# Find the unique MVUE

This question is from Robert Hogg's Introduction to Mathematical Statistics 6th Version problem 7.4.9 at page 388.

Let $$X_1,...,X_n$$ be iid with pdf $$f(x;\theta)=1/3\theta,-\theta zero elsewhere, where $$\theta>0$$.

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

(b) Is $$\hat{\theta}$$ a sufficient statistics for $$\theta$$ ? Why ?

(c) Is $$(n+1)\hat{\theta}/n$$ the unique MVUE of $$\theta$$? Why ?

## I think I can solve (a) and (b), but I am confused by (c).

For (a):

Let $$Y_1 be the order statistics.

$$L(\theta;x)=\frac{1}{3\theta}\times\frac{1}{3\theta}\times...\times\frac{1}{3\theta}=\frac{1}{(3\theta)^n}$$ when $$-\theta< y_1$$ and $$y_n < 2\theta$$;elsewhere $$L(\theta;x)=0$$

$$\frac{dL(\theta;x)}{d\theta}=-n(3\theta)^{n-1}$$, since $$\theta>0$$, we can see this derivative is negative,

so likelihood function $$L(\theta;x)$$ is decreasing.

From $$(-\theta< y_1$$ and $$y_n < 2\theta)$$, $$\Rightarrow$$ $$(\theta>-y_1$$ and $$\theta>y_n/2), \Rightarrow \theta>max(-y_1,y_n/2)$$

$$L(\theta,x)$$ is decreasing, so when $$\theta$$ has the samllest value the likelihood function will achieve maximum, since $$\theta>max(-y_1,y_n/2)$$, when $$\theta=max(-y1,y_n/2)$$, the likelihood function will achieve the maximum value.

$$\therefore$$ mle $$\hat{\theta}=max(-y_1,y_n/2)$$

For (b):

$$f(x_1;\theta)f(x_2;\theta)...f(x_n;\theta)=\frac{1}{(3\theta)^n}\prod_{i}^{n} I(-\theta

$$\therefore$$ By factorization theorem of Neyman, $$y_n=max(x_i)$$ is a sufficient statistic for $$\theta$$. Therefore, $$y_n/2$$ is also a sufficient statisitc

Samely,

$$f(x_1;\theta)f(x_2;\theta)...f(x_n;\theta)=\frac{1}{(3\theta)^n}\prod_{i}^{n} I(-\theta-\theta)\times 1$$

$$\therefore$$ By factorization theorem of Neyman, $$y_1=min(x_i)$$ is a sufficient statistic for $$\theta$$. Therefore, $$-y_1$$ is also a sufficient statisitc.

For (c):

First, we find the CDF of $$X$$

$$F(x)=\int_{-\theta}^{x}\frac{1}{3\theta}dt=\frac{x+\theta}{3\theta},-\theta

Next, we can find pdf for both $$Y_1$$ and $$Y_n$$ from the formula of the book for the order statistics.

$$f(y_1)=\frac{n!}{(1-1)!(n-1)!}[F(y_1)]^{1-1}[1-F(y_1)]^{n-1}f(y_1)=n[1-\frac{y_1+\theta}{3\theta}]^{n-1}\frac{1}{3\theta}=n\frac{1}{(3\theta)^n}(2\theta-y_1)^{n-1}$$

Samely,

$$f(y_n)=n(\frac{y_n+\theta}{3\theta})^{n-1}\frac{1}{3\theta}=n\frac{1}{(3\theta)^n}(y_n+\theta)^{n-1}$$

Next, we show the completeness of family of pdf for $$f(y_1)$$ and $$f(y_n)$$

$$E[u(Y_1)]=\int_{-\theta}^{2\theta}u(y_1)n\frac{1}{(3\theta)^n}(2\theta-y_1)^{n-1}dy_1=0 \Rightarrow \int_{-\theta}^{2\theta}u(y_1)(2\theta-y_1)dy_1=0$$. By $$FTC$$ (derivate the integral) we can show $$u(\theta)=0$$ for all $$\theta>0$$.

Therefore, family of pdf $$Y_1$$ is complete..

Samely, still by $$FTC$$, we can show that family of pdf $$Y_n$$ is complete.

## The problem is now we need to show that $$\frac{(n+1)\hat{\theta}}{n}$$ is unbiased.

When $$\hat{\theta}=-y_1$$

$$E(-y_1)=\int_{-\theta}^{2\theta}(-y_1)\frac{n}{(3\theta)^n}(2\theta-y_1)^{n-1}dy_1=\frac{1}{(3\theta)^n}\int_{-\theta}^{2\theta}y_1d(2\theta-y_1)^n$$

We can solve the integral by integrate by parts

$$E(-y_1)=\frac{1}{(3\theta)^n}[y_1(2\theta-y_1)^n\mid_{-\theta}^{2\theta}-\int_{-\theta}^{2\theta}(2\theta-y_1)^ndy_1]=\frac{1}{(3\theta)^n}[\theta (3\theta)^n-\frac{(3\theta)^{n+1}}{n+1}]=\theta-\frac{3\theta}{n+1}=\frac{(n-2)\theta}{n+1}$$

$$\therefore E(\frac{(n+1)\hat{\theta}}{n})=\frac{n+1}{n}E(-y_1)=\frac{n+1}{n}\frac{(n-2)\theta}{n+1}=\frac{n-2}{n}\theta$$

