kaoyan1basic 概率论与数理统计 第12题
📝 题目
### 【强化篇】第12题（解答题） 12．设总体 $\displaystyle X \sim\left(\begin{array}{ccc}0 & 1 & 2 \\ \frac{\theta}{4 N} & \frac{\theta}{2 N} & \frac{4 N-3 \theta}{4 N}\end{array}\right)$ ，其中 $N>0$ 已知，$\theta>0$ 未知，设 $X_{1}, X_{2}, \cdots, X_{n}$ 是来自总体 $X$ 的简单随机样本，取到 0 的个数为 $n_{0}$ ，取到 1 的个数为 $n_{1}$ ，取到 2 的个数为 $n_{2}$ ，即 $n_{0}+n_{1}+n_{2}=n$ ． （1）求 $\theta$ 的矩估计量 $\hat{\theta}_{1}$ 和最大似然估计量 $\hat{\theta}_{2}$ ； （2）求 $\dot{\theta}_{1}$ 和 $\dot{\theta}_{2}$ 的数学期望； （3）求 $\hat{\theta}_{1}$ 和 $\hat{\theta}_{2}$ 的方差。
💡 答案解析
**答案**：（1）$\displaystyle \hat{\theta}_1 = \frac{4N}{3n}(n_1+2n_2)$，$\displaystyle \hat{\theta}_2 = \frac{4N}{3n}(n_1+2n_2)$；（2）$E(\hat{\theta}_1)=E(\hat{\theta}_2)=\theta$；（3）$\displaystyle D(\hat{\theta}_1)=D(\hat{\theta}_2)=\frac{16N^2}{9n}[\frac{\theta}{2N} + \frac{4(4N-3\theta)}{4N} - (\frac{\theta}{2N}+\frac{2(4N-3\theta)}{4N})^2] = \frac{16N^2}{9n}[\frac{\theta}{2N} + \frac{4N-3\theta}{N} - (\frac{\theta}{2N}+\frac{4N-3\theta}{2N})^2] = \frac{16N^2}{9n}[\frac{\theta+8N-6\theta}{2N} - (\frac{4N-2\theta}{2N})^2] = \frac{16N^2}{9n}[\frac{8N-5\theta}{2N} - (\frac{2N-\theta}{N})^2]$ **解析**：步骤1：总体分布$\displaystyle P(X=0)=\frac{\theta}{4N}, P(X=1)=\frac{\theta}{2N}, P(X=2)=\frac{4N-3\theta}{4N}$。$\displaystyle E(X)=0\cdot\frac{\theta}{4N}+1\cdot\frac{\theta}{2N}+2\cdot\frac{4N-3\theta}{4N}=\frac{\theta}{2N}+\frac{4N-3\theta}{2N}=\frac{4N-2\theta}{2N}=\frac{2N-\theta}{N}$。由矩估计法，$\displaystyle \bar{X}=\frac{2N-\hat{\theta}_1}{N}$，解得$\displaystyle \hat{\theta}_1 = 2N - N\bar{X} = 2N - \frac{N}{n}(n_1+2n_2) = \frac{2Nn - Nn_1 - 2Nn_2}{n} = \frac{N(2n - n_1 - 2n_2)}{n}$。由于$n=n_0+n_1+n_2$，$2n - n_1 - 2n_2 = 2n_0+2n_1+2n_2 - n_1 - 2n_2 = 2n_0+n_1$，故$\displaystyle \hat{\theta}_1 = \frac{N(2n_0+n_1)}{n}$。 步骤2：似然函数$\displaystyle L(\theta)=(\frac{\theta}{4N})^{n_0} (\frac{\theta}{2N})^{n_1} (\frac{4N-3\theta}{4N})^{n_2}$，取对数$\ln L = n_0\ln\theta + n_1\ln\theta + n_2\ln(4N-3\theta) - (n_0+n_1+n_2)\ln(4N) - n_1\ln2$，求导$\displaystyle \frac{d\ln L}{d\theta} = \frac{n_0+n_1}{\theta} - \frac{3n_2}{4N-3\theta}=0$，解得$\displaystyle \hat{\theta}_2 = \frac{4N(n_0+n_1)}{3(n_0+n_1)+3n_2} = \frac{4N(n_0+n_1)}{3n}$。 步骤3：$\displaystyle E(\hat{\theta}_1)=\frac{N}{n}E(2n_0+n_1)=\frac{N}{n}[2nP(X=0)+nP(X=1)] = \frac{N}{n}[2n\cdot\frac{\theta}{4N}+n\cdot\frac{\theta}{2N}] = \frac{N}{n}[\frac{n\theta}{2N}+\frac{n\theta}{2N}] = \theta$。$\displaystyle E(\hat{\theta}_2)=\frac{4N}{3n}E(n_0+n_1)=\frac{4N}{3n}[nP(X=0)+nP(X=1)] = \frac{4N}{3n}[n\cdot\frac{\theta}{4N}+n\cdot\frac{\theta}{2N}] = \frac{4N}{3n}\cdot\frac{3n\theta}{4N} = \theta$。 步骤4：$\displaystyle D(\hat{\theta}_1)=\frac{N^2}{n^2}D(2n_0+n_1)=\frac{N^2}{n^2}[4D(n_0)+D(n_1)+4Cov(n_0,n_1)]$，其中$\displaystyle n_0\sim B(n,\frac{\theta}{4N}), n_1\sim B(n,\frac{\theta}{2N}), Cov(n_0,n_1)=-n\cdot\frac{\theta}{4N}\cdot\frac{\theta}{2N}$。$\displaystyle D(\hat{\theta}_1)=\frac{N^2}{n^2}[4n\frac{\theta}{4N}(1-\frac{\theta}{4N})+n\frac{\theta}{2N}(1-\frac{\theta}{2N})-4n\frac{\theta^2}{8N^2}] = \frac{N^2}{n^2}[n\frac{\theta}{N}(1-\frac{\theta}{4N})+n\frac{\theta}{2N}(1-\frac{\theta}{2N})-n\frac{\theta^2}{2N^2}] = \frac{N^2}{n}[\frac{\theta}{N}-\frac{\theta^2}{4N^2}+\frac{\theta}{2N}-\frac{\theta^2}{4N^2}-\frac{\theta^2}{2N^2}] = \frac{N^2}{n}[\frac{3\theta}{2N}-\frac{\theta^2}{N^2}] = \frac{3N\theta}{2n} - \frac{\theta^2}{n}$。 $\displaystyle D(\hat{\theta}_2)=\frac{16N^2}{9n^2}D(n_0+n_1)=\frac{16N^2}{9n^2}[D(n_0)+D(n_1)+2Cov(n_0,n_1)] = \frac{16N^2}{9n^2}[n\frac{\theta}{4N}(1-\frac{\theta}{4N})+n\frac{\theta}{2N}(1-\frac{\theta}{2N})-2n\frac{\theta^2}{8N^2}] = \frac{16N^2}{9n^2}[n\frac{\theta}{4N}-n\frac{\theta^2}{16N^2}+n\frac{\theta}{2N}-n\frac{\theta^2}{4N^2}-n\frac{\theta^2}{4N^2}] = \frac{16N^2}{9n^2}[n\frac{3\theta}{4N}-n\frac{9\theta^2}{16N^2}] = \frac{16N^2}{9n}[\frac{3\theta}{4N}-\frac{9\theta^2}{16N^2}] = \frac{4N\theta}{3n} - \frac{\theta^2}{n}$。 **难度**：★★★★☆