Suppose we have a uniform normal distribution $ \mathcal{N}(\mu,\,\sigma^{2})$ where $\mu = 0$ and $\sigma = 1$. $V$ is a set of $d$ dimensional vector such that $V=[v_1,v_2,..,v_\inf]$ $v_i \in R^d$. Each $v_i$ is drawn independently from the unifirm normal distribution. I would like to know the pdf the length of these vectors. I ran a simulation with $d=100$ using 10000 samples and I got the following histogram:
It looks like the mean of the length distribution at $d=100$ is around 10. How do I calculate this distribution directly without running the simulation?
The lengths have a chi-distribution with parameter $d$, since their squares have a chi-squared distribution as the sum of squares of $d$ standard normal random variables
so the mean is $\mu=\sqrt{2}\,\frac{\Gamma((d+1)/2)}{\Gamma(d/2)}$ and the variance is $\sigma^2=d-\mu^2$
with $d=100$ this gives a mean of about $9.9750$ and a variance of about $0.4987$ so a standard deviation of about $0.7062$
For large $d$ you can alternatively use an approximation result similar to the central limit theorem but for square roots of sums; the underlying squares of standard normals have mean $1$ and variance $2$ so the square-root of the sum of $d$ of them has a distribution which is approximately normal with mean approximately $\sqrt{d-\frac12}$ (again about $9.9750$ when $d=100$), and variance approximately $\frac12$