If the sum of two i.i.d. random variables is normal, must the variables themselves be normal?

Question

It is well known that if two i.i.d. random variables are normally distributed, their sum is also normally distributed.

Is the converse also true? That is, 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?

Thoughts: If the pdfs of $X$ and $Y$ are both $f$, the pdf of $X+Y$ is $f*f$, and its Fourier transform is $\hat f^2$. So if $\hat g=\hat f^2$ is a Gaussian function, then $\hat f$ could be any pointwise product of $\sqrt{\hat g}$ and an even function $\mathbb R\mapsto\{-1,1\}$. But it is not clear that any of those correspond to probability distribution functions, which must be nonnegative everywhere.

Answer 1

Suppose that $$\hat{f}(\xi) = \sqrt{\hat{g}(\xi)} \cdot h(\xi)$$ for some function $h: \mathbb{R} \to \{-1,1\}$. Since $\hat{f}$ is itself a characteristic function, we know that $\tilde{f}(0)=1>0$. Hence, $h(0)=1$. On the other hand,

$$h(\xi) = \frac{\hat{f}(\xi)}{\sqrt{\hat{g(\xi)}}}$$

is a continuous function as $\hat{f}$, $\hat{g}$ are continuous (note that $\hat{g}(\xi)>0$ for all $\xi \in \mathbb{R}$). Therefore, the intermediate value theorem shows that there cannot exist $\xi \in \mathbb{R}$ such that $h(\xi)=-1$. Hence,

$$\hat{f}(\xi) = \sqrt{\hat{g}(\xi)} \cdot 1.$$

Answer 2

The OP asked

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?

The answer is Yes as @saz has already said. A slightly more general result has been known for quite some time. According to a theorem conjectured by P. Lévy and proved by H. Cramér (see Feller, Chapter XV.8, Theorem 1),

If $X$ and $Y$ are independent random variables and $X+Y$ is normally distributed, then both $X$ and $Y$ are normally distributed.

Note that $X$ and $Y$ need not be identically distributed; just being independent suffices. Feller's book does not include Cramér's proof of this theorem on the grounds that it is based on analytic function theory and thus quite different from Feller's treatment of characteristic functions.