kaoyan1basic 概率论与数理统计 第3题
📝 题目
### 【强化篇】第3题（选择题） 3．设总体 $\displaystyle X \sim f(x ; \theta)=\left\{\begin{array}{ll}\frac{x}{\theta} \mathrm{e}^{\frac{x^{2}}{20}}, & x>0, \\ 0, & x \leqslant 0,\end{array} \theta>0\right.$ 未知，$X_{1}, X_{2}, \cdots, X_{n}$ 为来自总体 $X$ 的简单随机样本，记 $\hat{\theta}_{M}$ 与 $\hat{\theta}_{L}$ 分别是 $\theta$ 的矩估计量和最大似然估计量，则（）。 （A）$\displaystyle \hat{\theta}_{M}=\frac{2}{\pi} \bar{X}^{2}, E\left(\hat{\theta}_{M}\right)=\theta$ （B）$\displaystyle \hat{\theta}_{M}=\frac{1}{\pi} \bar{X}^{2}, E\left(\hat{\theta}_{M}\right)=\theta$ （C）$\displaystyle \hat{\theta}_{L}=\frac{1}{n} \sum_{i=1}^{n} X_{i}^{2}, E\left(\hat{\theta}_{L}\right)=\theta$ （D）$\displaystyle \hat{\theta}_{L}=\frac{1}{2 n} \sum_{i=1}^{n} X_{i}^{2}, E\left(\hat{\theta}_{L}\right)=\theta$
💡 答案解析
**答案**：D **解析**：步骤1：计算总体矩。$\displaystyle E(X)=\int_0^{+\infty} x \cdot \frac{x}{\theta} e^{-\frac{x^2}{2\theta}} dx = \int_0^{+\infty} \frac{x^2}{\theta} e^{-\frac{x^2}{2\theta}} dx$，令$\displaystyle t=\frac{x^2}{2\theta}$，则$x=\sqrt{2\theta t}$，$\displaystyle dx=\frac{\sqrt{2\theta}}{2\sqrt{t}}dt$，$\displaystyle E(X)=\int_0^{+\infty} \frac{2\theta t}{\theta} e^{-t} \cdot \frac{\sqrt{2\theta}}{2\sqrt{t}} dt = \sqrt{2\theta} \int_0^{+\infty} t^{\frac{1}{2}} e^{-t} dt = \sqrt{2\theta} \Gamma(\frac{3}{2}) = \sqrt{2\theta} \cdot \frac{\sqrt{\pi}}{2} = \sqrt{\frac{\pi\theta}{2}}$。由矩估计法，$\displaystyle \bar{X} = \sqrt{\frac{\pi\hat{\theta}_M}{2}}$，解得$\displaystyle \hat{\theta}_M = \frac{2}{\pi}\bar{X}^2$。$\displaystyle E(\hat{\theta}_M) = \frac{2}{\pi} E(\bar{X}^2) = \frac{2}{\pi} [D(\bar{X}) + (E(\bar{X}))^2] = \frac{2}{\pi} [\frac{D(X)}{n} + \frac{\pi\theta}{2}] \neq \theta$，故A、B错误。 步骤2：求最大似然估计。似然函数$\displaystyle L(\theta)=\prod_{i=1}^n \frac{x_i}{\theta} e^{-\frac{x_i^2}{2\theta}} = \frac{\prod x_i}{\theta^n} e^{-\frac{\sum x_i^2}{2\theta}}$，取对数得$\displaystyle \ln L = \sum \ln x_i - n\ln\theta - \frac{\sum x_i^2}{2\theta}$，对$\theta$求导并令其为0：$\displaystyle \frac{d\ln L}{d\theta} = -\frac{n}{\theta} + \frac{\sum x_i^2}{2\theta^2}=0$，解得$\displaystyle \hat{\theta}_L = \frac{1}{2n}\sum_{i=1}^n X_i^2$。$\displaystyle E(\hat{\theta}_L) = \frac{1}{2n} \sum E(X_i^2) = \frac{1}{2} E(X^2)$。计算$\displaystyle E(X^2)=\int_0^{+\infty} x^2 \cdot \frac{x}{\theta} e^{-\frac{x^2}{2\theta}} dx = \int_0^{+\infty} \frac{x^3}{\theta} e^{-\frac{x^2}{2\theta}} dx$，令$\displaystyle t=\frac{x^2}{2\theta}$，则$x^2=2\theta t$，$x dx = \theta dt$，$\displaystyle E(X^2)=\int_0^{+\infty} \frac{2\theta t}{\theta} e^{-t} \cdot \theta dt = 2\theta \int_0^{+\infty} t e^{-t} dt = 2\theta \Gamma(2)=2\theta$，故$\displaystyle E(\hat{\theta}_L)=\frac{1}{2} \cdot 2\theta = \theta$。 **难度**：★★★☆☆