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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 $f(x;\theta)=1/3\theta,-\theta<x<2\theta,$ 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 $Y_1<Y_2<...Y_n$ 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):

$f(x_1;\theta)f(x_2;\theta)...f(x_n;\theta)=\frac{1}{(3\theta)^n}\prod_{i}^{n} I(-\theta<x_i<2\theta)=\frac{1}{(3\theta)^n}I(max(x_i)<2\theta)\times 1$

$\therefore$ 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,

$f(x_1;\theta)f(x_2;\theta)...f(x_n;\theta)=\frac{1}{(3\theta)^n}\prod_{i}^{n} I(-\theta<x_i<2\theta)=\frac{1}{(3\theta)^n}I(min(x_i)>-\theta)\times 1$

$\therefore$ 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 $X$

$F(x)=\int_{-\theta}^{x}\frac{1}{3\theta}dt=\frac{x+\theta}{3\theta},-\theta<x<2\theta$

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

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

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

Still, $\frac{(n+1)\hat{\theta}}{n}$ is not an unbiased estimator of $\theta$ when $\hat{\theta}=y_n/2$

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.

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  • $\begingroup$ I do not see any computation of the distribution of $\hat\theta$. $\endgroup$
    – whuber
    Aug 3, 2015 at 17:17
  • $\begingroup$ 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. $\endgroup$
    – Deep North
    Aug 4, 2015 at 0:31
  • $\begingroup$ And from the two above distributions, I calculated $E(\hat{\theta})=E(-Y_1)$ and $E(\hat{\theta})=E(Y_n/2)$ then $E(\frac{n+1}{n}\hat{\theta})$ $\endgroup$
    – Deep North
    Aug 4, 2015 at 0:42

2 Answers 2

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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.

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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.

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  • 1
    $\begingroup$ In what way, specifically, is this a "similar" case and why would that imply lack of sufficiency? $\endgroup$
    – whuber
    Jan 13 at 16:39
  • $\begingroup$ 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 $\endgroup$
    – Jan Nowak
    Jan 13 at 17:09
  • $\begingroup$ This might be an answer to another question! Did you post at the wrong place? $\endgroup$ Jan 13 at 19:29
  • $\begingroup$ No, I'm talking about b) $\endgroup$
    – Jan Nowak
    Jan 13 at 19:43
  • 1
    $\begingroup$ Your are right, but that error was corrected in my answer, which you haven't addressed. $\endgroup$
    – whuber
    Jan 14 at 16:29

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