What can be said about i.i.d. $X$ and $Y$ such that $XY=(X+Y)/2$ in distribution?

Question

Let $X$ and $Y$ be i.i.d.

If $(X+Y)/2$ is equal in distribution to $XY$, then what do we know about the distributions of $X$ and $Y$?

I feel like I can't say much about these distributions. I can only think of degenerate examples, but I have a feeling that there is something crafty going on here that I am missing.

Answer 1

I assume the first two moments of $X$ exist.

$$E[X] = \frac{E[X] + E[X]}{2} = E\left[\frac{X+Y}{2}\right] = E[XY] = E[X]E[Y] = E[X]^2$$ so $E[X]$ is either $0$ or $1$.

Similarly $$\frac{E[X^2]+E[X]^2}{2} = E \frac{(X+Y)^2}{4} = E[X^2 Y^2] = E[X^2] E[Y^2] = E[X^2]^2.$$

• In the case $E[X]=0$ we have $E[X^2] = 2 E[X^2]^2$ so $E[X^2] = \text{Var}(X)$ is either $0$ or $1/2$.
• In the case $E[X]=1$ we have $E[X^2] = 2 E[X^2]^2 - 1$, so $E[X^2] = 1$. But this means $\text{Var}(X) = 0$.

The only non-degenerate case is $E[X] = 0$ and $\text{Var}(X) = 1/2$. I am not sure how to construct an example for this case or show that this case is impossible.

Answer 2

Try something like $X=\sin 2\pi T$ with $T\sim U(0,1)$