题目
4.设总体X的密度函数为 f(x)= ^3)(theta -x),0lt xlt theta 0, . X1,-|||-X2,···,Nn是取自X的简单随机样本.-|||-(1)求θ的矩估计量θ;-|||-(2)求θ的方差D(θ);-|||-(3)讨论θ的无偏性和一致性(相合性).
题目解答
答案
解析
步骤 1：求总体X的期望E(X)
根据密度函数，计算总体X的期望E(X)。
\[ E(X) = \int_{0}^{\theta} x \cdot \frac{6x}{\theta^3} (\theta - x) dx \]
\[ = \frac{6}{\theta^3} \int_{0}^{\theta} x^2 (\theta - x) dx \]
\[ = \frac{6}{\theta^3} \left[ \int_{0}^{\theta} x^2 \theta dx - \int_{0}^{\theta} x^3 dx \right] \]
\[ = \frac{6}{\theta^3} \left[ \theta \int_{0}^{\theta} x^2 dx - \int_{0}^{\theta} x^3 dx \right] \]
\[ = \frac{6}{\theta^3} \left[ \theta \cdot \frac{\theta^3}{3} - \frac{\theta^4}{4} \right] \]
\[ = \frac{6}{\theta^3} \cdot \frac{\theta^4}{12} \]
\[ = \frac{\theta}{2} \]
步骤 2：求θ的矩估计量
根据矩估计法，令总体X的期望E(X)等于样本均值$\overline{X}$，即
\[ E(X) = \frac{\theta}{2} = \overline{X} \]
\[ \theta = 2\overline{X} \]
所以，θ的矩估计量为${\hat {\theta }}_{圆}=2\overline {X}$。
步骤 3：求θ的方差D(θ)
根据矩估计量${\hat {\theta }}_{圆}=2\overline {X}$，计算其方差D(θ)。
\[ D(\theta) = D(2\overline{X}) = 4D(\overline{X}) \]
\[ D(\overline{X}) = \frac{D(X)}{n} \]
其中，D(X)为总体X的方差，计算如下：
\[ D(X) = E(X^2) - [E(X)]^2 \]
\[ E(X^2) = \int_{0}^{\theta} x^2 \cdot \frac{6x}{\theta^3} (\theta - x) dx \]
\[ = \frac{6}{\theta^3} \int_{0}^{\theta} x^3 (\theta - x) dx \]
\[ = \frac{6}{\theta^3} \left[ \int_{0}^{\theta} x^3 \theta dx - \int_{0}^{\theta} x^4 dx \right] \]
\[ = \frac{6}{\theta^3} \left[ \theta \int_{0}^{\theta} x^3 dx - \int_{0}^{\theta} x^4 dx \right] \]
\[ = \frac{6}{\theta^3} \left[ \theta \cdot \frac{\theta^4}{4} - \frac{\theta^5}{5} \right] \]
\[ = \frac{6}{\theta^3} \cdot \frac{\theta^5}{20} \]
\[ = \frac{3\theta^2}{10} \]
\[ D(X) = \frac{3\theta^2}{10} - \left( \frac{\theta}{2} \right)^2 = \frac{3\theta^2}{10} - \frac{\theta^2}{4} = \frac{\theta^2}{20} \]
\[ D(\overline{X}) = \frac{\theta^2}{20n} \]
\[ D(\theta) = 4D(\overline{X}) = \frac{4\theta^2}{20n} = \frac{\theta^2}{5n} \]
步骤 4：讨论θ的无偏性和一致性
无偏性：根据矩估计量${\hat {\theta }}_{圆}=2\overline {X}$，计算其期望E(θ)。
\[ E(\theta) = E(2\overline{X}) = 2E(\overline{X}) = 2E(X) = 2 \cdot \frac{\theta}{2} = \theta \]
所以，θ的矩估计量是无偏的。
一致性：根据矩估计量${\hat {\theta }}_{圆}=2\overline {X}$，计算其方差D(θ)。
