HW12(习题七)
A3(只算极大似然估计)
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
\[L(\theta) = \prod\limits_{i=1}^n f(x_i;\theta) = (2^{-\theta}\theta)^n \prod\limits_{i=1}^n x_i^{\theta - 1}\]
\[\frac{\partial \ln L(\theta)}{\partial \theta} = -n \ln 2 + \frac{n}{\theta} + \sum\limits_{i=1}^n \ln x_i = 0\]
\[\hat{\theta} = \frac{1}{\ln 2 - \frac{1}{n} \sum\limits_{i=1}^n \ln x_i}\]
代入数据得极大似然估计值为 0.577
\[L(\theta) = \prod\limits_{i=1}^n f(x_i;\theta) = (\frac{1}{2\theta})^n e^{-\frac{\sum\limits_{i=1}^n |x_i|}{\theta}}\]
\[\frac{\partial \ln L(\theta)}{\partial \theta} = - \frac{n}{\theta} + \frac{\sum\limits_{i=1}^n |x_i|}{\theta^2} = 0\]
\[\hat{\theta} = \frac{\sum\limits_{i=1}^n |x_i|}{n}\]
代入数据得极大似然估计值为 0.468
\[L(\theta) = \prod\limits_{i=1}^n f(x_i;\theta) = (\frac{1}{2-\theta})^n\]
随着 \(\theta\) 的增大，\(L(\theta)\) 的值增大
\[\therefore \hat{\theta} = min\{X_1,X_2,\cdots,X_n\}\]
代入数据得极大似然估计值为 0.35
A6
Question
设总体 \(X\) 具有如下概率分布律：
| \(X\) |
0 |
1 |
2 |
| \(P\) |
\(\theta\) |
\(\lambda\) |
\(1 - \theta - \lambda\) |
其中 \(0 < \theta < 1\)，\(0 < \lambda < 1\)，且 \(0 < \theta + \lambda < 1\)，从上述总体中抽取样本容量为 9 的简单随机样本，观察值：2，0，2，1，0，0，1，2，1，求参数 \(\theta\) 和 \(\lambda\) 的矩估计值和极大似然估计值。
Answer
- 矩估计
\[E(X) = 2 - 2\theta - \lambda\]
\[E(X^2) = 4 - 4\theta - 2\lambda\]
\[\begin{cases} \theta = 1 - \frac{3}{2} E(X) + \frac{1}{2} E(X^2) \\ \lambda = 2 E(X) - E(X^2) \end{cases}\]
\[\therefore \begin{cases} \hat{\theta} = 1 - \frac{3}{2} \cdot \overline{X} + \frac{1}{2} \overline{X^2} \\ \hat{\lambda} = 2 \cdot \overline{X} - \overline{X^2} \end{cases}\]
代入数据得矩估计值为 \(\hat{\theta} = \frac{1}{3}\)，\(\hat{\lambda} = \frac{1}{3}\)
- 极大似然估计
\[L(\theta,\lambda) = \theta^3 \cdot \lambda^3 \cdot (1-\theta-\lambda)^3\]
\[\frac{\partial L(\theta,\lambda)}{\partial \theta} = 3\lambda^2 \theta^3 (1 - \lambda - \theta)^3 - 3 \lambda^3 \theta^3 (1 - \lambda - \theta)^2 = 0\]
\[\frac{\partial L(\theta,\lambda)}{\partial \lambda} = 3 \lambda^3 \theta^2 (1 - \lambda - \theta)^3 - 3 \lambda^3 \theta^3 (1 - \lambda - \theta)^2 = 0\]
\[\therefore \begin{cases} \hat{\theta} = \frac{1}{3} \\ \hat{\lambda} = \frac{1}{3} \end{cases}\]
A7
Question
设总体 \(X\) 的密度函数为
\[f(x;\theta) = \begin{cases} \frac{x}{\theta}e^{-\frac{x^2}{2\theta}}, & x > 0 \\ 0, & \text{otherwise} \end{cases}\]
其中 \(\theta > 0\) 未知，记 \(\mu_2 = E(X^2)\) ， \(p = P\{X > 1\}\) . \(X_1, \cdots, X_n\) 是来自总体 \(X\) 的简单随机样本，求参数 \(\theta , \mu_2, p\) 的极大似然估计量
Answer
\[L(\theta) = \prod\limits_{i=1}^n(\frac{x_i}{\theta}e^{-\frac{x_i^2}{2\theta}})\]
