HW11(习题六、七(只计算矩估计))
A11
Question
设总体 \(X\) 服从标准正态分布， \(X_1, X_2, \cdots, X_{16}\) 是来自总体 \(X\) 的简单随机样本，写出下列统计量的分布：
(1)样本均值 \(\overline{X}\) ；
(2) \(\sum\limits_{i=1}^{16}X_i^2\) ；
(3) \(\frac{3X_1}{\sqrt{\sum\limits_{i=2}^{10}X_i^2}}\) ；
(4) \(\frac{X_1+X_2}{\sqrt{X_3^2+X_4^2}}\) ；
(5) \(\overline{X} - X_1\) ；
Answer
\[N(0,\frac{1}{16})\]
\[\chi^2(16)\]
\[\frac{3X_1}{\sqrt{\sum\limits_{i=2}^{10}X_i^2}} = \frac{X_1}{\sqrt{\frac{\sum\limits_{i=2}^{10}X_i^2}{9}}}\]
\[\therefore t(9)\]
\[X_1+X_2 \sim N(0,2)\]
\[\therefore \frac{1}{\sqrt{2}}(X_1+X_2) \sim N(0,1)\]
\[\sqrt{X_3^2+X_4^2} \sim \chi^2(2)\]
\[\therefore t(2)\]
\[N(0,\frac{15}{16})\]
\[E(\overline{X} - X_1) = E(\overline{X}) - E(X_1) = 0\]
\[D(\overline{X} - X_1) = D(\overline{X}) + D(X_1) - 2Cov(\overline{X}, X_1)\]
\[D(\overline{X}) = \frac{1}{16}, D(X_1) = 1\]
\[Cov(\overline{X}, X_1) = Cov(\frac{1}{16}\sum\limits_{i=1}^{16}X_i, X_1) = \frac{1}{16}Cov(X_1, X_1) = \frac{1}{16}\]
\[\therefore D(\overline{X} - X_1) = \frac{15}{16}\]
\[\therefore N(0,\frac{15}{16})\]
B2
Question
设总体 \(X \sim N(\mu, \sigma^2)\)， \(X_1, X_2, \cdots, X_{9}\) 是来自总体 \(X\) 的简单随机样本， \(\overline{X}\) 是样本均值， \(\sigma^2\) 是样本方差，写出下列抽样分布：
(1) \(\frac{3(\overline{X} - \mu)}{\sigma}\) ；
(2) \(\frac{3(\overline{X} - \mu)}{S}\) ；
(3) \(\frac{\sum\limits_{i=1}^{9}(X_i - \overline{X})^2}{\sigma^2}\) ；
(4) \(\frac{\sum\limits_{i=1}^{9}(X_i - \mu)^2}{S^2}\) ；
(5) \(\frac{9(\overline{X} - \mu)^2}{\sigma^2}\) ；
(6) \(\frac{9(\overline{X} - \mu)^2}{S^2}\) ；
(7) \(\frac{2(X_1 - X_2)^2}{(X_3 - X_4)^2 + (X_5 - X_6)^2}\) ；
(8) \(\frac{(X_1 - Y_1)^2 + (X_2 - Y_1)^2 + (X_3 - Y_1)^2}{(X_4 - Y_2)^2 + (X_5 - Y_2)^2 + (X_6 - Y_2)^2}\) ，其中 \(Y_1 = \frac{X_1 + X_2 + X_3}{3}\) ， \(Y_2 = \frac{X_4 + X_5 + X_6}{3}\) ；
Answer
\[N(0,1)\]
\[S^2 = \frac{1}{8}\sum\limits_{i=1}^{9}(X_i - \overline{X})^2\]
\[ \frac{8S^2}{\sigma^2} \sim \chi^2(8) \]
\[\frac{\frac{3(\overline{X} - \mu)}{\sigma}}{\sqrt{\frac{8S^2}{8\sigma^2}}} \sim t(8)\]
\[\sum\limits_{i=1}^{9}(\frac{X_i - \overline{X}}{\sigma})^2 \sim \chi^2(8)\]
\[\sum\limits_{i=1}^{9}(\frac{X_i - \mu}{\sigma})^2 \sim \chi^2(9)\]
\[\text{记} Z = \frac{3(\overline{X} - \mu)}{\sigma} \sim N(0,1)\]
\[\therefore Z^2 \sim \chi^2(9)\]
\[\frac{\frac{9(\overline{X} - \mu)^2}{\sigma^2 \cdot 1}}{\frac{8S^2}{8\sigma^2}} = \frac{9(\overline{X} - \mu)^2}{S^2} \sim F(1,8)\]
\[X_1 - X_2 \sim N(0,2\sigma^2)\]
