# Approximate confidence interval; Rayleigh Distribution

This is a HW problem in my undergrad Statistics course. I am unsure of how to proceed. We were given a text file of data. I assumed that we would need $\overline{X}$; which I found to be $1.312670963$. In a past assignment we showed that as
\begin{equation} n\rightarrow \infty, \tilde{\theta} = \sqrt{\dfrac{1}{2n}\sum_{i=1}^nX_i^2}\xrightarrow{d}N\bigg(\theta,\dfrac{\theta^2}{4n}\bigg) \end{equation} We are asked to construct an approximate $95\%$ confidence interval for $\theta$. If I have the Fisher function as $J(\theta)=\dfrac{4n}{\theta^2}$. Can I state
\begin{equation} \bigg(\hat{\theta}-\dfrac{1.96\hat{\theta}}{2\sqrt{n}},\hat{\theta}-\dfrac{1.96\hat{\theta}}{2\sqrt{n}}\bigg) \end{equation} Using the score function and setting it to $0$. I found: \begin{equation} \hat{\theta}=\sqrt{\dfrac{1}{2n}\sum_{i=1}^nX_i^2} \end{equation}

Would I be correct in using this in the above equation? I am not sure how to start this problem. Any help would be appreciated. Thanks

• How does the Rayleigh distribution enter into this problem? Apr 24, 2017 at 17:32
• Only in the sense that it was the distribution that we started from. Apr 24, 2017 at 17:38

If your CI is "approximate", then assuming your $n$ is sufficiently large, you can use the asymptotic distribution of $\tilde\theta$. ie, for 95% confidence,
$$CI = \hat{\tilde{\theta}} \pm 1.96\times\sqrt\frac{\tilde{\hat{\theta}}^2}{4n}$$
• How do I find $\theta$? Apr 24, 2017 at 17:36
• So, is $\hat{\theta}=\overline{X}$? Or do I use the score equation and set $S(\theta)=0$ as I did above? Apr 24, 2017 at 17:48
• Sorry, my mistake. $\hat\theta$ will be the empirical version of your $\tilde\theta$. ie, $$\hat\theta = \sqrt{\frac{1}{2n}\sum x^2_i}$$. Apr 24, 2017 at 17:51