题目
设总体X的概率密度为-|||-(x;theta )= ,0lt xlt theta dfrac {1)(2(1-theta )),theta leqslant xlt 1 0,) 是否为θ^2)
题目解答
答案
解析
步骤 1：求总体X的期望
根据给定的概率密度函数，计算总体X的期望值E(X)。
\[ E(X) = \int_{0}^{\theta} x \cdot \frac{1}{2\theta} dx + \int_{\theta}^{1} x \cdot \frac{1}{2(1-\theta)} dx \]
\[ = \frac{1}{2\theta} \int_{0}^{\theta} x dx + \frac{1}{2(1-\theta)} \int_{\theta}^{1} x dx \]
\[ = \frac{1}{2\theta} \left[ \frac{x^2}{2} \right]_{0}^{\theta} + \frac{1}{2(1-\theta)} \left[ \frac{x^2}{2} \right]_{\theta}^{1} \]
\[ = \frac{1}{2\theta} \cdot \frac{\theta^2}{2} + \frac{1}{2(1-\theta)} \cdot \left( \frac{1}{2} - \frac{\theta^2}{2} \right) \]
\[ = \frac{\theta}{4} + \frac{1}{4(1-\theta)} - \frac{\theta^2}{4(1-\theta)} \]
\[ = \frac{\theta}{4} + \frac{1 - \theta^2}{4(1-\theta)} \]
\[ = \frac{\theta}{4} + \frac{1 + \theta}{4} \]
\[ = \frac{1}{2} \]
步骤 2：求参数θ的矩估计量
由于样本均值 $\overline{X}$ 是总体均值E(X)的无偏估计量，因此有 $\overline{X} = E(X) = \frac{1}{2}$。根据步骤1中的计算结果，可以得到 $\theta$ 的矩估计量 $\hat{\theta}$。
\[ \hat{\theta} = 2\overline{X} - \frac{1}{2} \]
步骤 3：判断4x^2是否为θ^2的无偏估计量
计算 $E(4X^2)$ 并与 $\theta^2$ 比较。
\[ E(4X^2) = 4E(X^2) \]
\[ E(X^2) = \int_{0}^{\theta} x^2 \cdot \frac{1}{2\theta} dx + \int_{\theta}^{1} x^2 \cdot \frac{1}{2(1-\theta)} dx \]
\[ = \frac{1}{2\theta} \int_{0}^{\theta} x^2 dx + \frac{1}{2(1-\theta)} \int_{\theta}^{1} x^2 dx \]
\[ = \frac{1}{2\theta} \left[ \frac{x^3}{3} \right]_{0}^{\theta} + \frac{1}{2(1-\theta)} \left[ \frac{x^3}{3} \right]_{\theta}^{1} \]
\[ = \frac{1}{2\theta} \cdot \frac{\theta^3}{3} + \frac{1}{2(1-\theta)} \cdot \left( \frac{1}{3} - \frac{\theta^3}{3} \right) \]
\[ = \frac{\theta^2}{6} + \frac{1}{6(1-\theta)} - \frac{\theta^3}{6(1-\theta)} \]
\[ = \frac{\theta^2}{6} + \frac{1 - \theta^3}{6(1-\theta)} \]
\[ = \frac{\theta^2}{6} + \frac{1 + \theta + \theta^2}{6} \]
\[ = \frac{1 + \theta + 2\theta^2}{6} \]
\[ E(4X^2) = 4 \cdot \frac{1 + \theta + 2\theta^2}{6} = \frac{2 + 2\theta + 4\theta^2}{3} \]
\[ E(4X^2) \neq \theta^2 \]
因此，4x^2不是θ^2的无偏估计量。
根据给定的概率密度函数，计算总体X的期望值E(X)。
\[ E(X) = \int_{0}^{\theta} x \cdot \frac{1}{2\theta} dx + \int_{\theta}^{1} x \cdot \frac{1}{2(1-\theta)} dx \]
\[ = \frac{1}{2\theta} \int_{0}^{\theta} x dx + \frac{1}{2(1-\theta)} \int_{\theta}^{1} x dx \]
\[ = \frac{1}{2\theta} \left[ \frac{x^2}{2} \right]_{0}^{\theta} + \frac{1}{2(1-\theta)} \left[ \frac{x^2}{2} \right]_{\theta}^{1} \]
\[ = \frac{1}{2\theta} \cdot \frac{\theta^2}{2} + \frac{1}{2(1-\theta)} \cdot \left( \frac{1}{2} - \frac{\theta^2}{2} \right) \]
\[ = \frac{\theta}{4} + \frac{1}{4(1-\theta)} - \frac{\theta^2}{4(1-\theta)} \]
\[ = \frac{\theta}{4} + \frac{1 - \theta^2}{4(1-\theta)} \]
\[ = \frac{\theta}{4} + \frac{1 + \theta}{4} \]
\[ = \frac{1}{2} \]
步骤 2：求参数θ的矩估计量
由于样本均值 $\overline{X}$ 是总体均值E(X)的无偏估计量，因此有 $\overline{X} = E(X) = \frac{1}{2}$。根据步骤1中的计算结果，可以得到 $\theta$ 的矩估计量 $\hat{\theta}$。
\[ \hat{\theta} = 2\overline{X} - \frac{1}{2} \]
步骤 3：判断4x^2是否为θ^2的无偏估计量
计算 $E(4X^2)$ 并与 $\theta^2$ 比较。
\[ E(4X^2) = 4E(X^2) \]
\[ E(X^2) = \int_{0}^{\theta} x^2 \cdot \frac{1}{2\theta} dx + \int_{\theta}^{1} x^2 \cdot \frac{1}{2(1-\theta)} dx \]
\[ = \frac{1}{2\theta} \int_{0}^{\theta} x^2 dx + \frac{1}{2(1-\theta)} \int_{\theta}^{1} x^2 dx \]
\[ = \frac{1}{2\theta} \left[ \frac{x^3}{3} \right]_{0}^{\theta} + \frac{1}{2(1-\theta)} \left[ \frac{x^3}{3} \right]_{\theta}^{1} \]
\[ = \frac{1}{2\theta} \cdot \frac{\theta^3}{3} + \frac{1}{2(1-\theta)} \cdot \left( \frac{1}{3} - \frac{\theta^3}{3} \right) \]
\[ = \frac{\theta^2}{6} + \frac{1}{6(1-\theta)} - \frac{\theta^3}{6(1-\theta)} \]
\[ = \frac{\theta^2}{6} + \frac{1 - \theta^3}{6(1-\theta)} \]
\[ = \frac{\theta^2}{6} + \frac{1 + \theta + \theta^2}{6} \]
\[ = \frac{1 + \theta + 2\theta^2}{6} \]
\[ E(4X^2) = 4 \cdot \frac{1 + \theta + 2\theta^2}{6} = \frac{2 + 2\theta + 4\theta^2}{3} \]
\[ E(4X^2) \neq \theta^2 \]
因此，4x^2不是θ^2的无偏估计量。