\[\frac{\partial \ln L(\theta)}{\partial \theta} = -\frac{n}{\theta} + \frac{1}{2\theta^2}\sum\limits_{i=1}^n x_i^2 = 0\]
\[\therefore \hat{\theta} = \frac{1}{2n}\sum\limits_{i=1}^n x_i^2 = \frac{1}{2}A_2\]
\[\mu_2 = E(X^2) = \int_0^\infty x^2 \cdot \frac{x}{\theta}e^{-\frac{x^2}{2\theta}}dx = 2\theta\]
\[\therefore \hat{\mu_2} = 2\hat{\theta} = \frac{1}{n} \sum\limits_{i=1}^n x_i^2 =A_2\]
\[p = P\{X > 1\} = \int_1^\infty \frac{x}{\theta}e^{-\frac{x^2}{2\theta}}dx\]
\[u = \frac{x^2}{2\theta} \Rightarrow dx = \frac{\sqrt{\theta}}{\sqrt{2u}}du\]
\[p = \int_{\frac{1}{2\theta}}^\infty \sqrt{2 \theta u} \cdot \frac{1}{\theta}e^{-u} \cdot \frac{\sqrt{\theta}}{\sqrt{2u}}du = \int_{\frac{1}{2\theta}}^\infty e^{-u}du = e^{-\frac{1}{2\theta}}\]
\[\therefore \hat{p} = e^{-\frac{n}{\sum\limits_{i=1}^n x_i^2}}\]
A8
Question
设总体 \(X\) 的均值为 \(\mu\) ，方差为 \(\sigma^2\) ， \(X_1,X_2,\cdots,X_{10}\) 为来自总体 \(X\) 的简单随机样本，问 \(a\) 取何值时， \(a \sum\limits_{i=1}^{9}(X_{i+1} - X_i)^2\) 是 \(\sigma^2\) 的无偏估计量？
Answer
\[\therefore E(a \sum\limits_{i=1}^{9}(X_{i+1} - X_i)^2) = \sigma^2\]
\[\therefore a \sum\limits_{i=1}^{9}E((X_{i+1} - X_i)^2) = \sigma^2\]
\[\therefore a \sum\limits_{i=1}^{9}(Var(X_{i+1} - X_i) + E^2(X_{i+1} - X_i))= \sigma^2\]
\[\therefore a \sum\limits_{i=1}^{9}(2 \cdot \sigma^2 + 0) = \sigma^2\]
\[\therefore a = \frac{1}{18}\]
A9
Question
设总体 \(X \sim N(\mu , \sigma^2)\) ， \(\mu,\sigma^2\) 未知， \(X_1,X_2,X_3\) 是总体 \(X\) 的简单随机样本，用
\[\hat{\mu_1} = \frac{1}{2}X_1 + \frac{1}{4}X_2 + \frac{1}{4}X_3\]
\[\hat{\mu_2} = 2X_1 - 2X_2 + X_3\]
\[\hat{\mu_3} = \frac{1}{3}X_1 + \frac{1}{3}X_2 + \frac{1}{3}X_3\]
估计参数 \(\mu\) ，它们都是无偏估计量吗？如果是，哪个更有效？
Answer
\[E(\hat{\mu_1}) = \frac{1}{2}E(X_1) + \frac{1}{4}E(X_2) + \frac{1}{4}E(X_3) = \mu\]
\[E(\hat{\mu_2}) = 2E(X_1) - 2E(X_2) + E(X_3) = \mu\]
\[E(\hat{\mu_3}) = \frac{1}{3}E(X_1) + \frac{1}{3}E(X_2) + \frac{1}{3}E(X_3) = \mu\]
所以 \(\hat{\mu_1},\hat{\mu_2},\hat{\mu_3}\) 都是 \(\mu\) 的无偏估计量。
\[D(\hat{\mu_1}) = (\frac{1}{2})^2Var(X_1) + (\frac{1}{4})^2Var(X_2) + (\frac{1}{4})^2Var(X_3) = \frac{3}{8}\sigma^2\]
\[D(\hat{\mu_2}) = (2)^2Var(X_1) + (-2)^2Var(X_2) + Var(X_3) = 9\sigma^2\]
\[D(\hat{\mu_3}) = (\frac{1}{3})^2Var(X_1) + (\frac{1}{3})^2Var(X_2) + (\frac{1}{3})^2Var(X_3) = \frac{1}{3}\sigma^2\]
\[ D(\hat{\mu_3}) < D(\hat{\mu_1}) < D(\hat{\mu_2}) \]
所以 \(\hat{\mu_3}\) 是最有效的估计量。
B3
Question
设总体 \(X\) 具有如下概率分布律：
| \(x\) |
\(a_1\) |
\(a_2\) |
\(a_3\) |
| \(p\) |
\(\theta\) |
\(\frac{1 - \theta}{2}\) |
\(\frac{1 - \theta}{2}\) |
