Computing the UMVUE for $U(0,2\theta+1)$

Question

This question is from one of the introductory books in Mathematical Statistics.

Let $X_1,...,X_n$ be a random sample from a pdf $f(x;\theta)=\frac{1}{2\theta+1},0<x<2\theta+1,$ zero elsewhere.

(a) Find the MLE $\hat{\theta}$ of $\theta$

(b) Find a complete and sufficient statistics for $\theta$.

(c) Find the UMVUE of $\theta$

I think I can solve (a) and a half (b)

For (a):

Let $Y_1<Y_2<...Y_n$ be the order statistics.

$L(\theta)=\frac{1}{(2\theta+1)^{n}} I(0 \leq Y_{1}) I(Y_{n} \leq 2\theta+1)$

$\therefore$ MLE $\hat{\theta}=\frac{1}{2}(Y_{n}-1)$

For (b):

$f(x_1;\theta)f(x_2;\theta)...f(x_n;\theta)=\frac{1}{(2\theta+1)^{n}} I(0 \leq Y_{1}) I(Y_{n} \leq 2\theta+1)$

$\therefore$ By factorization theorem of Neyman, $Y_n=\max(x_i)$ is a sufficient statistic for $\theta$. Therefore, $\frac{1}{2}(Y_{n}-1)$ is also a sufficient statisitc

I was wondering if someone would help me out in showing it is complete and find UMVUE of $\theta$.

Answer 1

If you can find a function of $Y_n$ that is an unbiased estimator of $\theta$, then the Lehmann-Scheffe Theorem tells you that it will be the UMVUE.
So, first find $EY_n$ and then see if you can find a function of it that is unbiased.
The density function for $Y_n$ is $n[F(x)]^{n-1}f(x)$ for $x \in [0,2\theta+1]$.
Thus, $$EY_n=\int_0^{2\theta+1} xn \left[\frac{x}{2 \theta+1} \right]^{n-1}\frac{1}{2 \theta+1}dx=\frac{n}{(2 \theta+1)^n}\frac{(2 \theta+1)^{n+1}}{n+1}$$ Finally, $E\left[\frac{\frac{n+1}{n}Y_n-1}{2} \right]=\frac{\frac{n+1}{n}EY_n-1}{2}=\theta$.
So, $\frac{\frac{n+1}{n}Y_n-1}{2} $ is the UMVUE.

Answer 2

Would you give me an idea of how would I show completeness?

You can use the definition.

The density of $Y_{(n)}$ is

$$f_{Y_{(n)}}(t)=\frac{n t^{n-1}}{(2\theta+1)^n}$$

By definition $T$ is complete if

$$\mathbb{E}[g(T)]=0$$

implies

$$\mathbb{P}[g(T)=0]=1$$

In your case you have

$$\int_0^{2\theta+1}g(t)\frac{n t^{n-1}}{(2\theta+1)^n}dt=0$$

now derivating both sides and using Fundamental Theorem of Calculus you get

$$\frac{2n}{2\theta+1}\cdot g(2\theta+1)=0$$

which is true iff $g=0$

This shows the completeness of $Y_{(n)}$