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Let $X_1 , X_2 , ... , X_n$ be a random sample from a distribution with probability density function $\frac{1}{\theta}x^{\frac{1- \theta}{\theta}}I_{[0,1]}, \theta>0$. I want to find the 100$\alpha \% $ confidence interval for $\theta$.

This is what I have done:

$F_X(x;\theta) = x^{\frac{1}{\theta}}I_{[0,1]} + I_{(1, \infty)}$

We can pick $0<q_1, q_2<1$ such that $P(q_1 < \prod_{i=1}^{n}F_{x}(X_i ; \theta) < q_2) = \alpha$

Now we have that

$P(q_1 < \prod_{i=1}^{n}F_{x}(X_i ; \theta) < q_2) = P(q_1 < \prod_{i=1}^{n}X_i^{\frac{1}{\theta}} < q_2) = P(log(q_1) < \frac{1}{\theta}log(\prod_{i=1}^{n}X_i) < log(q_2)) = P(\frac{log(\prod_{i=1}^{n}X_i)}{log(q_1)} < \theta < \frac{log(\prod_{i=1}^{n}X_i)}{log(q_2)})$

Therefore a confidence interval for $\theta$ is $[\frac{log(\prod_{i=1}^{n}X_i)}{log(q_1)},\frac{log(\prod_{i=1}^{n}X_i)}{log(q_2)}]$. I believe this is correct. What I find unusual is, the expected length of the confidence interval seems to tend to infinity as n tends to infinity, I find this very counterintuitive, can this really be correct:

The length of the interval is $log(\prod_{i=1}^{n}X_i)(\frac{1}{log(q_2)} - \frac{1}{log(q_1)})$, this has expected value $nE(log(X_1))(\frac{1}{log(q_2)} - \frac{1}{log(q_1)}) = -n\theta(\frac{1}{log(q_2)} - \frac{1}{log(q_1)})$ (because $E(log(X_1)) = \int_{0}^{1} \frac{1}{\theta}log(x)x^{\frac{1}{\theta} - 1} dx = -\theta$). We can see that as n tends to $\infty$ so does the length of the confidence interval.

Can this be correct?

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Your method for finding a confidence interval is correct. But there is a small mistake:

$$q_1<\prod_{i=1}^n X_i^{1/\theta}<q_2\iff \frac{\sum\ln X_i}{\ln \color\red{q_2}}<\theta<\frac{\sum\ln X_i}{\ln \color\red{q_1}}$$


You might also notice that $X_i\stackrel{\text{i.i.d}}\sim \text{Beta}\left(\frac{1}{\theta},1\right)$, so that $-\frac{2}{\theta}\ln X_i\stackrel{\text{i.i.d}}\sim\chi^2_2$.

Hence a pivot for $\theta$ is $$T(\mathbf X,\theta)=-\frac{2}{\theta}\sum_{i=1}^n\ln X_i\sim\chi^2_{2n}$$

So, $$P_{\theta}\left(\chi^2_{1-(1-\alpha)/2,2n}\leqslant T\leqslant \chi^2_{(1-\alpha)/2,2n}\right)=\alpha\quad\forall\,\theta$$

Equivalently, $$P_{\theta}\left(\frac{-2\sum\ln X_i}{\chi^2_{(1-\alpha)/2,2n}}\leqslant\theta\leqslant\frac{-2\sum\ln X_i}{\chi^2_{1-(1-\alpha)/2,2n}}\right)=\alpha\quad\forall\,\theta$$

So a $100\alpha\%$ C.I. for $\theta$ is

$$\left[\frac{-2\sum\limits_{i=1}^n\ln X_i}{\chi^2_{(1-\alpha)/2,2n}},\frac{-2\sum\limits_{i=1}^n\ln X_i}{\chi^2_{1-(1-\alpha)/2,2n}}\right]$$

Here, $\chi^2_{\alpha,2n}$ is of course the $(1-\alpha)$th fractile of a chi-square distribution with $2n$ degrees of freedom.

Length of this C.I is $$L=-2\sum_{i=1}^n \ln X_i\left[\frac{1}{\chi^2_{1-(1-\alpha)/2,2n}}-\frac{1}{\chi^2_{(1-\alpha)/2,2n}}\right]$$

When you take the expected length, the quantity within parenthesis is surely a constant, but it is dependent on $n$ (say, $c(n)$). That is, expected length is of the form $$E(L)=-2c(n)E\left[\sum_{i=1}^n \ln X_i\right]=2nc(n)\theta$$

I think you are ignoring the limiting behaviour of $c(n)$ (your $q_1,q_2$ also depend on $n$) when you are taking the limit $n\to\infty$.

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