题目
已知总体sim N(mu ,(sigma )^2),sim N(mu ,(sigma )^2)是其一组样本,证明:sim N(mu ,(sigma )^2)的估计量sim N(mu ,(sigma )^2)比sim N(mu ,(sigma )^2)有效。
已知总体
,
是其一组样本,证明:
的估计量
比
有效。
题目解答
答案
首先求出两个估计量的方差，
的方差为:
的方差为:
因为
，即
，所以估计量
比
有效.
解析
步骤 1：计算${\hat {\mu }}_{1}$的方差
由于$X_{1}, X_{2}, X_{3}$是独立同分布的随机变量，且$X_{i} \sim N(\mu, \sigma^{2})$，则${\hat {\mu }}_{1} = \dfrac {1}{3}X_{1} + \dfrac {1}{3}X_{2} + \dfrac {1}{3}X_{3}$的方差为：
$$D({\hat {\mu }}_{1}) = D\left(\dfrac {1}{3}X_{1} + \dfrac {1}{3}X_{2} + \dfrac {1}{3}X_{3}\right) = \dfrac {1}{9}D(X_{1}) + \dfrac {1}{9}D(X_{2}) + \dfrac {1}{9}D(X_{3}) = \dfrac {1}{9}\sigma^{2} + \dfrac {1}{9}\sigma^{2} + \dfrac {1}{9}\sigma^{2} = \dfrac {1}{3}\sigma^{2}$$
步骤 2：计算${\hat {\mu }}_{2}$的方差
同理，${\hat {\mu }}_{2} = \dfrac {1}{3}X_{1} + \dfrac {1}{2}X_{2} + \dfrac {1}{6}X_{3}$的方差为：
$$D({\hat {\mu }}_{2}) = D\left(\dfrac {1}{3}X_{1} + \dfrac {1}{2}X_{2} + \dfrac {1}{6}X_{3}\right) = \dfrac {1}{9}D(X_{1}) + \dfrac {1}{4}D(X_{2}) + \dfrac {1}{36}D(X_{3}) = \dfrac {1}{9}\sigma^{2} + \dfrac {1}{4}\sigma^{2} + \dfrac {1}{36}\sigma^{2} = \dfrac {7}{18}\sigma^{2}$$
步骤 3：比较两个估计量的方差
由于$\dfrac {1}{3}\sigma^{2} < \dfrac {7}{18}\sigma^{2}$，即$D({\hat {\mu }}_{1}) < D({\hat {\mu }}_{2})$，所以估计量${\hat {\mu }}_{1}$比${\hat {\mu }}_{2}$有效。
由于$X_{1}, X_{2}, X_{3}$是独立同分布的随机变量，且$X_{i} \sim N(\mu, \sigma^{2})$，则${\hat {\mu }}_{1} = \dfrac {1}{3}X_{1} + \dfrac {1}{3}X_{2} + \dfrac {1}{3}X_{3}$的方差为：
$$D({\hat {\mu }}_{1}) = D\left(\dfrac {1}{3}X_{1} + \dfrac {1}{3}X_{2} + \dfrac {1}{3}X_{3}\right) = \dfrac {1}{9}D(X_{1}) + \dfrac {1}{9}D(X_{2}) + \dfrac {1}{9}D(X_{3}) = \dfrac {1}{9}\sigma^{2} + \dfrac {1}{9}\sigma^{2} + \dfrac {1}{9}\sigma^{2} = \dfrac {1}{3}\sigma^{2}$$
步骤 2：计算${\hat {\mu }}_{2}$的方差
同理，${\hat {\mu }}_{2} = \dfrac {1}{3}X_{1} + \dfrac {1}{2}X_{2} + \dfrac {1}{6}X_{3}$的方差为：
$$D({\hat {\mu }}_{2}) = D\left(\dfrac {1}{3}X_{1} + \dfrac {1}{2}X_{2} + \dfrac {1}{6}X_{3}\right) = \dfrac {1}{9}D(X_{1}) + \dfrac {1}{4}D(X_{2}) + \dfrac {1}{36}D(X_{3}) = \dfrac {1}{9}\sigma^{2} + \dfrac {1}{4}\sigma^{2} + \dfrac {1}{36}\sigma^{2} = \dfrac {7}{18}\sigma^{2}$$
步骤 3：比较两个估计量的方差
由于$\dfrac {1}{3}\sigma^{2} < \dfrac {7}{18}\sigma^{2}$，即$D({\hat {\mu }}_{1}) < D({\hat {\mu }}_{2})$，所以估计量${\hat {\mu }}_{1}$比${\hat {\mu }}_{2}$有效。