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

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.

• Good question. My guess is that $X$ must be degenerate. As you have observed some clever argument may give this. – Kavi Rama Murthy Dec 10 '18 at 6:30

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