pdf of a maximum order statistic with a uniform distribution

Question

I have the following problem,

$X_1,X_2,...,X_n$~$U[\theta,2\theta]$ I am tasked with finding the pdf of the mle. First off, I know that $f(x)=\frac{1}{\theta}$ just by definition and after some calculations, I have found the following,

$$lik(\theta)=\frac{1}{\theta^n} \implies \hat{\theta}_{mle}=\frac{X_{(n)}}{2}$$

Where I'm struggling is that I now want to find the pdf of $\hat{\theta}_{mle}$ using the order statistic $X_{(n)}$ by using the formula of $f_u(u)=n[F(u)]^{n-1}f(u)$ where $U=\hat{\theta}_{mle}=X_{(n)}/2$.

Using some calculations I get the following, $$f(u)=\frac{1}{\hat{\theta}_{mle}}$$ $$F(u)=\frac{x-\hat{\theta}_{mle}}{\hat{\theta}_{mle}}$$ $$\implies f_u(u)=n[\frac{x-\hat{\theta}_{mle}}{\hat{\theta}_{mle}}]^{n-1} \frac{1}{\hat{\theta}_{mle}}$$ Are my assertions correct on this? The reason I ask is because when I take the expectation of that function I should get $$E(X_{(n)})=\frac{2n+1}{n+1} \theta$$

However, that's not the outcome that I am getting. Thanks

Answer 1

Given $\theta$,

• the probability all $n$ values in a sample are less than or equal to $y$ is $\left(\frac{y-\theta}{\theta}\right)^n$ for $\theta \le y \le 2 \theta$

• so $\mathbb P(\hat{\theta}_{mle} \le x) = \left(\frac{2x-\theta}{\theta}\right)^n$ for $\frac\theta2 \le x \le \theta$

• so the density of the distribution of the maximum likelihood estimator is $\frac{2n}{\theta}\left(\frac{2x-\theta}{\theta}\right)^{n-1}$ when $\frac\theta2 \le x \le \theta$; (your density is for $X_{(n)}$ when $\theta \le x \le 2\theta$ rather than for $\hat{\theta}_{mle}$ and subtracts the estimator rather than the actual parameter)

and that gives a mean of $\mathbb E[\hat{\theta}_{mle}] = \frac{2n+1}{2n+2} \theta = \theta - \frac{1}{2n+2}\theta$ and $\mathbb E[X_{(n)}] =\mathbb E[2\hat{\theta}_{mle}]= \frac{2n+1}{n+1} \theta = 2\theta - \frac{1}{n+1}\theta$ as you might expect