Variance of the Maximum Likelihood Estimator of the parameter of a Rayleigh distribution

Question

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 !

Answer 1

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.

Answer 2

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 $\large \frac{s_2}{2n}$:

where:

• 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.