Determining a confidence interval for $\sigma$ from a Rayleigh distribution Hello stackexchangers,
Suppose we have $n$ Rayleigh distributions defined by
$$f_X(x)=\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}.$$
How would you go about determining an approximative confidence interval for $\sigma$?
I got this problem from a dear friend, and he suggested that I use the method of least-squares on $\sigma$.
Guided by his wisdom, I found the least-squares prediction to be
$$\sigma^*=\bar{x}\sqrt{\frac{2}{\pi}},$$
where $\bar{x}$ is the mean value, but I am completely lost as to how to proceed.
Any hints would be greatly appreciated.
Your Mayor, 
Ron Ford
 A: Since you asked this we have updated the Rayleigh wikipedia entry with the unbiased MLE for the Rayleigh parameter as well as confidence intervals, which are conveniently functions of the $\chi^2$ distribution.
In particular:
Given a sample of N i.i.d. samples $x_i$ from the Rayleigh distribution with parameter $\sigma$, $\widehat{\sigma^2}\approx \!\,\frac{1}{2N}\sum_{i=1}^N x_i^2$ is an unbiased maximum likelihood estimate.
To find the (1 −  α) confidence interval, first find $\chi_1^2, \ \chi_2^2$ where $Pr(\chi^2(2n) \leq \chi_1^2) = \alpha/2, \quad Pr(\chi^2(2n) \leq \chi_2^2) = 1 - \alpha/2$, then $\frac{2n\overline{x^2}}{\chi_2^2} \leq \widehat{\sigma}^2 \leq \frac{2n\overline{x^2}}{\chi_1^2}$
A: I will give you a rough approximation. Because we have to start somewhere, we boldly assume that $n$ is large enough for the central limit theorem to hold for $\sigma^*=\bar{X}\sqrt{\frac{2}{\pi}}$; note that your point estimator $\sigma_{\text{obs}}^*=\bar{x}\sqrt{\frac{2}{\pi}}$ of the scale parameter $\sigma$ is an observation of the random variable $\sigma^*$.
Now, since $\sigma^*$ is approximately normally distributed with expected value $\sigma$ and standard deviation $D[\sigma^*]=f(\sigma)$, an approximate confidence interval for the estimator is given by $I_\sigma=(\sigma_{\text{obs}}^*\pm\underbrace{\lambda_{0.025}}_{1.96}d)$, where we calculate $d$ as follows:
$$D[\sigma^*]=\sqrt{V[\sigma^*]}=\sqrt{V[\bar{X}\sqrt{\frac{2}{\pi}}]}=\sqrt{\frac{2}{\pi}V[\bar{X}]}=\sqrt{\frac{2}{\pi}\frac{1}{n}V[X]}=\sqrt{\frac{2}{\pi}\frac{1}{n}\frac{4-\pi}{2}\sigma^2}\\\implies d=\sqrt{\frac{2}{\pi}\frac{1}{n}\frac{4-\pi}{2}{\sigma_{\text{obs}}^*}^2}$$
