If $X$ and $Y$ are independent random variables, each having the normal distribution with mean $0$ and variance $1$, find the distribution of $X+Y$.
I thought that I could use the convolution formula, as $X$ and $Y$ are independent random variables and we know their density functions, where we call $Z = X+Y$.
$$
f_{z}(z) = \int_{-\infty}^\infty f_x(x) \cdot f_y(z-x)dx.
$$
Then we can just fill in both density functions and the boundaries for z, but we know that there are no boundaries for $x$ and $y$, because they are distributed normally, so therefore z has no boundaries. (Is this correct? And is it a correct way of reasoning?)
$$
f_z(z) = \frac{1}{2\pi} \int_{-\infty}^{\infty}e^{-\frac{1}{2}x^2} e^{-\frac{1}{2}(z-x)^2}dx,
$$
$$
f_z(z) = \frac{1}{2\pi}e^{-\frac{1}{2}z^2} \int_{-\infty}^{\infty}e^{-x^2}e^{xz}dx.
$$
This is where I can't seem to get any further. I know that my answer should be a normal distribution with mean $0$ and variance $2$, but I can't seem to figure out how to solve the integral. I know the standard integral of $ \int^{\infty}_{-\infty}e^{-x^2}dx = \sqrt{\pi}$. Can I use this somehow?
Can someone give me a nudge in the right direction and confirm whether my reasoning until this point is correct and whether I could attack such problems more efficiently?
Thanks for reading,
K.
Best Answer
$$\int_{-\infty}^\infty e^{-x^2}e^{xz}\,dx =e^{z^2/4}\int_{-\infty}^\infty e^{-(x-z/2)^2}\,dx =e^{z^2/4}\int_{-\infty}^\infty e^{-u^2}\,du $$ etc.