1
$\begingroup$

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

|cite|improve this question
$\endgroup$
  • $\begingroup$ How does the Rayleigh distribution enter into this problem? $\endgroup$ – Michael Chernick Apr 24 '17 at 17:32
  • $\begingroup$ Only in the sense that it was the distribution that we started from. $\endgroup$ – Lanous Apr 24 '17 at 17:38
1
$\begingroup$

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,

$$CI = \hat{\tilde{\theta}} \pm 1.96\times\sqrt\frac{\tilde{\hat{\theta}}^2}{4n}$$

|cite|improve this answer
$\endgroup$
  • $\begingroup$ How do I find $\theta$? $\endgroup$ – Lanous Apr 24 '17 at 17:36
  • $\begingroup$ So, is $\hat{\theta}=\overline{X}$? Or do I use the score equation and set $S(\theta)=0$ as I did above? $\endgroup$ – Lanous Apr 24 '17 at 17:48
  • $\begingroup$ Sorry, my mistake. $\hat\theta$ will be the empirical version of your $\tilde\theta$. ie, $$\hat\theta = \sqrt{\frac{1}{2n}\sum x^2_i}$$. $\endgroup$ – Tim Atreides Apr 24 '17 at 17:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.