I'm studying the Rayleigh distribution given by $$ f_v(x) = \frac{x}{v} \ \exp(-\frac{x^2}{2v}) \ \mathbb{1}\{x \geq 0\}$$

and found an estimator (using method of moments) to be $$ \tilde{v} = \frac{2}{\pi n^2} \left(\sum_{i=1}^{n} X_i \right)^2$$

Now I want to know whether it's biased or not, but I get stuck when I try to compute the expectation of $\tilde{v}$. I have $$ \mathbb{E}[\tilde{v}] = \frac{2}{\pi n^2} \mathbb{E}\left[\left(\sum_{i=1}^{n} X_i \right)^2\right]$$

I tried stating that $\mathbb{E}[X] = \frac{1}{n}\sum_{i=1}^{n} X_i$, which yields $\left(\sum_{i=1}^{n} X_i \right)^2 = n^2 \mathbb{E}[X]^2$ and an unbiased estimator.

However, that seems wrong, as that statement is exactly what I used to get the estimator in the first place, so it seems like I'm reasoning in circles here. Is it wrong? If so, how else could I get the bias?

Thanks in advance!

  • $\begingroup$ Yes, the exercise insisted on using $v$ instead of $\sigma^2$, which complicated matters somewhat. However, taking the square root of my estimator results in the estimator given in that link, so I assumed it was correct (as $v = \sigma^2$ in this parametrization), . $\endgroup$ – Tiamo P. Oct 5 '16 at 12:21
  • $\begingroup$ Revision of earlier comment, to which your responded: To start, please see http://math.stackexchange.com/questions/738519/method-of-moments-and-maximum-likelihood-question and the Wikipedia article on 'Rayleigh distribution'. You are using parameter vv while the usual parameterization is in terms of $v = \sigma^2$. $\endgroup$ – BruceET Oct 7 '16 at 16:23

Recall that $var(X)=\mathbb{E}X^2-\mathbb{E}^2X$, so \begin{align} \mathbb{E}\tilde{v} &= \frac{2}{\pi n^2} \mathbb{E}\left(\sum_{i=1}^{n} X_i \right)^2 = \frac{2}{\pi n^2}var(\sum_{i=1}^{n} X_i )+\frac{2}{\pi n^2}\mathbb{E}^2(\sum_{i=1}^{n} X_i)\\ &= \frac{2}{\pi n^2}n\sigma_X^2+\frac{2}{\pi n^2}(n\mu_X)^2\\ &= \frac{4-\pi}{\pi n}v+ v \neq v,\,\,\, \forall n\in\mathbb{N} \end{align}


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