In this question, I explored the Rayleigh distribution, with PDF

$$f_{\sigma}(x) = \dfrac{x}{\sigma^2} e^{-\dfrac{x^2}{2\sigma^2}},$$

where $x \ge 0$.

I calculated that the MLE is $\hat{\sigma^2} = \sqrt{\dfrac{1}{n} \sum_{i = 1}^n \dfrac{x_i^2}{2}}$. I now want to check whether it's unbiased. Is my calculation for the MLE correct? How do I check whether it's biased?

Thanks to COOLSerdash's comments, I realised that the MLE is actually $\hat{\sigma^2} = \dfrac{1}{n} \sum_{i = 1}^n \dfrac{x_i^2}{2}$, since we're finding the MLE for $\hat{\sigma^2}$ rather than $\hat{\sigma}$.

I now want to check whether the MLE is biased. I begin by taking the expected value of the MLE:

$$E\left(\hat{\sigma^2}\right) = \dfrac{1}{2n} E\left(\sum_{i = 1}^n x_i^2 \right)$$

We bring the expected value inside the summation:

$$E\left(\hat{\sigma^2}\right) = \dfrac{1}{2n} \sum_{i = 1}^n E(x_i^2)$$

So how do I find $E(x_i^2)$?

  • $\begingroup$ The MLE is wrong for $\hat{\sigma^2}$ (hint: ditch the square root). To check whether it's biased, calculate the expectation of the MLE. $\endgroup$ Apr 16 at 21:29
  • $\begingroup$ @COOLSerdash Oh, oops. You're right, it should be $$\hat{\sigma^2} = \dfrac{1}{n} \sum_{i = 1}^n \dfrac{x_i^2}{2},$$ since we're find the MLE of $\hat{\sigma^2}$ rather than $\hat{\sigma}$. $\endgroup$ Apr 16 at 21:32
  • $\begingroup$ @COOLSerdash $$E\left(\hat{\sigma^2}\right) = \dfrac{1}{2n} E\left(\sum_{i = 1}^n x_i^2 \right)$$ Can we move the expected value into the summation to get $$E\left(\hat{\sigma^2}\right) = \dfrac{1}{2n} \sum_{i = 1}^n E(x_i^2)$$? $\endgroup$ Apr 16 at 21:37
  • 1
    $\begingroup$ That's right, but please update your question rather than posting more comments. $\endgroup$ Apr 16 at 21:40
  • $\begingroup$ @COOLSerdash I have updated the question. I think now all I need to do is find $E(x_i^2)$. Any ideas? $\endgroup$ Apr 16 at 21:45

In response to:

So how do I find $E(x^2_i)$?

Assuming there are no issues with the steps in the derivation up to that point, you can use the following standard result relating expectation, variance and 2nd moments

$$\mathbb{E}[X^2_i] = \text{Var}(X_i) + \mathbb{E}[X_i]^2$$

I know nothing of the Rayleigh distribution, but according to wikipedia, it has mean

$$\mathbb{E}[X_i] = \sigma \sqrt{\frac{\pi}{2}}$$

and variance

$$\text{Var}(X_i) = \frac{4 - \pi}{2} \sigma^2$$

Using these facts we have that the expectation of the estimator $\hat{\sigma}^2$ is

\begin{align*} \mathbb{E}[\hat{\sigma}^2] &= \frac{1}{2n} \sum_{i=1}^n \mathbb{E}[X_i^2] \\ &= \frac{1}{2n} \sum_{i=1}^n \left( \frac{4 - \pi}{2} \sigma^2 + \frac{\pi}{2} \sigma^2 \right) \\ &= \frac{2n \sigma^2}{2n} \\ &= \sigma^2 \end{align*}

Therefore the estimator $\hat{\sigma}^2(X_1, ..., X_n)$ is unbiased as $\mathbb{E}[\hat{\sigma}^2] = \sigma^2$.


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