Proof that the sum of two Gaussian variables is another Gaussian The sum of two Gaussian variables is another Gaussian.
It seems natural, but I could not find a proof using Google.
What's a short way to prove this?
Thanks!
Edit: Provided the two variables are independent.

I prepared the following as an answer to a question which happened to 
close just as I was putting the finishing touches on my work.  I posted it as a different (self-answered) question but following suggestions from Srivatsan Narayanan and Mike Spivey, I am putting it here and deleting my so-called question.
If $X$ and $Y$ are independent standard Gaussian random variables, what is
the cumulative distribution function of $\alpha X + \beta Y$?
Let $Z = \alpha X + \beta Y$.  We assume without loss of generality that $\alpha$ and $\beta$ are positive real numbers since if, say, $\alpha < 0$, then we can replace $X$ by $-X$ and $\alpha$ by $\vert\alpha\vert$. Then, the cumulative probability distribution function of $Z$ is
$$
F_Z(z) = P\{Z \leq z\} = P\{\alpha X + \beta Y \leq z\} = \int\int_{\alpha x + \beta y \leq z} \phi(x)\phi(y) dx dy
$$
where $\phi(\cdot)$ is the unit Gaussian density function. But, since the integrand $(2\pi)^{-1}\exp(-(x^2 + y^2)/2)$ has circular symmetry, the value of the integral depends only on the distance of the origin from the line $\alpha x + \beta y = z$.
 Indeed, by a rotation of coordinates, we can write
the integral as
$$
F_Z(z) = \int_{x=-\infty}^d \int_{y=-\infty}^{\infty}\phi(x)\phi(y) dx dy
= \Phi(d)
$$
where $\Phi(\cdot)$ is the standard Gaussian cumulative distribution function.
But, 
$$d = \frac{z}{\sqrt{\alpha^2 + \beta^2}}$$
and thus the cumulative distribution function of $Z$ is that of a zero-mean Gaussian random variable with variance $\alpha^2 + \beta^2$.
