题目
6.12(仅数学一)设X1,X2,X3取自存在有限数学期望μ和方差σ^2的总体X,下列统计量中不为总-|||-体X数学期望μ的无偏估计量的是 () .-|||-(A) (hat {mu )}_(1)=dfrac (1)(5)(X)_(1)+dfrac (3)(10)(X)_(2)+dfrac (1)(2)(X)_(3) (B) (mu )_(2)=dfrac (1)(3)(X)_(1)+dfrac (1)(4)(X)_(2)+dfrac (5)(12)(X)_(3) .-|||-(C) (hat {mu )}_(3)=dfrac (1)(4)(X)_(1)+dfrac (1)(4)(X)_(2)+dfrac (1)(4)(X)_(3) . (D) (hat {mu )}_(4)=dfrac (1)(3)(X)_(1)+dfrac (3)(4)(X)_(2)-dfrac (1)(12)(X)_(3) .
题目解答
答案
解析
步骤 1：计算 ${\hat {\mu }}_{1}$ 的期望
根据题设，$E{X}_{i}=\mu (i=1,2,3)$，则
$E{\hat {\mu }}_{1}=\dfrac {1}{5}E{X}_{1}+\dfrac {3}{10}E{X}_{2}+\dfrac {1}{2}E{X}_{3}=\dfrac {1}{5}\mu +\dfrac {3}{10}\mu +\dfrac {1}{2}\mu =\mu$。
步骤 2：计算 ${\hat {\mu }}_{2}$ 的期望
$E{\hat {\mu }}_{2}=\dfrac {1}{3}E{X}_{1}+\dfrac {1}{4}E{X}_{2}+\dfrac {5}{12}E{X}_{3}=\dfrac {1}{3}\mu +\dfrac {1}{4}\mu +\dfrac {5}{12}\mu =\mu$。
步骤 3：计算 ${\hat {\mu }}_{3}$ 的期望
$E{\hat {\mu }}_{3}=\dfrac {1}{4}E{X}_{1}+\dfrac {1}{4}E{X}_{2}+\dfrac {1}{4}E{X}_{3}=\dfrac {1}{4}\mu +\dfrac {1}{4}\mu +\dfrac {1}{4}\mu =\dfrac {3}{4}\mu$。
步骤 4：计算 ${\hat {\mu }}_{4}$ 的期望
$E{\hat {\mu }}_{4}=\dfrac {1}{3}E{X}_{1}+\dfrac {3}{4}E{X}_{2}-\dfrac {1}{12}E{X}_{3}=\dfrac {1}{3}\mu +\dfrac {3}{4}\mu -\dfrac {1}{12}\mu =\mu$。
根据题设，$E{X}_{i}=\mu (i=1,2,3)$，则
$E{\hat {\mu }}_{1}=\dfrac {1}{5}E{X}_{1}+\dfrac {3}{10}E{X}_{2}+\dfrac {1}{2}E{X}_{3}=\dfrac {1}{5}\mu +\dfrac {3}{10}\mu +\dfrac {1}{2}\mu =\mu$。
步骤 2：计算 ${\hat {\mu }}_{2}$ 的期望
$E{\hat {\mu }}_{2}=\dfrac {1}{3}E{X}_{1}+\dfrac {1}{4}E{X}_{2}+\dfrac {5}{12}E{X}_{3}=\dfrac {1}{3}\mu +\dfrac {1}{4}\mu +\dfrac {5}{12}\mu =\mu$。
步骤 3：计算 ${\hat {\mu }}_{3}$ 的期望
$E{\hat {\mu }}_{3}=\dfrac {1}{4}E{X}_{1}+\dfrac {1}{4}E{X}_{2}+\dfrac {1}{4}E{X}_{3}=\dfrac {1}{4}\mu +\dfrac {1}{4}\mu +\dfrac {1}{4}\mu =\dfrac {3}{4}\mu$。
步骤 4：计算 ${\hat {\mu }}_{4}$ 的期望
$E{\hat {\mu }}_{4}=\dfrac {1}{3}E{X}_{1}+\dfrac {3}{4}E{X}_{2}-\dfrac {1}{12}E{X}_{3}=\dfrac {1}{3}\mu +\dfrac {3}{4}\mu -\dfrac {1}{12}\mu =\mu$。