题目

3.(2021,Ⅲ)设(X1 Y1),(X2,Y2),···,(Xn,Yn)为来自总体N(μ1,-|||-μ2;σ1^2,σ2^2,ρ)的简单随机样本,令 theta =(mu )_(1)-(mu )_(2), overline (X)=dfrac (1)(n)sum _(i=1)^nX overline (Y)=dfrac (1)(n)sum _(i=1)^n(Y)_(i),-|||-hat (theta )=overline (X)-overline (Y), 则 () .-|||-(A) (hat (theta ))=theta ,D(hat (theta ))=dfrac ({{sigma )_(1)}^2+({sigma )_(2)}^2}(n)-|||-(B) (hat (theta ))=theta ,D(hat (theta ))=dfrac ({{sigma )_(1)}^2+({sigma )_(2)}^2-2(rho )_(2)(Q)_(2)}(n)-|||-(C) (theta )neq theta , (hat (theta ))=dfrac ({{sigma )_(1)}^2+({sigma )_(2)}^2}(n)-|||-(D) (theta )neq theta , (hat (theta ))=dfrac ({{sigma )_(1)}^2+({sigma )_(2)}^2-2rho (sigma )_(1)(sigma )_(2)}}(n)

题目解答

答案

解析

考查要点：本题主要考查二元正态分布下样本均值差的期望与方差计算，重点在于理解协方差在独立样本中的作用。

解题核心思路：

  1. 期望计算：利用线性性质，直接求出$\hat{\theta} = \overline{X} - \overline{Y}$的期望。
  2. 方差计算：需考虑$\overline{X}$与$\overline{Y}$之间的协方差，因为原始变量$X_i$与$Y_i$存在相关性$\rho$。

破题关键点：

  • 期望的线性性：$E(\hat{\theta}) = E(\overline{X}) - E(\overline{Y}) = \mu_1 - \mu_2 = \theta$。
  • 方差的协方差项：由于$X_i$与$Y_i$相关，$D(\hat{\theta})$需包含$-2\rho\sigma_1\sigma_2/n$项，而选项A忽略了这一项。

期望计算

由$\overline{X} = \frac{1}{n}\sum X_i$和$\overline{Y} = \frac{1}{n}\sum Y_i$，根据期望的线性性：
$E(\hat{\theta}) = E(\overline{X} - \overline{Y}) = E(\overline{X}) - E(\overline{Y}) = \mu_1 - \mu_2 = \theta.$

方差计算

  1. 方差展开：
    $D(\hat{\theta}) = D(\overline{X} - \overline{Y}) = D(\overline{X}) + D(\overline{Y}) - 2\text{Cov}(\overline{X}, \overline{Y}).$

  2. 计算各部分：

  • $D(\overline{X}) = \frac{\sigma_1^2}{n}$，$D(\overline{Y}) = \frac{\sigma_2^2}{n}$。
  • 协方差计算：
    $\text{Cov}(\overline{X}, \overline{Y}) = \text{Cov}\left(\frac{1}{n}\sum X_i, \frac{1}{n}\sum Y_j\right) = \frac{1}{n^2}\sum_{i=1}^n \sum_{j=1}^n \text{Cov}(X_i, Y_j).$
    由于当$i \neq j$时，$X_i$与$Y_j$独立，$\text{Cov}(X_i, Y_j) = 0$；当$i = j$时，$\text{Cov}(X_i, Y_i) = \rho\sigma_1\sigma_2$。因此：
    $\text{Cov}(\overline{X}, \overline{Y}) = \frac{1}{n^2} \cdot n \cdot \rho\sigma_1\sigma_2 = \frac{\rho\sigma_1\sigma_2}{n}.$
  1. 代入方差公式：
    $D(\hat{\theta}) = \frac{\sigma_1^2}{n} + \frac{\sigma_2^2}{n} - 2 \cdot \frac{\rho\sigma_1\sigma_2}{n} = \frac{\sigma_1^2 + \sigma_2^2 - 2\rho\sigma_1\sigma_2}{n}.$

结论：选项B正确。