Find the MVUE for a uniform parameter

Question

Let $Y_{(n)}=\max(Y_{1},Y_{2},\ldots,Y_{n})$, where $Y_{1},Y_{2},\ldots,Y_{n}$ is a sample uniform distribution in $(0,\theta)$. Find the MVUE for $\theta$.

My approach:

1. I know that $U=Y_{(n)}$ is a sufficient statistic for $\theta$. I could prove this using the factorization theorem by Neymann.
2. I know that $\displaystyle h(U)=\left(\frac{n+1}{n}\right)U$ satisfy that $\mathbb{E}[h(U)]=\theta$.

So, $$\left(\frac{n+1}{n}\right)Y_{(n)}$$ is a (or the?) MVUE for $\theta$.

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My approach is correct? Can I find other MVUE for $\theta$?

Answer 1

As suggested, to confirm that $\frac{n+1}{n}Y_{(n)}$ is UMVUE you have only to prove that $Y_{(n)}$ is CSS (Complete and Sufficient Statistic)

• Sufficiency is easy proved observing that

$$L(\theta)=\underbrace{1}_{=h(\mathbf{y})}\times\underbrace{\frac{1}{\theta^n}\mathbb{1}_{(y_{(n)};\infty)}(\theta)}_{g[t(\mathbf{y}),\theta]}$$

• To prove completeness, as the model does not below to the Exponential family, you must use the definition. Say you have to prove that, for all $g$ measurable and for all $\theta$

$$\mathbb{E}_{\theta}g(T)=0\rightarrow \mathbb{P}_{\theta}[g(T)=0]=1$$

In your case you have

$$0=\mathbb{E}_{\theta}g(T)=\int_0^{\theta}g(t)\frac{nt^{n-1}}{\theta^n}dt=\frac{1}{\theta^n}\underbrace{\int_0^{\theta}g(t)nt^{n-1}dt}_{=0}$$

Now take the derivative w.r.t. $\theta$ obtaining

$$0=\frac{1}{\theta^n}g(\theta)n\theta^{n-1}=\frac{ng(\theta)}{\theta}$$

Thus actually expectation =0 implies also $g=0$ and $Y_{(n)}$ is complete

Now you have only to apply the following theorem