How to compute the confidence interval, given an unbiased estimator, for $\theta$, for the distribution $f(x)=2x/\theta^2$, $x\in [0,\theta]$?

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?

• 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. – David K Dec 18 '18 at 13:37