# Confidence intervall for Rayleigh distribution's parameter?

I manage to estimate $$\sigma$$ in the Rayleigh distribution, but how do I get a correct confidence interval for it?

The Rayleigh distribution is

$$f_X(x)=\frac{x}{\sigma^2}e^{-\frac{x^2}{2\sigma^2}}.$$

Let's assume we have $$n$$ Rayleigh distributed variables. With least-square-estimation of $$\sigma$$ we use

$$Q(\theta) = \sum_{i=1}^n \left[ x_i-\mu_i(\theta) \right]^2$$ so that it is as small as possible.

Asuming that $$\mu_i(\theta)$$ is the same for all variables, the value of $$\mu_i(\theta)$$ that will minimize $$Q(\theta)$$ is $$\overline x = \frac{1}{n}\sum_{i=1}^nx_i. \tag{1}$$

According to Wikipedia (1. Can this be derived somehow?) the expected value $$\mu$$ for a Rayleigh distributed random variable $$X$$ is

$$\mu_X = E(X)=\sigma \sqrt{\frac{\pi}{2}}. \tag{2}$$

With equation $$(1)$$ and $$(2)$$ we get that $$\sigma$$ can be estimated as

$$\sigma_{obs}^* = \frac{1}{\sqrt{\pi/2}} \overline x = \frac{1}{n\sqrt{\pi/2}}\sum_{i=1}^nx_i. \tag{estimated value}$$

Now, to derive a confidence interval for $$\sigma$$ with confidence level $$1-\alpha$$ as let's say $$95\%$$ we assume that the amount $$n$$ of Rayleigh distributed variables is large enough so that normal approximation can be used. 2. Is this a correct assumption?

Since $$\sigma$$ is unknown, we use the Student's $$t$$-distribution for the confidence interval $$I_{\sigma}$$ such that

$$I_{\sigma} = \sigma_{obs}^* \pm t_{\alpha/2}(f) \cdot d \tag{confidence intervall}$$

where $$f=n-1$$, $$d = s/\sqrt{n}$$ and $$s$$ is given by

$$s=\sqrt{\frac{1}{n-1}\sum_{i=1}^n(x_i-\overline x)^2}$$

3. Is the above reasoning correct? Is this a correct confidence interval for $$\sigma$$ in a Rayleigh distribution?