef(Y)I_{\{Y\leq t\}}$ for all $t \in \mathbb R$. This gives $\int_0 ^{\sqrt t} x \, dP(x)=\int_{-\sqrt t}^{\sqrt t} f(x^{2}) \, dP(x)=2\int_{0}^{\sqrt t} f(x^{2}) \, dP(x)$. Put $u=x^{2}$ in the last integral. we get $\int_{0}^{\sqrt t} f(x^{2}) \, dP(x)=\int_0^{t} f(u) \frac 1 {2\sqrt u}\, dP(u)$. Thus $\int_0 ^{\sqrt t} x \, dP(x)=2\int_0^{t} f(u) \frac 1 {2\sqrt u}\, dP(u)$ for all $t$. Now differentiate both sides with respect to $t$ to find $f$. You will get $f(t)=\frac {\sqrt t} 2$. Hence $E(X|Y)=\frac {\sqrt Y} 2$. This procedure is quite general and you can use the same method for $E(Y|X)$.

  • $\begingroup$ Thank you! I have difficulty with the theory of conditional expectation. Have you a good reference where I can learn from? Something with intuition behind it $\endgroup$ – user562844 May 18 '18 at 15:43
  • $\begingroup$ can you please also compute $\mathbb{E}(Y|X)$ $\endgroup$ – user562844 May 23 '18 at 20:25
  • $\begingroup$ my Approach is the following: $\int_{1}^{t}x^{2}dx = \int_{-\sqrt{t}}^{\sqrt{t}}f(max[0,x])dx$. But now I have no idea how to proceed $\endgroup$ – user562844 May 23 '18 at 20:38
  • $\begingroup$ @gregor The answer is $E(Y|X)=\frac 1 3 I_{X=0} +X^{2}$ $\endgroup$ – Kavi Rama Murthy May 24 '18 at 5:34
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    $\begingroup$ Put $t=0$ in the equation you have derived to get $g(0)=\frac 1 3$. For $t>0$ differentiate the equation to get $g(t)=t^{2}$. Hence $g(X)=\frac 1 3$ if $X=0$ and $X^{2}$ if $X>0$. These two facts can be expressed in one equation as $g(X)=\frac I 3 I_{X=0}+X^{2}$. $\endgroup$ – Kavi Rama Murthy May 25 '18 at 9:10

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