题目

Problem

A discrete random variable $X$ has probability distribution

P(X = x) = \frac{1}{\alpha}, \quad x = 1, 2, \dots, \alpha

The mean of a random sample of size $n$ , taken from this distribution, is denoted by $\bar X$ .

(a) Show that $2\bar{X}$ is a biased estimator of $\alpha$ .

(2)

A random sample of 6 observations of $X$ is taken and the results are given below.

8,\ 7,\ 3,\ 7,\ 2,\ 9

(b) Use the sample mean to estimate $\alpha$ .

(2)

(Total 4 marks)

题目中文翻译

离散随机变量 $X$ 的概率分布为

P(X = x) = \frac{1}{\alpha}, \quad x = 1, 2, \dots, \alpha

从这个分布中抽取容量为 $n$ 的随机样本，其样本均值记为 $\bar X$ 。

(a) 证明 $2\bar{X}$ 是 $\alpha$ 的有偏估计量。

从 $X$ 的分布中抽取 6 个观测值，结果如下：

8,\ 7,\ 3,\ 7,\ 2,\ 9

(b) 用样本均值估计 $\alpha$ 。

解答

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离散均匀分布在 $1,2,\ldots,\alpha$ 上取值，因此先求 $X$ 的期望。样本均值 $\bar X$ 的期望等于总体均值，再与参数 $\alpha$ 比较；若估计量的期望不等于参数，它就是有偏估计量。

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The expectation of $X$ is

\begin{align*} E(X)=&\,\sum_{x=1}^{\alpha}xP(X=x)\\ =&\,\frac{1}{\alpha}\sum_{x=1}^{\alpha}x\\ =&\,\frac{\alpha(\alpha+1)}{2\alpha}\\ =&\,\frac{\alpha+1}{2}. \end{align*}

Since $E(\bar X)=E(X)$ ,

\begin{align*} E(2\bar X)=&\,2E(\bar X)\\ =&\,2\bigg(\frac{\alpha+1}{2}\bigg)\\ =&\,\alpha+1\\ \neq&\,\alpha. \end{align*}

Therefore, $2\bar X$ is a biased estimator of $\alpha$ .

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先由六个观测值求样本均值。因为 $E(\bar X)=(\alpha+1)/2$ ，用矩估计思想令理论均值等于样本均值，再解出 $\alpha$ 。

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The sample mean is

\begin{align*} \bar{x}=&\,\frac{8+7+3+7+2+9}{6}\\ =&\,6. \end{align*}

Using

E(\bar X)=\frac{\alpha+1}{2},

an estimate of $\alpha$ is

\begin{align*} \hat{\alpha}=&\,2\bar{x}-1\\ =&\,2(6)-1\\ =&\,\boxed{11}. \end{align*}