Let $X, Y$ be independent standard normal random variables. We already know that $X+Y$ is normal with mean 0 and variance 2. However, I am trying to prove this result in a slightly different way than usual, i.e., using convolutions.
We can address the problem in the following way: the sum of the two variables has a density $h(x)$ given by the convolution product of the densities of the two variables, say $f$ and $g$, so that
$$ h(x) = \int_{-\infty}^{\infty} f(x-y)g(y)dy. $$
However, $$ f(y) = g(y) = \frac{1}{\sqrt {2\pi}}\exp(-y^2/2). $$
Hence now the integral becomes:
$$ h(x) =\frac{1}{2\pi}\int_{-\infty}^{\infty}\exp(-(x-y)^2/2)\exp(-y^2/2)dy. $$
However, I am puzzled here. How should we go on from here? I cannot see how to transform this integral into the integral of a normal r.v of type $N(0,2)$. Thanks in advance for your help.
The integral becomes
$$ h(x) =\frac{1}{2\pi}\int_{-\infty}^{\infty}\exp(-(\color{red}z-y)^2/2)\exp(-y^2/2)dy$$
due to $x=z-y$
$$h(x) =\frac{1}{2\pi}\int_{-\infty}^{\infty}\exp(-(z^2-2zy+y^2)/2)\exp(-y^2/2)dy$$
Focussing on the exponents
$-(y^2/2+y^2/2-zy)-z^2/2$
Completing the square
$-(y^2-zy+z^2/4)+z^2/4-z^2/2$
$-(y-z/2)^2-z^2/4$
$$h(x) =\frac{1}{2\pi}\cdot \exp(-z^2/4)\cdot \int_{-\infty}^{\infty}\exp(-(y-z/2)^2)dy$$
$$h(x) =\frac{1}{2\sqrt{\pi}}\cdot \exp(-z^2/4)\cdot \underbrace{\frac1{\sqrt \pi}\cdot \int_{-\infty}^{\infty}\exp(-(y-z/2)^2)dy}_{=1}$$
To see that the latter expression equals 1 you can substitute $y-z/2$ by $t$.