Vtyjkyuk

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 $X_1,X_2,ldots,X_M$?

$$X_isim N(0,sigma^2)$$

$$Y=sqrtsum_i=1^M X_i^2$$

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 $1$.

$$f_Y(y) = text?$$

$$E[Y] = text?$$

The fact that the variance of the normals is not one is immaterial. You could just compute everything for $sigma=1$ and then rescale $Yto sigma Y$ at the end. Recall that if $X$ is $N(0,sigma^2)$ then $X/sigma$ is $N(0,1).$ So $Y/sigma = sqrtsum _i(X_i/sigma)^2$ 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 $Y = sigmasqrtZ$ where I included the scaling factor. Then you can write the cumulative distribution for $Y$ as $$ F_Y(y) = P(Yle y) = P(sigmasqrt Z < y) = P(Z<y^2/sigma^2) = F_Z(y^2/sigma^2)$$ where $F_Z$ is the CDF for a $chi^2(M).$ To get the PDF, take the derivative: $$ f_Y(y) = F'_Y(y) = frac2ysigma^2F'_Z(y^2/sigma^2) = frac2ysigma^2f_Z(y^2/sigma^2)$$ 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.