题目
点估计7.1A-12 设(X_(1),X_(2),...,X_(n))是取自总体X的一个样本，X的密度函数为f(x)=}(theta+1)x^theta,0<1,0,其他.其中theta>0未知，求theta的矩估计和最大似然估计.
点估计
7.1A-12 设$(X_{1},X_{2},\cdots,X_{n})$是取自总体X的一个样本，X的密度函数为
$f(x)=\begin{cases}(\theta+1)x^{\theta},0<1,\\0,其他.\end{cases}$
其中$\theta>0$未知，求$\theta$的矩估计和最大似然估计.
题目解答
答案
**矩估计：**
计算总体期望 $E(X) = \frac{\theta + 1}{\theta + 2}$，令其等于样本均值 $\overline{X}$，解得
\[
\theta = \frac{1 - 2\overline{X}}{\overline{X} - 1}.
\]
**最大似然估计：**
似然函数 $L(\theta) = (\theta + 1)^n \prod_{i=1}^n X_i^\theta$，取对数并求导得
\[
\frac{d \ln L(\theta)}{d \theta} = \frac{n}{\theta + 1} + \sum_{i=1}^n \ln X_i = 0,
\]
解得
\[
\theta = -1 - \frac{n}{\sum_{i=1}^n \ln X_i} = -1 - \frac{1}{\overline{\ln X}}.
\]
**答案：**
矩估计量：$\boxed{\frac{1 - 2\overline{X}}{\overline{X} - 1}}$
最大似然估计量：$\boxed{-1 - \frac{n}{\sum_{i=1}^n \ln X_i}}$ 或 $\boxed{-1 - \frac{1}{\overline{\ln X}}}$
解析
步骤 1：计算总体期望
总体X的密度函数为$f(x)=\begin{cases}(\theta+1)x^{\theta},00$未知。计算总体期望$E(X)$，即
\[ E(X) = \int_0^1 x f(x) dx = \int_0^1 x (\theta+1)x^{\theta} dx = (\theta+1) \int_0^1 x^{\theta+1} dx. \]
步骤 2：计算积分
计算积分$\int_0^1 x^{\theta+1} dx$，得到
\[ \int_0^1 x^{\theta+1} dx = \left. \frac{x^{\theta+2}}{\theta+2} \right|_0^1 = \frac{1}{\theta+2}. \]
步骤 3：求解矩估计
将步骤2的结果代入步骤1的期望公式中，得到
\[ E(X) = (\theta+1) \frac{1}{\theta+2} = \frac{\theta+1}{\theta+2}. \]
令总体期望$E(X)$等于样本均值$\overline{X}$，即
\[ \frac{\theta+1}{\theta+2} = \overline{X}, \]
解得
\[ \theta = \frac{1 - 2\overline{X}}{\overline{X} - 1}. \]
步骤 4：计算最大似然估计
似然函数$L(\theta) = (\theta + 1)^n \prod_{i=1}^n X_i^\theta$，取对数并求导得
\[ \ln L(\theta) = n \ln (\theta + 1) + \theta \sum_{i=1}^n \ln X_i, \]
\[ \frac{d \ln L(\theta)}{d \theta} = \frac{n}{\theta + 1} + \sum_{i=1}^n \ln X_i = 0, \]
解得
\[ \theta = -1 - \frac{n}{\sum_{i=1}^n \ln X_i} = -1 - \frac{1}{\overline{\ln X}}. \]
总体X的密度函数为$f(x)=\begin{cases}(\theta+1)x^{\theta},0
\[ E(X) = \int_0^1 x f(x) dx = \int_0^1 x (\theta+1)x^{\theta} dx = (\theta+1) \int_0^1 x^{\theta+1} dx. \]
步骤 2：计算积分
计算积分$\int_0^1 x^{\theta+1} dx$，得到
\[ \int_0^1 x^{\theta+1} dx = \left. \frac{x^{\theta+2}}{\theta+2} \right|_0^1 = \frac{1}{\theta+2}. \]
步骤 3：求解矩估计
将步骤2的结果代入步骤1的期望公式中，得到
\[ E(X) = (\theta+1) \frac{1}{\theta+2} = \frac{\theta+1}{\theta+2}. \]
令总体期望$E(X)$等于样本均值$\overline{X}$，即
\[ \frac{\theta+1}{\theta+2} = \overline{X}, \]
解得
\[ \theta = \frac{1 - 2\overline{X}}{\overline{X} - 1}. \]
步骤 4：计算最大似然估计
似然函数$L(\theta) = (\theta + 1)^n \prod_{i=1}^n X_i^\theta$，取对数并求导得
\[ \ln L(\theta) = n \ln (\theta + 1) + \theta \sum_{i=1}^n \ln X_i, \]
\[ \frac{d \ln L(\theta)}{d \theta} = \frac{n}{\theta + 1} + \sum_{i=1}^n \ln X_i = 0, \]
解得
\[ \theta = -1 - \frac{n}{\sum_{i=1}^n \ln X_i} = -1 - \frac{1}{\overline{\ln X}}. \]