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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

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Since you are asking for an APPROXIMATE confidence interval, you could set it up like this: mean +/- z*SD where the SD is your formula above? This C.I. may not be the best one, but then, for an approximation it may work fine. How is life at the 38th perpendicular?? – imranfat May 16 '13 at 19:59
Are you sure you have $n$ Rayleigh distributions? If so, you need to say something about what samples you have from them, and you need to look for $n$ different parameters $\sigma$. It seems more likely that what you intended to say was that you have $n$ samples from one Rayleigh distribution. – joriki May 16 '13 at 21:12
Please try to keep discussions in the comments on topic. – Alexander Gruber May 25 '13 at 21:24
up vote 2 down vote accepted

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:

$$D[\sigma^*]=\sqrt{V[\sigma^*]}=\sqrt{V[\bar{X}\sqrt{\frac{2}{\pi}}]}=\sqrt{\frac{2}{\pi}V[\bar{X}]}=\sqrt{\frac{2}{\pi}\frac{1}{n}V[X]}=\sqrt{\frac{2}{\pi}\frac{1}{n}\frac{4-\pi}{2}\sigma^2}\\\implies d=\sqrt{\frac{2}{\pi}\frac{1}{n}\frac{4-\pi}{2}{\sigma_{\text{obs}}^*}^2}$$

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Excellent answer Tharsis! – Kortlek May 22 '13 at 11:39

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 $\frac{2n\overline{x^2}}{\chi_2^2} \leq \widehat{\sigma}^2 \leq \frac{2n\overline{x^2}}{\chi_1^2}$

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