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I need some help with the following problem:

I want to estimate $n$ of $X_i \sim U(1, n)$ with a $90\%$ confidence level. What is given is the sample size with $10$ and the maximum of the sample with $100$. $X_1,...,X_{10}$ are iid.

When calculating the MLE for this distribution I got $max\{X_1,...,X_{10}\}$, which is $100$. Now I want to know $Pr[U_1 \le 100 \le U_2] \ge 0.9$. I don't know how to calculate the confidence interval. Using the central limit theorem would be wrong since the estimator is not a sum of random variables, would it?

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    $\begingroup$ Seems similar to recent Question 1209163. Also, is n an integer? If so, see 'German tank problem' on this site and in Wikipedia. In any case, as you guessed, CLT not applicable. Can you find the distribution of the max? $\endgroup$ – BruceET Jul 3 '15 at 22:45
  • $\begingroup$ Thanks for your answer. Yes, n is an integer. I already read about the German tank problem, but it didn' t really help since there is never a confidence level used. I think the distribution of the max would be $x ^n$, so in this case $x ^{10}$, since the variables are independent. $\endgroup$ – guest Jul 3 '15 at 22:58
  • $\begingroup$ The distribution of the max is not hard. Let $Y$ be the max. Then $\Pr(Y\le y)=\frac{y^{10}}{(n-1)^{10}}$. FDifferentiate to get the density. $\endgroup$ – André Nicolas Jul 3 '15 at 23:06
  • $\begingroup$ Is it because each $y_i \le max\{y_1,...,y_i\}$ and since they are independent you can simply use $(F_Y(y) )^n$? Alright, just give me a second for the density. $\endgroup$ – guest Jul 3 '15 at 23:16
  • $\begingroup$ I got $\frac{10y^9}{(n-1)^{10}}$ for the density. Can I now use this for $U_2$? $\endgroup$ – guest Jul 3 '15 at 23:29
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What you need is a random variable depending on $n$ whose distribution does not depend on $n$. Also this random variable should only depend on the data through a sufficient statistic, which in this case is $X_{(10)}$.

One possible idea is to consider that $(X_i - 1) / (n - 1) \sim$ uniform$(0, 1)$ and so $- \log[(X_i - 1) / (n - 1)] \sim$ exponential$(1)$. Now $- \log[(X_{(10)} - 1) / (n - 1)]$ is the minimum of a random sample of $10$ unit exponential random variables, and therefore is itself exponential with rate $\lambda = 10$. So using the $0.9$ quantile of this distribution,

$$ \begin{align} P \left (0 \leq - \log \left [ \frac{X_{(10)} - 1}{n - 1} \right ] \leq 0.23 \right ) = 0.9. \end{align} $$

This event is the same as $X_{(10)} \leq n \leq (X_{(10)} - 1)e^{0.23} + 1$, and so these are the endpoints of a $90\%$ confidence interval for $n$.

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