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Suppose the random scalars $X$ and $Y$ are bivariate normally distributed. Then $E[X|Y=y]$ is linear in $y$ and $E[Y|X=x]$ is linear in $x$. In other words, both conditional expectation functions are linear. What I am wondering is whether the bivariate normal distribution is the only continuous distribution with this property? If not, then is it the only distribution with this property that also has say, a continuous pdf? I know that in higher dimensions the multivariate normal has the same liner conditional expectation property, but are there other examples in that case?

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    $\begingroup$ You can directly construct distributions with linear regressions: let $X$ have literally any distribution and let the distribution of $Y$ conditional on $X$ be any distribution with expectation $\alpha+\beta X$ for fixed $\alpha$ and $\beta.$ $\endgroup$
    – whuber
    Dec 17 '20 at 21:58
  • $\begingroup$ @whuber I agree that your construction gives a joint pdf for which $E[Y\mid X]=\alpha+\beta X$ for fixed $\alpha,\beta$, but I don't understand how this also proves that $E[X\mid Y]=\gamma+\delta X$ for some fixed $\gamma,\delta$ as the OP needs. Could you explain your construction just a little more to make it more obvious how both conditions are met by your constructed joint pdf? $\endgroup$ Dec 18 '20 at 13:21
  • $\begingroup$ @Dilip It only proves I didn't notice the reversal of $(X,Y)$ in the two conditionals! For a very general construction that I believe works for both $(X,Y)$ and $(Y,X)$ simultaneously, see stats.stackexchange.com/a/258389/919. $\endgroup$
    – whuber
    Dec 18 '20 at 13:26
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No, the bivariate normal is not the only distribution with the property that $E[X\mid Y=y]$ is a linear function of $y$ and also that $E[Y\mid X=x]$ is a linear function of $x$; many other distributions enjoy the same property.

For example, suppose that $(X,Y)$ is uniformly distributed on the triangle with vertices $(0,0), (1,1), (0,1)$ so that the joint density $f_{X,Y}(x,y)$ has value $2$ on the interior of the triangle. As motivation, note that this is the joint pdf of $\left(\min(U,V),\max(U,V)\right)$ where $U$ and $V$ are i.i.d. $\mathcal U(0,1)$ random variables. Now, notice that given $Y=y, y \in (0,1)$, the conditional distribution of $X$ is uniform on $(0,y)$ and so $E[X\mid Y=y] = y/2$ is a linear function of $y$. Similarly, given that $X=x, x \in (0,1)$, the conditional distribution of $Y$ is uniform on $(x,1)$ and so $E[Y\mid X=x] = \frac 12 + \frac x2$ is a linear function of $x$.

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No - it is not just a property of bivariate normals. For example

  • Let $A,B,C$ be i.i.d. with finite mean $\mu$. Then let $X=A+B$ and $Y=A+C$.

  • $E[A \mid X=x] =E[B \mid X=x] = \frac12 E[A+B \mid X=x]=\frac12 E[X \mid X=x]= \frac 12x$.

  • So $E[Y \mid X=x]=E[A \mid X=x] +E[C \mid X=x] = \frac 12x+\mu$ which is linear in $x$.

  • Similarly $E[X \mid Y=y]=E[A \mid Y=y] +E[B \mid Y=y] = \frac 12y+\mu$.

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