4.设总体密度函数如下,x1,···,xn是样本,试求未知参数的矩估计.-|||-(1) (x;theta )=dfrac (2)({theta )^2}(theta -x) lt xlt theta ,θ>0;-|||-(2) (x;theta )=(theta +1)(x)^theta , lt xlt 1 ,θ>0;-|||-(3) (x;theta )=sqrt (theta )(x)^sqrt (theta -1) lt xlt 1 ,θ>0;-|||-(4) (x;theta ,mu )=dfrac (1)(theta )(e)^-dfrac (x-mu {theta )} gt mu ,θ>0.
题目解答
答案
解析
矩估计法的核心思路是利用样本矩（如样本均值、样本方差）来估计总体矩，进而解出参数的估计值。本题四个小题均需计算总体均值，通过建立样本均值与总体均值的方程求解参数。对于含多个参数的情况（如第4题），需联立样本均值和样本方差方程求解。
关键步骤：
- 计算总体均值 $E(X)$；
- 建立方程 $\overline{X} = E(X)$（或联立方程）；
- 解方程得到参数的矩估计表达式。
(1) $P(x;\theta )=\dfrac{2}{\theta^{2}}(\theta -x)$，$0 计算总体均值
$E(X) = \int_{0}^{\theta} x \cdot \dfrac{2}{\theta^{2}} (\theta - x) \, dx = \dfrac{2}{\theta^{2}} \left( \dfrac{\theta^{3}}{2} - \dfrac{\theta^{3}}{3} \right) = \dfrac{\theta}{3}$
建立方程并求解 令 $\overline{X} = \dfrac{\theta}{3}$，解得 $\hat{\theta} = 3\overline{X}$。 (2) $p(x;\theta )=(\theta +1)x^{\theta}$，$0 计算总体均值
$E(X) = \int_{0}^{1} x \cdot (\theta +1)x^{\theta} \, dx = (\theta +1) \int_{0}^{1} x^{\theta +1} \, dx = \dfrac{\theta +1}{\theta +2}$
建立方程并求解 令 $\overline{X} = \dfrac{\theta +1}{\theta +2}$，解得 $\hat{\theta} = \dfrac{1 - 2\overline{X}}{\overline{X} - 1}$。 (3) $P(x;\theta )=\sqrt{\theta}x^{\sqrt{\theta} -1}$，$0 计算总体均值
$E(X) = \int_{0}^{1} x \cdot \sqrt{\theta}x^{\sqrt{\theta} -1} \, dx = \sqrt{\theta} \int_{0}^{1} x^{\sqrt{\theta}} \, dx = \dfrac{\sqrt{\theta}}{\sqrt{\theta} + 1}$
建立方程并求解 令 $\overline{X} = \dfrac{\sqrt{\theta}}{\sqrt{\theta} + 1}$，解得 $\hat{\theta} = \left( \dfrac{\overline{X}}{1 - \overline{X}} \right)^{2}$。 (4) $P(x;\theta ,\mu )=\dfrac{1}{\theta}e^{-\dfrac{x-\mu}{\theta}}$，$x>\mu$ 计算总体均值与方差
$E(X) = \mu + \theta, \quad \text{Var}(X) = \theta^{2}$
联立方程并求解 令 $\overline{X} = \mu + \theta$，$\text{Var}(X) = \theta^{2}$，解得 $\hat{\theta} = s$（样本标准差），$\hat{\mu} = \overline{X} - s$。