\[\frac{1}{2\sigma^2}(X_1 - X_2)^2 \sim \chi^2(1)\]
\[\frac{1}{2\sigma^2}((X_3 - X_4)^2 + (X_5 - X_6)^2) \sim \chi^2(2)\]
\[\frac{\frac{1}{2\sigma^2 \cdot 1}(X_1 - X_2)^2}{\frac{1}{2\sigma^2 \cdot 2}((X_3 - X_4)^2 + (X_5 - X_6)^2)} \sim F(1,2)\]
\[Y_1 \sim N(0, \frac{1}{3}\sigma^2) \quad Y_2 \sim N(0, \frac{1}{3}\sigma^2)\]
\[\frac{\sum\limits_{i=1}^{3}(X_i - Y_1)^2}{\sigma^2} \sim \chi^2(2)\]
\[\frac{\sum\limits_{i=1}^{3}(X_i - Y_1)^2}{\sum\limits_{i=1}^{3}(X_i - Y_2)^2} \sim F(2,2)\]
B5
Question
设总体 \(X\) 的密度函数
\[f(x) = \frac{1}{2}e^{-|x|},-\infty < x < +\infty\]
从总体中抽取样本容量为 10 的样本， \(\overline{X}\) 和 \(S^2\) 分别为样本均值和样本方差，求
(1) \(\overline{X}\) 的数学期望和方差；
(2) \(S^2\) 的数学期望；
Answer
\[E(X_i) = \int_{-\infty}^{+\infty} x \cdot \frac{1}{2}e^{-|x|} dx = 0\]
\[\therefore E(\overline{X}) = \frac{1}{n} \sum\limits_{i=1}^{n} E(X_i) = 0\]
\[E(X_i^2) = \int_{-\infty}^{+\infty} x^2 \cdot \frac{1}{2}e^{-|x|} dx = 2\]
\[\therefore D(X_i) = E(X_i^2) - [E(X_i)]^2 = 2 - 0^2 = 2\]
\[D(\overline{X}) = \frac{1}{n^2} \sum\limits_{i=1}^{n} D(X_i) = \frac{2}{n} = \frac{2}{10} = 0.2\]
\[S^2 = \frac{1}{n-1} \sum\limits_{i=1}^{n} (X_i - \overline{X})^2\]
\[\text{由于} S^2 \text{无偏}\]
\[E(S^2) = Var(X) = 2\]
B7
Question
设总体 \(X\) 的密度函数
\[f(x) = \begin{cases} \lambda e^{-\lambda x},x > 0 \\ 0,x \le 0 \end{cases}\]
从样本中抽取容量为 10 的样本
(1) 求样本均值的数学期望和方差；
(2) 记 \(X_{(1)} = min\{X_1,X_2,\cdots,X_{10}\}\)，求 \(X_{(1)}\) 的数学期望和方差。
Answer
\[E(X_i) = \int_{0}^{+\infty} x \cdot \lambda e^{-\lambda x} dx = \frac{1}{\lambda}\]
\[\therefore E(\overline{X}) = \frac{1}{n} \sum\limits_{i=1}^{n} E(X_i) = \frac{1}{\lambda}\]
\[E(X_i^2) = \int_{0}^{+\infty} x^2 \cdot \lambda e^{-\lambda x} dx = \frac{2}{\lambda^2}\]
\[\therefore D(X_i) = E(X_i^2) - [E(X_i)]^2 = \frac{1}{\lambda^2}\]
\[D(\overline{X}) = \frac{1}{n^2} \sum\limits_{i=1}^{n} D(X_i) = \frac{1}{10\lambda^2}\]
\[F_X(x) = P(X \le x) = 1 - e^{-\lambda x},x \ge 0\]
\[F_{X_{(1)}}(x) = P(X_{(1)} \le x) = 1 - P(X_1 > x,X_2 > x,\cdots,X_{10} > x) = 1 - (e^{-\lambda x})^{10},x \ge 0\]
\[f_{X_{(1)}}(x) = 10 \lambda e^{-10 \lambda x},x \ge 0\]
\[\therefore X_{(1)} \sim E(10 \lambda)\]
\[E(X_{(1)}) = \frac{1}{10 \lambda}\]
\[D(X_{(1)}) = \frac{1}{100 \lambda^2}\]
B10
Question
设总体 \(X \sim N(\mu,\sigma^2)\) ， \(X_1,X_2,\cdots,X_5\) 和 \(Y_1,Y_2,\cdots,Y_9\) 是来自总体 \(X\) 的两个独立样本， \(\overline{X}\) 和 \(\overline{Y}\) 分别是这两个样本的样本均值， \(S_1^2\) 和 \(S_2^2\) 分别是这两个样本的样本方差。
