# Finding mean and variance of $Y = \ln{\left(\sum_i X_{i}^{2}\right)}$ for $X_i \sim \mathrm{Rayleigh}(\sigma)$

For some set of $n$ i.i.d. variables $\{X \}$ which are Rayleigh-distributed such that

$$P(X|\sigma) = \frac{X}{\sigma^2}\exp{\left[-\frac{X^2}{2\sigma^2}\right]}$$

I'm interested in anything we can write down analytically about

$$Y = \ln{\left(\sum_{i=1}^{n} X_{i}^{2}\right)}.$$ Primarily I'm trying to get expressions for $\mathrm{E}[Y]$ and $\mathrm{Var}[Y]$ (clearly they can be calculated numerically via sampling but analytic results would be better). Obviously a closed form for $P(Y|n,\sigma)$ would be great, but I have no idea if one exists.

Im guessing for the mean we could use LOTUS such that $$\mathrm{E}[Y] = \int_{0}^{\infty}\dots \int_{0}^{\infty} \ln{\left(\sum_{i=1}^{n} X_{i}^{2}\right)} \prod_{i=1}^{n} P(X_i|\sigma) \; \mathrm{d}X_1 \dots \mathrm{d}X_n$$ but I have no idea how to evaluate that, or if there is a simpler way.

Thanks for any help!

• Since (by inspection) $X^2$ has a Gamma distribution, $Y$ has an exp-gamma distribution (sometimes misleadingly called a "log-gamma distribution"). – whuber Feb 23 '18 at 23:42
• @CBowman feel free to put that as an answer – Taylor Feb 24 '18 at 2:19

Whuber points out that if $X \sim \mathrm{Rayleigh}(\sigma)$ then $X^2 \sim \mathrm{Gamma}(1, 2\sigma^2)$, and because Gamma-distributed variables have the property that $$\sum_{i=1}^{n} g_i \sim \mathrm{Gamma}\left(\sum_{i=1}^{n} k_i, \theta\right)$$ for $g_i \sim \mathrm{Gamma}\left(k_i, \theta\right)$ it is the case that $$\sum_{i=1}^{n} X^{2}_i \sim \mathrm{Gamma}\left(n, 2\sigma^2\right).$$ We may therefore write $$P(Y|n,\sigma) = \frac{1}{2^n \sigma^{2n} \Gamma(n)} \exp{\left[n Y - \frac{1}{2 \sigma^{2}}e^{Y}\right]}.$$ Consequently the mean and variance are $$\mathrm{E}[Y] = \ln{(2\sigma^2)} + \psi_0 (n),$$ $$\mathrm{Var}[Y] = \psi_1 (n),$$ where $\psi_k$ is the polygamma function.