I previously asked a similar question about the sum of two i.i.d. random variables, thinking the two cases to be equivalent. But I can't see how to apply the proof of that case to this one.

It is well known that if two i.i.d. random variables are normally distributed, their difference is also normally distributed. Is the converse also true? That is to say:

Suppose $X$ and $Y$ are two i.i.d. random variables such that $X-Y$ is normal. Is it necessarily the case that $X$ and $Y$ are also normal?

Here's what I know so far. If the pdfs of $X$ and $Y$ are both $f(x)$, then the pdf of $X-Y$ is $f(x)*f(-x)$, and its characteristic function is $\hat f(t)\hat f(-t)=\hat f(t)\overline{\hat f(t)}=|\hat f(t)|^2$. So if $\hat g=|\hat f|^2$ is a Gaussian function, then $\hat f$ is the pointwise product of a Gaussian $\sqrt{\hat g}$ and a Hermitian function $h:\mathbb R\to\{z:|z|=1\}$. Certainly we can take $h(t)=e^{i\mu t}$, giving $f$ to be Gaussian with arbitrary mean $\mu$. But many other choices of $h$ are possible! Do any of those correspond to probability distributions, which must be nonnegative everywhere?


Cramér's Theorem even gives that if the sum $X+Y$ of any two independent (not necessarily identically distributed) random variables $X$ and $Y$ has a normal distribution, then the summands $X$ and $Y$ themselves must be normally distributed.

This is more than enough to cover your case: if $X$ is independent of $Y$, it is also independent of $-Y$, and $X-Y$ is the sum of $X$ and $-Y$. Now apply the theorem.

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    $\begingroup$ I'm gratified to learn that this is a nontrivial result and I wasn't missing something obvious! $\endgroup$ – user856 Dec 3 '14 at 18:36

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