6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Good question. My guess is that $X$ must be degenerate. As you have observed some clever argument may give this. $\endgroup$ – Kavi Rama Murthy Dec 10 '18 at 6:30
3
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.