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 posted the following in response to a question that got closed as a duplicate of this one:
It looks from your comment as if the meaning of your question is different from what I thought at first.  My first answer assumed you knew that the sum of independent normals is itself normal.
You have
$$
\exp\left(-\frac12 \left(\frac{x}{\alpha}\right)^2 \right) \exp\left(-\frac12 \left(\frac{z-x}{\beta}\right)^2 \right)
= \exp\left(-\frac12 \left( \frac{\beta^2x^2 + \alpha^2(z-x)^2}{\alpha^2\beta^2} \right) \right).
$$
Then the numerator is
$$
\begin{align}
& (\alpha^2+\beta^2)x^2 - 2\alpha^2 xz + \alpha^2 z^2 \\  \\
= {} & (\alpha^2+\beta^2)\left(x^2 - 2\frac{\alpha^2}{\alpha^2+\beta^2} xz\right) + \alpha^2 z^2 \\  \\
= {} & (\alpha^2+\beta^2)\left(x^2 - 2\frac{\alpha^2}{\alpha^2+\beta^2} xz + \frac{\alpha^4}{(\alpha^2+\beta^2)^2}z^2\right) + \alpha^2 z^2 - \frac{\alpha^4}{\alpha^2+\beta^2}z^2 \\  \\
= {} & (\alpha^2+\beta^2)\left(x - \frac{\alpha^2}{\alpha^2+\beta^2}z\right)^2 + \alpha^2 z^2 - \frac{\alpha^4}{\alpha^2+\beta^2}z^2,
\end{align}
$$
and then remember that you still have the $-1/2$ and the $\alpha^2\beta^2$ in the denominator, all inside the "exp" function.
(What was done above is completing the square.)
The factor of $\exp\left(\text{a function of }z\right)$ does not depend on $x$ and so is a "constant" that can be pulled out of the integral.
The remaining integral does not depend on "$z$" for a reason we will see below, and thus becomes part of the normalizing constant.
If $f$ is any probability density function, then
$$
\int_{-\infty}^\infty f(x - \text{something}) \; dx
$$
does not depend on "something", because one may write $u=x-\text{something}$ and then $du=dx$, and the bounds of integration are still $-\infty$ and $+\infty$, so the integral is equal to $1$.
Now look at
$$
\alpha^2z^2 - \frac{\alpha^4}{\alpha^2+\beta^2} z^2 = \frac{z^2}{\frac{1}{\beta^2} + \frac{1}{\alpha^2}}.
$$
This was to be divided by $\alpha^2\beta^2$, yielding
$$
\frac{z^2}{\alpha^2+\beta^2}=\left(\frac{z}{\sqrt{\alpha^2+\beta^2}}\right)^2.
$$
So the density is
$$
(\text{constant})\cdot \exp\left( -\frac12 \left(\frac{z}{\sqrt{\alpha^2+\beta^2}}\right)^2 \right) .
$$
Where the standard deviation belongs we now have $\sqrt{\alpha^2+\beta^2}$.
