Mean of root-sum-square of Suppose that I have several normally-distributed random variables xi, each with its own different variance.  All x's are zero-mean and independent.  If y is the root-sum-square of the xi's, how do I find the mean of y?
 A: This will involve application of the Law of the Unconscious Statistician here?  We know that $X_{i} \overset{id}{\thicksim} N(0,\sigma_{i}^{2})$ for $i=1,2,3,...,n$, so the joint probability density function of $X_1, X_2,...X_n$ is given by:
\begin{eqnarray*}
f_{X_{1},X_{2},...,X_{n}}(x_{,1},x_{,2},...x_{,n}) & = & \prod_{i=1}^{n}f_{X_{i}}(x_{i})\\
 & = & \left(\frac{1}{\sqrt{2\pi}}\right)^{n}\prod_{i=1}^{n}\frac{1}{\sigma_{i}}e^{-\frac{x_{i}^{2}}{2\sigma_{i}^{2}}}
\end{eqnarray*}
And by the Law of the Unconscious Statistician, if we set $Y=h(x_{1},x_{2},...,x_{n})=\sqrt{\sum_{i=1}^{n}x_{i}^{2}}$,
then:
\begin{eqnarray*}
E(Y) & = & \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty}h(x_{1},x_{2},...,x_{n})\left(\frac{1}{\sqrt{2\pi}}\right)^{n}\prod_{i=1}^{n}\frac{1}{\sigma_{i}}e^{-\frac{x_{i}^{2}}{2\sigma_{i}^{2}}}dx_{1}dx_{2}\cdots dx_{n}\\
 & = & \left(\frac{1}{\sqrt{2\pi}}\right)^{n}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty}\sqrt{\sum_{i=1}^{n}x_{i}^{2}}\prod_{i=1}^{n}\frac{1}{\sigma_{i}}e^{-\frac{x_{i}^{2}}{2\sigma_{i}^{2}}}dx_{1}dx_{2}\cdots dx_{n}
\end{eqnarray*}