\[ D(\theta) = \frac{\theta^2}{5n} \]
当样本容量n趋于无穷大时，D(θ)趋于0，所以θ的矩估计量是一致的。
根据密度函数，计算总体X的期望E(X)。
\[ E(X) = \int_{0}^{\theta} x \cdot \frac{6x}{\theta^3} (\theta - x) dx \]
\[ = \frac{6}{\theta^3} \int_{0}^{\theta} x^2 (\theta - x) dx \]
\[ = \frac{6}{\theta^3} \left[ \int_{0}^{\theta} x^2 \theta dx - \int_{0}^{\theta} x^3 dx \right] \]
\[ = \frac{6}{\theta^3} \left[ \theta \int_{0}^{\theta} x^2 dx - \int_{0}^{\theta} x^3 dx \right] \]
\[ = \frac{6}{\theta^3} \left[ \theta \cdot \frac{\theta^3}{3} - \frac{\theta^4}{4} \right] \]
\[ = \frac{6}{\theta^3} \cdot \frac{\theta^4}{12} \]
\[ = \frac{\theta}{2} \]
步骤 2：求θ的矩估计量
根据矩估计法，令总体X的期望E(X)等于样本均值$\overline{X}$，即
\[ E(X) = \frac{\theta}{2} = \overline{X} \]
\[ \theta = 2\overline{X} \]
所以，θ的矩估计量为${\hat {\theta }}_{圆}=2\overline {X}$。
步骤 3：求θ的方差D(θ)
根据矩估计量${\hat {\theta }}_{圆}=2\overline {X}$，计算其方差D(θ)。
\[ D(\theta) = D(2\overline{X}) = 4D(\overline{X}) \]
\[ D(\overline{X}) = \frac{D(X)}{n} \]
其中，D(X)为总体X的方差，计算如下：
\[ D(X) = E(X^2) - [E(X)]^2 \]
\[ E(X^2) = \int_{0}^{\theta} x^2 \cdot \frac{6x}{\theta^3} (\theta - x) dx \]
\[ = \frac{6}{\theta^3} \int_{0}^{\theta} x^3 (\theta - x) dx \]
\[ = \frac{6}{\theta^3} \left[ \int_{0}^{\theta} x^3 \theta dx - \int_{0}^{\theta} x^4 dx \right] \]
\[ = \frac{6}{\theta^3} \left[ \theta \int_{0}^{\theta} x^3 dx - \int_{0}^{\theta} x^4 dx \right] \]
\[ = \frac{6}{\theta^3} \left[ \theta \cdot \frac{\theta^4}{4} - \frac{\theta^5}{5} \right] \]
\[ = \frac{6}{\theta^3} \cdot \frac{\theta^5}{20} \]
\[ = \frac{3\theta^2}{10} \]
\[ D(X) = \frac{3\theta^2}{10} - \left( \frac{\theta}{2} \right)^2 = \frac{3\theta^2}{10} - \frac{\theta^2}{4} = \frac{\theta^2}{20} \]
\[ D(\overline{X}) = \frac{\theta^2}{20n} \]
\[ D(\theta) = 4D(\overline{X}) = \frac{4\theta^2}{20n} = \frac{\theta^2}{5n} \]
步骤 4：讨论θ的无偏性和一致性
无偏性：根据矩估计量${\hat {\theta }}_{圆}=2\overline {X}$，计算其期望E(θ)。
\[ E(\theta) = E(2\overline{X}) = 2E(\overline{X}) = 2E(X) = 2 \cdot \frac{\theta}{2} = \theta \]
所以，θ的矩估计量是无偏的。
一致性：根据矩估计量${\hat {\theta }}_{圆}=2\overline {X}$，计算其方差D(θ)。
\[ D(\theta) = \frac{\theta^2}{5n} \]
当样本容量n趋于无穷大时，D(θ)趋于0，所以θ的矩估计量是一致的。