2022年考研数学三第2题

**答案**: 见解析

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**解析**:

由已知 $E(X)=\theta=\displaystyle\frac{1}{\lambda_{1}} \Rightarrow \lambda_{1}=\displaystyle\frac{1}{\theta}, E(Y)=2 \theta=\displaystyle\frac{1}{\lambda_{2}} \Rightarrow \lambda_{2}=\displaystyle\frac{1}{2 \theta}$ ， 所以总体 $X \sim E\left(\displaystyle\frac{1}{\theta}\right), Y \sim E\left(\displaystyle\frac{1}{2 \theta}\right)$ ，从而可得

$$ f_{X}(x)=\left\{\begin{array}{ll} \frac{1}{\theta} \mathrm{e}^{-\frac{x}{\theta}}, & x\gt 0, \\ 0, & x \leq 0 . \end{array} \quad f_{Y}(y)= $\begin{cases}\frac{1}{2 \theta} \mathrm{e}^{-\frac{y}{2 \theta}}, & y\gt 0, \\ 0, & y \leq 0 .\end{cases}\right. $$

设 $x_{1}, x_{2}, \cdots, x_{n}, y_{1}, y_{2}, \cdots, y_{m}$ 为样本 $X_{1}, X_{2}, \cdots, X_{n}, Y_{1}, Y_{2}, \cdots, Y_{m}$ 的观测值，且样本相互独立，则似然函数为

$$ L(\theta)= $\begin{cases}\frac{1}{2^{m}} \frac{1}{\theta^{n+m}} \mathrm{e}^{-\frac{2 \sum_{i=1}^{n} x_{i}+\sum_{j=1}^{m} y_{j}}{2 \theta}}, & x_{i}, y_{j}\gt 0(i=1,2, \cdots, n ; j=1,2, \cdots, m), \\ 0, & \text { 其它. }\end{cases} $$

当 $x_{1}, x_{2}, \cdots, x_{n}, y_{1}, y_{2}, \cdots, y_{m}\gt 0$ 时，似然函数两边取对数

$$ \ln L(\theta)=-m \ln 2-(n+m) \ln \theta-\frac{2 \sum_{i=1}^{n} x_{i}+\sum_{j=1}^{m} y_{j}}{2 \theta}, $$

令 $\displaystyle\frac{\mathrm{d} \ln L(\theta)}{\mathrm{d} \theta}=-\displaystyle\frac{n+m}{\theta}+\displaystyle\frac{2 \displaystyle\sum_{i=1}^{n} x_{i}+\displaystyle\sum_{j=1}^{m} y_{j}}{2 \theta^{2}}=0$ ，解得 $\theta=\displaystyle\frac{2 \displaystyle\sum_{i=1}^{n} x_{i}+\displaystyle\sum_{j=1}^{m} y_{j}}{2(n+m)}$ ，

故 $\theta$ 的最大似然估计量为 $\hat{\theta}=\displaystyle\frac{2 \displaystyle\sum_{i=1}^{n} X_{i}+\displaystyle\sum_{j=1}^{m} Y_{j}}{2(n+m)}$ ． 由 $X \sim E\left(\displaystyle\frac{1}{\theta}\right), Y \sim E\left(\displaystyle\frac{1}{2 \theta}\right)$ ，则 $D(X)=\theta^{2}, D(Y)=4 \theta^{2}$ ， 则 $D(\hat{\theta})=\displaystyle\frac{1}{4(n+m)^{2}} D\left(2 \displaystyle\sum_{i=1}^{n} X_{i}+\displaystyle\sum_{j=1}^{m} Y_{j}\right)=\displaystyle\frac{1}{4(n+m)^{2}}\left(4 n \cdot \theta^{2}+m \cdot 4 \theta^{2}\right)=\displaystyle\frac{\theta^{2}}{n+m}$ ．