5
$\begingroup$

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!

$\endgroup$
2
  • 4
    $\begingroup$ Since (by inspection) $X^2$ has a Gamma distribution, $Y$ has an exp-gamma distribution (sometimes misleadingly called a "log-gamma distribution"). $\endgroup$ – whuber Feb 23 '18 at 23:42
  • 1
    $\begingroup$ @CBowman feel free to put that as an answer $\endgroup$ – Taylor Feb 24 '18 at 2:19
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.