(1)若 \(\frac{a(\overline{X} - \overline{Y})}{\sigma} \sim N(0,1)\) ，求 \(a\) 的值；
(2)若 \(\frac{b(\overline{X} - \overline{Y})}{\sqrt{S_1^2 + 2S_2^2}} \sim t(12)\) ，求 \(b\) 的值。
Answer
\[E(\overline{X} - \overline{Y}) = E(\overline{X}) - E(\overline{Y}) = \mu - \mu = 0\]
\[\text{由于独立}\]
\[D(\overline{X} - \overline{Y}) = D(\overline{X}) + D(\overline{Y}) = \frac{\sigma^2}{5} + \frac{\sigma^2}{9} = \frac{14\sigma^2}{45}\]
\[\therefore \overline{X} - \overline{Y} \sim N(0,\frac{14\sigma^2}{45})\]
\[\therefore a = \pm \sqrt{\frac{45}{14}}\]
\[S_1^2 = \frac{1}{4}\sum_{i=1}^5(X_i - \overline{X})^2\]
\[\frac{4S_1^2}{\sigma^2} \sim \chi^2(4)\]
\[S_2^2 = \frac{1}{8}\sum_{i=1}^9(Y_i - \overline{Y})^2\]
\[\frac{8S_2^2}{\sigma^2} \sim \chi^2(8)\]
\[\therefore \frac{4S_1^2 + 8S_2^2}{\sigma^2} \sim \chi^2(12)\]
\[\therefore \frac{\frac{\overline{X} - \overline{Y}}{\frac{14}{45}\sigma^2}}{{\sqrt{\frac{\frac{4S_1^2 + 8S_2^2}{\sigma^2}}{12}}}} \sim t(12)\]
\[\therefore b = \pm \sqrt{\frac{135}{14}}\]
A3(只算矩估计)
Question
设 \(X_1,X_2,\cdots,X_n\) 是来自下列总体 \(X\) 的简单随机样本，求各总体中未知参数 \(\theta\) 的矩估计量和极大似然估计量，并对所获得的样本值，求参数 \(\theta\) 的矩估计量和极大似然估计量。
(1)
\[f(x;\theta) = \begin{cases} 2^{-\theta}\theta x^{\theta-1}, & 0 < x < 2 \\ 0, & \text{其他} \end{cases} \theta > 0\]
样本值：0.45 0.2 0.5 0.47 0.35 1.63 0.14 0.06 0.89 0.34
(2)
\[f(x;\theta) = \frac{1}{2\theta}e^{-\frac{|x|}{\theta}}, -\infty < x < \infty, \theta > 0\]
样本值： -0.05 -0.47 0.01 -0.03 -0.18 1.65 -0.64 -1.05 0.41 -0.19
(3)
\[f(x;\theta) = \begin{cases} \frac{1}{2-\theta}, & \theta \le x < 2 \\ 0, & \text{其他} \end{cases} \theta < 2\]
样本值：0.95 0.63 1.69 1.97 0.84 1.81 0.53 0.35 1.34 0.82
Answer
\[E(X) = \int_0^2 x \cdot 2^{-\theta}\theta x^{\theta-1} dx = \frac{2\theta}{\theta + 1}\]
\[\overline{X} = 0.503\]
\[E(X) = \overline{X} \Rightarrow \hat{\theta} = \frac{\overline{X}}{2 - \overline{X}} \approx 0.336\]
\[E(X) = \int_{-\infty}^\infty x \cdot \frac{1}{2\theta}e^{-\frac{|x|}{\theta}} dx = 0\]
\[E(X^2) = \int_{-\infty}^\infty x^2 \cdot \frac{1}{2\theta}e^{-\frac{|x|}{\theta}} dx = 2\theta^2 = D(X)\]
\[A_2 = \frac{1}{n} \sum\limits_{i=1}^n X_i^2 = 0.46956\]
\[E(X^2) = A_2 \Rightarrow \hat{\theta} = \sqrt{\frac{1}{2}A_2} \approx 0.4845\]
3.
\[E(X) = \int_{\theta}^2 x \cdot \frac{1}{2-\theta} dx = \frac{\theta + 2}{2}\]
\[\overline{X} = 1.093\]
\[E(X) = \overline{X} \Rightarrow \hat{\theta} = 2\overline{X} - 2 = 0.186\]