题目
设总体X的概率密度-|||-(x;theta )= ^3)(theta -x), 0lt xlt theta 0,=-|||-__ -

题目解答
答案

解析
步骤 1:计算总体X的期望值
根据给定的概率密度函数,计算总体X的期望值E(X)。期望值E(X)的计算公式为:
$$E(X) = \int_{-\infty}^{\infty} x f(x) dx$$
对于给定的概率密度函数,期望值E(X)的计算公式为:
$$E(X) = \int_{0}^{\theta} x \cdot \frac{6x}{\theta^3}(\theta - x) dx$$
步骤 2:计算积分
计算上述积分,得到:
$$E(X) = \int_{0}^{\theta} \frac{6x^2}{\theta^3}(\theta - x) dx$$
$$E(X) = \frac{6}{\theta^3} \int_{0}^{\theta} x^2(\theta - x) dx$$
$$E(X) = \frac{6}{\theta^3} \int_{0}^{\theta} (x^2\theta - x^3) dx$$
$$E(X) = \frac{6}{\theta^3} \left[ \frac{x^3\theta}{3} - \frac{x^4}{4} \right]_{0}^{\theta}$$
$$E(X) = \frac{6}{\theta^3} \left( \frac{\theta^4}{3} - \frac{\theta^4}{4} \right)$$
$$E(X) = \frac{6}{\theta^3} \cdot \frac{\theta^4}{12}$$
$$E(X) = \frac{\theta}{2}$$
步骤 3:求解矩估计量
令总体X的期望值E(X)等于样本均值$\overline{X}$,即:
$$\frac{\theta}{2} = \overline{X}$$
解得:
$$\hat{\theta} = 2\overline{X}$$
根据给定的概率密度函数,计算总体X的期望值E(X)。期望值E(X)的计算公式为:
$$E(X) = \int_{-\infty}^{\infty} x f(x) dx$$
对于给定的概率密度函数,期望值E(X)的计算公式为:
$$E(X) = \int_{0}^{\theta} x \cdot \frac{6x}{\theta^3}(\theta - x) dx$$
步骤 2:计算积分
计算上述积分,得到:
$$E(X) = \int_{0}^{\theta} \frac{6x^2}{\theta^3}(\theta - x) dx$$
$$E(X) = \frac{6}{\theta^3} \int_{0}^{\theta} x^2(\theta - x) dx$$
$$E(X) = \frac{6}{\theta^3} \int_{0}^{\theta} (x^2\theta - x^3) dx$$
$$E(X) = \frac{6}{\theta^3} \left[ \frac{x^3\theta}{3} - \frac{x^4}{4} \right]_{0}^{\theta}$$
$$E(X) = \frac{6}{\theta^3} \left( \frac{\theta^4}{3} - \frac{\theta^4}{4} \right)$$
$$E(X) = \frac{6}{\theta^3} \cdot \frac{\theta^4}{12}$$
$$E(X) = \frac{\theta}{2}$$
步骤 3:求解矩估计量
令总体X的期望值E(X)等于样本均值$\overline{X}$,即:
$$\frac{\theta}{2} = \overline{X}$$
解得:
$$\hat{\theta} = 2\overline{X}$$