## Therefore, $$\frac{(n+1)\hat{\theta}}{n}$$ is not an unbiased estimator of $$\theta$$ when $$\hat{\theta}=-y_1$$

When $$\hat{\theta}=y_n/2$$

$$E(Y_n)=\int_{-\theta}^{2\theta}y_n\frac{n}{(3\theta)^n}(y_n+\theta)^{n-1}dy_n=\frac{1}{(3\theta)^n}\int_{-\theta}^{2\theta}y_nd(y_n+\theta)^n=\frac{1}{(3\theta)^n}[y_n(y_n+\theta)^n\mid_{-\theta}^{2\theta}-\int_{-\theta}^{2\theta}(y_n+\theta)^ndy_n]=\frac{1}{(3\theta)^n}[2\theta(3\theta)^-\frac{(3\theta)^{n+1}}{n+1}]=2\theta-\frac{3\theta}{n+1}=\frac{2n-1}{n+1}\theta$$

$$\therefore E(\frac{(n+1)\hat{\theta}}{n})=\frac{n+1}{n}E(Y_n/2)=\frac{n+1}{2n}E(Y_n)=\frac{n+1}{2n}\frac{2n-1}{n+1}\theta=\frac{2n-1}{2n}\theta$$

## But the book's answer is that $$\frac{(n+1)\hat{\theta}}{n}$$ is an unique MVUE. I don't understand why it is a MVUE if it is a biased estimator.

Or my calculations are wrong, please help me to find the mistakes, I can give you more detailed calculations.

• I do not see any computation of the distribution of $\hat\theta$.
– whuber
Aug 3, 2015 at 17:17
• Thanks, whuber, the $\hat{\theta}=max(-y_1,y_n/2)$. It is either $-y_1$ or $y_n/2$ depends on which one is bigger. I calculated the distributions both for $y_1$ and $y_n$. You can see $f(y_1)=n\frac{1}{(3\theta)^n}(2\theta-y_1)^{n-1}$and $f(y_n)=n\frac{1}{(3\theta)^n}(y_n+\theta)^{n-1}$ in the text. Aug 4, 2015 at 0:31
• And from the two above distributions, I calculated $E(\hat{\theta})=E(-Y_1)$ and $E(\hat{\theta})=E(Y_n/2)$ then $E(\frac{n+1}{n}\hat{\theta})$ Aug 4, 2015 at 0:42

Working with extrema requires care, but it doesn't have to be difficult. The crucial question, found near the middle of the post, is

... we need to show that $\frac{n+1}{n}{\hat \theta}^{n}$ is unbiased.

Earlier you obtained

$$\hat\theta = \max(-y_1, y_n/2) = \max\{-\min\{y_i\}, \max\{y_i\}/2\}.$$

Although that looks messy, the calculations become elementary when you consider the cumulative distribution function $F$. To get started with this, note that $0\le \hat\theta \le \theta$. Let $t$ be a number in this range. By definition,

\eqalign{ F(t) &= \Pr(\hat\theta\le t) \\&= \Pr(-y_1 \lt t\text{ and }y_n/2 \le t) \\ &= \Pr(-t \le y_1 \le y_2 \cdots \le y_n \le 2t). }

This is the chance that all $n$ values lie between $-t$ and $2t$. Those values bound an interval of length $3t$. Because the distribution is uniform, the probability that any specific $y_i$ lies in this interval is proportional to its length:

$$\Pr(y_i \in [-t, 2t]) = \frac{3t}{3\theta} = \frac{t}{\theta}.$$

Because the $y_i$ are independent, these probabilities multiply, giving

$$F(t) = \left(\frac{t}{\theta}\right)^n.$$

The expectation can immediately be found by integrating the survival function $1-F$ over the interval of possible values for $\hat\theta$, $[0, \theta]$, using $y=t/\theta$ for the variable:

$$\mathbb{E}(\hat\theta) = \int_0^\theta \left(1 - \left(\frac{t}{\theta}\right)^n\right)dt = \int_0^1 (1-y^n)\theta dy = \frac{n}{n+1}\theta.$$

(This formula for the expectation is derived from the usual integral via integration by parts. Details are given at the end of https://stats.stackexchange.com/a/105464.)

Rescaling by $(n+1)/n$ gives

$$\mathbb{E}\left(\frac{n+1}{n}\,{\hat \theta}\right) = \theta,$$

QED.

Why can we say that MLE is sufficient? I think it is not.

This is similar case to Example 7.8.1. from Robert Hogg's Introduction to Mathematical Statistics : U[θ,θ+1] - uniform distribution on [θ,θ+1]. It's known that (X(1),X(n)) is minimal sufficient statistics. Hence there are no one-dimensional sufficient statistics. Thus MLE isn't sufficient.

• In what way, specifically, is this a "similar" case and why would that imply lack of sufficiency?
– whuber
Jan 13 at 16:39
• In question formulation above we can conclude that (-X(1),X(n)/2) is minimal sufficient statistic. It is two dimentional, so there are no one-dimensional sufficient statistics Jan 13 at 17:09
• This might be an answer to another question! Did you post at the wrong place? Jan 13 at 19:29
• No, I'm talking about b) Jan 13 at 19:43