Given an unbiased estimator, how can one compute confidence interval? Consider $Y$, with distribution given by the largest order statistic for a sample of size $2$ with $X_{i}$ sampled from the uniform distribution on $[0,\theta]$, that is, pdf of $Y$ is $f_Y(t)=2t/\theta^2$. How to compute the $.05$ confidence interval for $\theta$, using the unbiased estimator $\hat\theta=3(Y_1+...+Y_n)/2n$? The example in Wikipedia is limited to normal distribution, so it's not very helpful.

EDIT: Ok, here's some thoughts I have on how to proceed, thanks to https://math.stackexchange.com/a/568579/627534. By central limit theorem, $\text{pdf }(2\sqrt{2n}(\hat\theta-\theta)/\theta)=\mathcal{N}(0,1)$ for large enough $n$, so $$.95=P(-1.96\leq 2\sqrt{2n}(\hat\theta-\theta)/\theta\leq 1.96)=P(-1.96\leq 2\sqrt{2n}\hat\theta/\theta-2\sqrt{2n}\leq 1.96)=P(-1.96+2\sqrt{2n}\leq 2\sqrt{2n}\hat\theta/\theta\leq 1.96+2\sqrt{2n})=P(1/(-1.96+2\sqrt{2n})\geq \theta/2\sqrt{2n}\hat\theta\geq 1/(1.96+2\sqrt{2n}))=P(2\sqrt{2n}\hat\theta/(-1.96+2\sqrt{2n})\geq \theta\geq 2\sqrt{2n}\hat\theta/(1.96+2\sqrt{2n})).$$ Is this right?

  • $\begingroup$ There's a big difference between "what is the definition of X" and "how do you compute X". You could give a much better title. $\endgroup$ – David K Dec 18 '18 at 13:37

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