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I am looking for a concrete example of the following: $(X,Y)$ are real-valued random variables such that:

  • $E[Y|X]$ is smooth
  • $E[X|Y]$ is discontinuous

Even better, I'd like to see an example where all densities exist: $p(x,y)$, $p(x|y)$, etc.

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    $\begingroup$ Take $(X,Y)$ with Gaussian density restricted to the region $0<y/x<1$. $\endgroup$ Commented Jun 2, 2021 at 21:48

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Suppose $T$ is uniformly distributed on $(-\pi/2,\pi/2)$.

Let $X=\tan(T)$, $Y=\sin(2T)$, so that the parametric graph of $(X,Y)$ looks like:

enter image description here

Then $Y = 2X/(1+X^2)$, so $E[Y|X]$ is smooth. But

$$E[X|Y]= \begin{cases} 1/Y \text{ if }Y \neq 0\\ \ \ 0 \ \ \text{ if } Y = 0 \end{cases}$$ which is discontinuous.

This comes from calculating $E[X|Y]$ is (the limit of) a weighted average of the two solutions for $y=2x/(1+x^2)$: $$\left\{\frac{1-\sqrt{1+y^2}}y,\frac{1+\sqrt{1-y^2}}y\right\},$$ where the weights are the relative probabilities that $(\tan(T),\sin(2T))$ is in the range $((1-\sqrt{1+y^2})/y\pm\delta,y\pm\epsilon)$ or in the range $((1+\sqrt{1+y^2})/y\pm\delta,y\pm\epsilon)$. The second range of $T$'s is just $\pm\pi/2$ minus the first range, so the probabilities are equal, and the average is equally weighted, which yields the simple formula.

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  • $\begingroup$ Can you explain your calculation for $E[X|Y]$? I did a quick calculation and I am getting a different result (which is continuous). (FWIW, my calculation of $E[Y|X]$ agrees with yours.) $\endgroup$
    – user19200
    Commented Jun 2, 2021 at 21:34
  • $\begingroup$ I revised the answer to explain the calculation, but I expect I won't make any further comments. $\endgroup$
    – user44143
    Commented Jun 2, 2021 at 22:21

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