从总体 \(X\) 中取得样本容量为 \(n\) 的样本 \(X_1, X_2, \cdots, X_n\) ，记其中取 \(a_1,a_2,a_3\) 的个数分别为 \(n_1,n_2,n_3\) ，其中 \(n_1 + n_2 + n_3 = n\) ，求参数 \(\theta\) 的矩估计量和极大似然估计量。
Answer
矩估计量：
\[E(X) = a_1p + a_2\frac{1 - \theta}{2} + a_3\frac{1 - \theta}{2} = \overline{X}\]
\[\therefore \hat{\theta} = \frac{2\overline{X} - a_2 - a_3}{2a_1 - a_2 - a_3}\]
极大似然估计量：
\[L(\theta) = \theta^{n_1}(1 - \theta)^{n_2 + n_3}\]
\[\frac{\partial \ln L(\theta)}{\partial \theta} = \frac{n_1}{\theta} - \frac{n_2 + n_3}{1 - \theta} = 0\]
\[\therefore \hat{\theta} = \frac{n_1}{n}\]
B6
Question
设总体 \(X\) 的密度函数为
\[f(x;\theta) = \begin{cases} \frac{1}{\theta}e^{-\frac{x}{\theta}}, & x > 0 \\ 0, & x \leq 0 \end{cases}\]
\(\theta > 0\) 未知, \(X_1, X_2, \cdots, X_n\) 是来自总体 \(X\) 的简单随机样本.
(1)证明：样本均值是 \(\theta\) 的矩估计量，也是极大似然估计量；
(2)在形如 \(c\sum\limits_{i = 1}^n X_i\) 的估计中求 \(c\) ，使其在均方误差准则下最优；
(3)判断由 (2) 得到的估计量是否是 \(\theta\) 的相合估计量.
Answer
(1)矩估计量：
\[E(X) = \int_0^\infty x \frac{1}{\theta}e^{-\frac{x}{\theta}} dx = \theta\]
\[\therefore \hat{\theta} = \overline{X}\]
极大似然估计量：
\[L(\theta) = \prod\limits_{i=1}^n (\frac{1}{\theta}e^{-\frac{x_i}{\theta}})\]
\[\frac{\partial \ln L(\theta)}{\partial \theta} = -\frac{n}{\theta} + \sum\limits_{i=1}^n \frac{x_i}{\theta^2} = 0\]
\[\therefore \hat{\theta} = \frac{1}{n} \sum\limits_{i=1}^n X_i = \overline{X}\]
(2)均方误差准则下最优的估计量：
\[Mse(c \sum\limits_{i=1}^n X_i) = E[((c \sum\limits_{i=1}^n X_i) - \theta)^2] = Var((c \sum\limits_{i=1}^n X_i) - \theta) + E^2((c \sum\limits_{i=1}^n X_i) - \theta)\]
\[Var[(c \sum\limits_{i=1}^n X_i) - \theta] = Var(c \sum\limits_{i=1}^n X_i) = c^2 \sum\limits_{i=1}^n Var(X_i)\]
\[Var(X_i) = E(X_i^2) - [E(X_i)]^2 = \int_0^\infty x^2 \frac{1}{\theta}e^{-\frac{x}{\theta}} dx - \theta^2 = \theta^2\]
\[\therefore Var(c \sum\limits_{i=1}^n X_i) = n \theta^2 c^2\]
\[E^2[(c \sum\limits_{i=1}^n X_i) - \theta] = [(c \sum\limits_{i=1}^n E(X_i)) - \theta]^2 = [(c n \theta) - \theta]^2\]
\[\therefore Mse(c \sum\limits_{i=1}^n X_i) = \theta^2 [(n^2+n)c^2 - 2nc + 1]\]
\[\therefore c= \frac{n}{n^2 + n} = \frac{1}{n+1}\]
(3)
\[\hat{\theta} = \frac{1}{n+1} \sum\limits_{i=1}^n X_i = \frac{n}{n+1} \overline{X}\]
\[P\{|\hat{\theta} - \theta| \ge \varepsilon\} \le \frac{Var(\hat{\theta})}{\varepsilon^2} = \frac{n^2}{(n+1)^2} \frac{Var(\overline{X})}{\varepsilon^2} = \frac{n^2}{(n+1)^2} \frac{1}{n} \frac{Var(X)}{\varepsilon^2} = \frac{n}{(n+1)^2} \frac{\theta^2}{\varepsilon^2}\]
\[\mathop{lim}\limits_{n \to \infty} P\{|\hat{\theta} - \theta| \ge \varepsilon\} \rightarrow 0\]
\[\therefore \hat{\theta} \text{是相合估计量.}\]