Suppose $X_1,...,X_d\sim\mathcal{N}(0,1)$ are i.i.d.'s, each distributed normally around 0 with variation 1. It looks like $\mathbb{E}\left(\sum X_i^2\right)=d$. Why is that true? And how $Y=\sum X_i^2$ is distributed?
Perhaps a bit more interesting is the following: it looks like $d-1 < \left(\mathbb{E}\left(\sqrt{\sum X_i^2}\right)\right)^2 < d$, but I am not sure if it's true or why. How is $Z=\sqrt{\sum X_i^2}$ distributed? For large $d$'s, it looks a little bit normal.
The motivation behind this question is the distribution of distance of a point in $\mathbb{R}^d$ from 0, when a point has a Gaussian distribution.
as remarked above, $Y$ is distributed as chi-sqaured distribution with $d$ degrees of freedom.
for more info, see:
http://en.wikipedia.org/wiki/Chi-squared_distribution
$Z = \sqrt{Y}$, so a simple transformation tells you $Z$ has density function
$f(z)= \frac{1}{2^{d/2-1}\Gamma(d/2)}z^{d-3}\exp(-\frac{z^2}{2})1_{\{z>0\}}$.
whatever the hell this is.