# Is multivariate normal the only distribution with this property?

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?

• 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 $\beta.$
– whuber
Dec 17 '20 at 21:58
• @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? Dec 18 '20 at 13:21
• @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.
– whuber
Dec 18 '20 at 13:26

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 $$x$$.

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