Let $X_1,..,X_n \sim \text{IID U}[\theta,\theta+1]$ be a random sample from a uniform distribution with the stipulated bounds depending on the parameter $\theta$. Given a sample of $n=100$ observations from that distribution, the following statistic was calculated:
$$\sum_{i=1}^n x_i = 350.492.$$
I need to calculate 90% asymptotic confidence interval for the unknown parameter $\theta$ by using the method of moments.
By the method of moments I have obtained that $\hat{\theta}_\text{MOM} = \bar{X_n}-1/2$. However, I am not sure how to proceed from here. Do I need to show that the obtained estimator is asymptotically normal and how can I find the asymptotic confidence interval?
Best Answer
You managed to find the first part, that is the unbiased estimator for $\theta$, by noting:
$$E[X] = \theta + \frac{1}{2} \implies \hat{\theta} = \bar{X} - \frac{1}{2}$$
Now you want a confidence interval at the $\beta$ level.
This means that you want $$P_{\bar{X}}[a\leq\bar{X}-\frac{1}{2} \leq b] = 1-\beta$$
So what you really need is to find the distribution of $\bar{X}$.
Now you can find that $$t = \frac{\bar{X} - \mu}{(\frac{s}{\sqrt{n}})}$$
Where $t$ follows a $t-distribution$ with $n-1$ degrees of freedom. Where $s$ is your sample standard deviation (coming from your second moment).
Generally when you find a confidence interval you want it to be of minimum measure/cardinality/length, however in this case the distribution is symmetric, so you don't need to struggle to find the range as it is symmetric around your estimator.
I encourage you to find out why the above t distribution is true!
Hope this helps.
______EDIT_____
As noted by jbowman in the comments, we already know the standard deviation, in which case we can use :
$$ z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$$
Where $\sigma$ is the "known" standard deviation and $z$ follows a standard normal distribution