设 $\left(X_{1}, Y_{1}\right),\left(X_{2}, Y_{2}\right), \cdots,\left(X_{n}, Y_{n}\right)$ 为来自总体 $N\left(\mu_{1}, \mu_{2} ; \sigma_{1}^{2}, \sigma_{2}^{2} ; \rho\right)$ 的简单随机样本, 令 $\theta=\mu_{1}-\mu_{2}, \bar{X}=\frac{1}{n} \sum_{i=1}^{n} X_{i}, \bar{Y}=\frac{1}{n} \sum_{i=1}^{n} Y_{i}, \hat{\theta}=\bar{X}-\bar{Y}$, 则
(A) $\hat{\theta}$ 是 $\theta$ 的无偏估计, $D(\hat{\theta})=\frac{\sigma_{1}^{2}+\sigma_{2}^{2}}{n}$
(B) $\hat{\theta}$ 不是 $\theta$ 的无偏估计, $D(\hat{\theta})=\frac{\sigma_{1}^{2}+\sigma_{2}^{2}}{n}$
(C) $\hat{\theta}$ 是 $\theta$ 的无偏估计, $D(\hat{\theta})=\frac{\sigma_{1}^{2}+\sigma_{2}^{2}-2 \rho \sigma_{1} \sigma_{2}}{n}$
(D) $\hat{\theta}$ 不是 $\theta$ 的无偏估计, $D(\hat{\theta})=\frac{\sigma_{1}^{2}+\sigma_{2}^{2}-2 \rho \sigma_{1} \sigma_{2}}{n}$
因为 $X, Y$ 是二维正态分布, 所以 $\bar{X}$ 与 $\bar{Y}$ 也服从二维正态分布, 则 $\bar{X}-\bar{Y}$ 也服从二维正态分布, 即 $E(\hat{\theta})=E(\bar{X}-\bar{Y})=E(\bar{X})-E(\bar{Y})=\mu_{1}-\mu_{2}=\theta$,
$D(\hat{\theta})=D(\bar{X}-\bar{Y})=D(\bar{X})+D(\bar{Y})-\operatorname{cov}(\bar{X}, \bar{Y})=\frac{\sigma_{1}^{2}+\sigma_{2}^{2}-2 \rho \sigma_{1} \sigma_{2}}{n}$, 故正确答案为C $\mathrm{C}$.