The sum of two independent random variables (X, Y) is normal iff X and Y are normally distributed.
I came across this fact recently, and it totally baffles me. I am not convinced it is true. Is there a proof for this somewhere?
UPDATE:
The direction of starting with two independent random variables that are normally distributed ($X$, $Y$) to proofing that $X+Y$ is also normally distributed is not confusing. I am more confused on the other direction of the proof. It seems like we could have two random variables that are independent and NOT normally distributed were the sum could be normally distributed. Suppose we have a random variable $Y$ that is standard normal, and $U$, $V$ which are iid with finite mean and variance from a non-normal distribution. If we have another variable $W = Y -U$, and then take $W + V$ is it really the case that $W + V$ would not be normally distributed? I have done a few arbitrary examples that concur with the statement. I am curious what the proof would look like so I know it is true always.
UPDATE #2
I think I found a counter example. Suppose that $X\sim I(2)$ that is, $X$ is a degenerate random variable, and $Y \sim N(0,1)$. Then the MGF of $Y$ is $exp(t^2/2)$ and the MGF of $X$ is $exp(2t)$, hence the MGF of $X+Y$ is $exp(2t + t^2/2)$, which implies that $X+Y \sim N(2, 1)$
Suppose $X+Y$ is normal and $X,Y$ are independent.
$X+Y \sim N(\mu_X+\mu_Y,\sigma_X^2+\sigma_Y^2)$ since we know its mean and variance.
Thus the MGF of $X+Y$ is $e^{(\mu_X+\mu_Y)t+\frac{1}{2}t^2(\sigma_X^2 + \sigma_Y^2)}$.
By independence this is equal to $E[e^{tX}]E[e^{tY}]$. I can now deduce $E[e^{tX}]$ is proportional to $e^{\mu_X t+\frac{1}{2}t^2 \sigma_X^2 }$. Hence it is a normal random variate. The same can be said of $Y$.
The other direction is quite easy (see comments).