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?

  • $\begingroup$ As noted, the problem is very simple when the variances are equal (look up Nakagami Distribution). When they are not, the problem is fairly complicated. See, e.g. this post. $\endgroup$ – steveo'america Jan 30 '19 at 21:21
  • $\begingroup$ It seems like there might be a link to the Nakagami distribution even when the variances aren't the same? From Prof. Wikipedia, " ...accordingly, Nakagami-m is viewed as a generalization of Chi-distribution, similar to a gamma distribution being considered as a generalization of Chi-squared distributions." I've done a couple of test cases, and the fit to a Nakagami seems pretty good--now I just need a good explanation of the fit parameters. $\endgroup$ – Brian Jan 30 '19 at 21:56
  • $\begingroup$ @steveo'america --I guess I don't understand how the linked post applies. I'm forming the square root of a sum of gamma-distributed RV's, not just the sum of gamma-distributed RV's. Or am I missing something? $\endgroup$ – Brian Jan 31 '19 at 19:21
  • $\begingroup$ @brian : the linked post is for a simpler problem, but still very complicated. You have the further complication of taking the expected value of the square root of a sum of gammas. $\endgroup$ – steveo'america Jan 31 '19 at 20:15

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*}

  • $\begingroup$ This seems true enough, but I was hoping to find E(Y) without resorting to Monte Carlo or numerical integration (unless Gaussian Quadrature can be brought to bear). This calculation is going to occur inside an optimization loop, so I need something fast, even if it's approximate. Just for organizing thinking, I'll order the sigma_i's in decreasing magnitude (i.e., sigma_1>sigma_2>sigma_3, etc., and set sigma_1=1. I don't lose anything by doing this. $\endgroup$ – Brian Jan 31 '19 at 2:00

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