题号:1248    题型:解答题    来源:2001年全国硕士研究生招生考试试题
类型:考研真题
设总体 $X$ 服从正态分布 $N\left(\mu, \sigma^{2}\right)(\sigma > 0)$, 从该总体中抽取简单随机样本 $X_{1}, X_{2}, \cdots, X_{2 n}(n \geqslant 2)$, 其样本均值为 $\bar{X}=\frac{1}{2 n} \sum_{i=1}^{2 n} X_{i}$, 求统计量 $Y=\sum_{i=1}^{n}\left(X_{i}+X_{n+i}-2 \bar{X}\right)^{2}$ 的数学期望 $E(Y)$.
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答案:
记 $\overline{X_{1}}=\frac{1}{n} \sum_{t=1}^{n} X_{i}, \overline{X_{2}}=\frac{1}{n} \sum_{t=1}^{n} X_{n+i}$, 则 $\bar{X}=\frac{1}{2}\left(\overline{X_{1}}+\overline{X_{2}}\right)$, 即 $2 \bar{X}=\overline{X_{1}}+\overline{X_{2}}$
且 $E \overline{X_{1}}=E\left(\frac{1}{n} \sum_{i=1}^{n} X_{i}\right)=\frac{\sum_{i=1}^{n} E X_{i}}{n}=\frac{n u}{n}=u, E \overline{X_{2}}=E\left(\frac{1}{n} \sum_{i=1}^{n} X_{n+i}\right)=u$
因此
$$
\begin{aligned}
&E(Y)=E\left[\sum_{i=1}^{n}\left(X_{i}+X_{n+i}-2 \bar{X}\right)^{2}\right]=E\left\{\sum_{i=1}^{n}\left[\left(X_{i}-\overline{X_{1}}\right)+\left(X_{n+i}-\overline{X_{2}}\right)\right]^{2}\right\} \\
&=E\left\{\sum_{i=1}^{n}\left[\left(X_{i}-\overline{X_{1}}\right)^{2}+2\left(X_{i}-\overline{X_{1}}\right)\left(X_{n+i}-\overline{X_{2}}\right)+\left(X_{n+i}-\overline{X_{2}}\right)^{2}\right]\right\} \\
&=E\left[\sum_{i=1}^{n}\left(X_{i}-\overline{X_{1}}\right)^{2}\right]+E\left\{\sum_{i=1}^{n}\left[2\left(X_{i}-\overline{X_{1}}\right)\left(X_{n+i}-\overline{X_{2}}\right)\right]\right\}+E\left[\sum_{i=1}^{n}\left(X_{n+i}-\overline{X_{2}}\right)^{2}\right]
\end{aligned}
$$
因为样本方差 $S^{2}=\frac{1}{n-1}\left[\sum_{i=1}^{n}\left(X_{i}-\bar{X}_{1}\right)^{2}\right]$ 是总体方差的无偏估计, 则 $E S^{2}=\sigma^{2}$, 即

$$
E S^{2}=E\left[\frac{1}{n-1} \sum_{i=1}^{n}\left(X_{i}-\overline{X_{1}}\right)^{2}\right]=\sigma^{2}
$$
所以 $E\left[\sum_{i=1}^{n}\left(X_{i}-\overline{X_{1}}\right)^{2}\right]=(n-1) \sigma^{2}$, 同理 $E\left[\sum_{i=1}^{n}\left(X_{n+i}-\overline{X_{2}}\right)^{2}\right]=(n-1) \sigma^{2}$
而 $E\left\{\sum_{i=1}^{n}\left[2\left(X_{i}-\overline{X_{1}}\right)\left(X_{n+i}-\overline{X_{2}}\right)\right]\right\}=2 E\left\{\sum_{i=1}^{n}\left[\left(X_{i}-\overline{X_{1}}\right)\left(X_{n+i}-\overline{X_{2}}\right)\right]\right\}$
$$
\begin{aligned}
&=2 \sum_{i=1}^{n} E\left[\left(X_{i}-\overline{X_{1}}\right)\left(X_{n+i}-\overline{X_{2}}\right)\right]=\sum_{i=1}^{n} E\left(X_{i} X_{n+i}-X_{i} \overline{X_{2}}-\overline{X_{1}} X_{n+i}+\overline{X_{1}} \overline{X_{2}}\right) \\
&=\sum_{i=1}^{n}\left(E X_{i} X_{n+i}-E X_{i} \overline{X_{2}}-E \overline{X_{1}} X_{n+i}+E \overline{X_{1}} \overline{X_{2}}\right)
\end{aligned}
$$

由于 $X_{1}, X_{2}, \cdots, X_{2 n}(n \geq 2)$ 相互独立同分布, 则 $X_{i}$ 与 $\overline{X_{2}}, \overline{X_{1}}$ 与 $X_{n+i}, \overline{X_{1}}$ 与 $\overline{X_{2}}$ 也独 立 $(i=1,2 \cdots n)$. 而由独立随机变量期望的性质(若随机变量 $X, Y$ 独立, 且 $E X, E Y$ 都存在,
则 $E X Y=E X E Y)$, 所以
$$
\begin{aligned}
&E X_{i} X_{n+i}=E X_{i} E X_{n+i}=u^{2}, E X_{i} \overline{X_{2}}=E X_{i} E \overline{X_{2}}=u^{2} \\
&E \overline{X_{1}} X_{n+i}=E \overline{X_{1}} E X_{n+i}=u^{2}, E \overline{X_{1}} \overline{X_{2}}=E \overline{X_{1}} E \overline{X_{2}}=u^{2}
\end{aligned}
$$
故有 $E\left\{\sum_{i=1}^{n}\left[\left(X_{i}-\overline{X_{1}}\right)\left(X_{n+i}-\overline{X_{2}}\right)\right]\right\}$
$$
=\sum_{i=1}^{n}\left(E X_{i} X_{n+i}-E X_{i} \overline{X_{2}}-E \overline{X_{1}} X_{n+i}+E \overline{X_{1}} \overline{X_{2}}\right)=\sum_{i=1}^{n}\left(u^{2}-u^{2}-u^{2}+u^{2}\right)=0
$$

$$
\begin{aligned}
E(Y) &=E\left[\sum_{i=1}^{n}\left(X_{i}-\overline{X_{1}}\right)^{2}\right]+E\left\{\sum_{i=1}^{n}\left[2\left(X_{i}-\overline{X_{1}}\right)\left(X_{n+i}-\overline{X_{2}}\right)\right]\right\}+E\left[\sum_{i=1}^{n}\left(X_{n+i}-\overline{X_{2}}\right)^{2}\right] \\
&=(n-1) \sigma^{2}+(n-1) \sigma^{2}=2(n-1) \sigma^{2}
\end{aligned}
$$

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