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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?

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  • $\begingroup$ Your title misrepresents your question. You're not computing an interval for the distribution, your post is actually about an interval for a parameter (specifically a location parameter for a unit-width uniform). $\endgroup$
    – Glen_b
    Commented Apr 11, 2018 at 3:56

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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

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    $\begingroup$ The $t$ distribution isn't true, it's an approximation that becomes quite accurate as the sample size grows - in fact, it's already fairly accurate (in some informal sense of the word) for $N = 30$ in the case of the Uniform distribution. Note also that we know the population variance exactly, it's $1/12$, so don't need to rely on the sample standard deviation (nor, consequently, on the $t$ distribution at all.) $\endgroup$
    – jbowman
    Commented Apr 10, 2018 at 19:42
  • $\begingroup$ @jbowman In that case can't we simply use normal? But overall approach still holds correct? $\endgroup$
    – rannoudanames
    Commented Apr 10, 2018 at 19:55
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    $\begingroup$ Yes, we can, but it's still an asymptotic approximation. With $N=100$, it's going to be a really, really good approximation! And, since the OP asks for a 90% asymptotic confidence interval, I'd just modify my answer slightly and go with it. $\endgroup$
    – jbowman
    Commented Apr 10, 2018 at 19:58
  • $\begingroup$ @jbowman Thx for insight! $\endgroup$
    – rannoudanames
    Commented Apr 10, 2018 at 20:00
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    $\begingroup$ The CLT itself proves asymptotic normality of a standardized $\bar{X}$. If the conditions of the CLT hold, surely you can simply invoke it rather than prove it. If you didn't want to rely on an asymptotic argument (since you don't actually have $n\to\infty$, but some particular $n$), you could use one or another inequality to bound the error in the normal approximation of the cdf. However with n=100 the sample mean of uniforms will work very well if you don't go into the extreme tail. $\endgroup$
    – Glen_b
    Commented Apr 11, 2018 at 3:59

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