I want to calculate the variance of the maximum likelihood estimator of a Rayleigh distribution using $N$ observations.

The density probability function of this distribution is :

$$ f(\sigma,y_i) = \frac{y_i}{\sigma^2} e^{-\frac{y_i^2}{2\sigma^2}} $$

I also know that the mean is $\sigma \sqrt{\frac{\pi}{2}}$, its variance is $\frac{4 - \pi}{2}\sigma^2$ and its raw moments are $E[Y_i^k] = \sigma^k 2^{\frac{k}{2}}\Gamma(1+\frac{k}{2})$. Knowing this, I was able to calculate the maximum likelihood estimator $\hat{\sigma}^{2,ML} = \frac{\sum_{i=1}^{N} y_i^2}{2N}$

I calculated the mean of this estimator : $m_{\hat{\sigma}^{2,ML}} = E[\frac{\sum_{i=1}^{N} y_i^2}{2N}] = \frac{2N \sigma^2}{2N} = \sigma^2$ knowing that $E[y_i^2] = \sigma^2 2 \Gamma(2) = 2\sigma^2$.

For the variance, however, I do not see how to do it. I have tried to do as follow:

$$ Var(Z) = E[Z^2] - E[Z]^2 = E[(\frac{\sum_{i=1}^{N} y_i^2}{2N})^2] - E[\frac{\sum_{i=1}^{N} y_i^2}{2N}]^2 = \frac{1}{4N^2} E[(\sum_{i=1}^{N}y_i^2)^2] - \sigma^4 $$

My problem is that I do not know how to calculate $E[(\sum_{i=1}^{N}y_i^2)^2]$. Could someone give me a hint ?

Thanks !


Since the independence of $Y_i$ and $Y_j, j \neq i$, implies that $Y_i^2$ and $Y_j^2$ also are independent random variables, $$E\left[\left(\sum_{i=1}^N Y_i^2\right)^2\right] = \left(\sum_{i=1}^N E[Y_i^4]\right) + 2\left(\sum_{i=1}^N\sum_{j=2}^N E[Y_i^2]E[Y_j^2]\right)$$ all of which expectations on the right have values that you know already.

  • $\begingroup$ Thanks, I think I figure it out how to finish ! $\endgroup$ – Dust009 Apr 9 '15 at 16:00

This is just a special case of a much more general problem known as finding 'moments of moments'. Define the power sum $$s_r = \sum _{i=1}^n Y_i^r$$ Then you seek $Var\big(\large \frac{s_2}{2n}\big)$ ... i.e. the $2^{nd}$ Central Moment of $\large \frac{s_2}{2n}$:

enter image description here


  • CentralMomentToRaw is a function from the mathStatica package for Mathematica,

  • $\acute{\mu}_k = E[Y^k]$ which you already know.

In the case of a Rayleigh parent, the solution simplifies to $\frac{\sigma^4}{n}$. I should perhaps add that I am one of the authors of the function/software used.


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.