PDF of the Euclidean norm of $M$ normal distributions If $X_1, X_2, \dots, X_M$ are independent identically distributed (i.i.d.) normal random variables, how to calculate the PDF and mean of the Euclidean norm of $X_1,X_2,\ldots,X_M$?
$$X_i\sim N(0,\sigma^2)$$
$$Y=\sqrt{\sum_{i=1}^M X_i^2}$$
There is a similarity between $Y$ and the square root of the $\chi$-squared distribution, i.e., the sum of the squares of $k$ independent standard normal random variables; however, in the case of $Y$, $\sigma$ may not necessarily be $1$.
$$f_Y(y) = \text{?}$$
$$E[Y] = \text{?}$$
 A: The fact that the variance of the normals is not one is immaterial. You could just compute everything for $\sigma=1$ and then rescale $Y\to \sigma Y$ at the end. Recall that if $X$ is $N(0,\sigma^2)$ then $X/\sigma$ is $N(0,1).$ So $Y/\sigma = \sqrt{\sum _i(X_i/\sigma)^2}$ which is $\chi^2(M).$
So you just need to compute square root of a $\chi^2(M).$ This is a usual change of variables. So let $Z$ be a $\chi^2(M)$ and let $Y = \sigma\sqrt{Z}$ where I included the scaling factor. Then you can write the cumulative distribution for $Y$ as $$ F_Y(y) = P(Y\le y) = P(\sigma\sqrt Z < y) = P(Z<y^2/\sigma^2) = F_Z(y^2/\sigma^2)$$ where $F_Z$ is the CDF for a $\chi^2(M).$ To get the PDF, take the derivative: $$ f_Y(y) = F'_Y(y) = \frac{2y}{\sigma^2}F'_Z(y^2/\sigma^2) = \frac{2y}{\sigma^2}f_Z(y^2/\sigma^2)$$ where $f_Z$ is the PDF for a $\chi^2(M).$
The expectation can be computed by either integrating the above PDF or computing the expectation of the square root of a chi squared. Remember, you don't need to worry about the $\sigma.$ The answer's going to be proportional to it... just do the computation with $\sigma = 1$ and then scale by $\sigma$ at the end.
