product of two independent Gaussian vectors?

Question

I saw one equation, $x,y \in \mathbb{R}^n$, each entry $\sim N(0,1)$, and then

$|x^Ty| \approx \sqrt{n}$.

First I thought it is zero but it diverges as $n$. How can we prove this?

Answer 1

The density of the distribution of the product of two independent standard normals is essentially a modified Bessel function of the second kind, a symmetric distribution with mean and mode $0$ and variance $1$.

Your $x^Ty$ is then the sum of $n$ independent copies of these and so has a symmetric distribution with mean and mode $0$ and variance $n$. By the central limit theorem, $\frac1{\sqrt n}x^Ty$ converges in distribution to a standard normal as $n$ increases.

You then take the absolute value of this. So for increasing $n$ you should expect $\frac1{\sqrt n}|x^Ty|$ to converge in distribution to a standard half-normal with mode $0$, mean $\sqrt{\frac2\pi}\approx 0.79788$ and variance $1- \frac2\pi \approx 0.36338$, so in a sense for large $n$ you have the mean of $|x^Ty|$ approaching $$0.79788 \sqrt{n}$$ and this may be what you are suggesting in your question; the most likely value of $|x^Ty|$ is still $0$, but the distribution of absolute values is necessarily non-negative. The actual mean is slightly less than this and for $n=1$ is slightly less than $0.64$.