设x1,x2,x3,x4是来自总体X的样本,且 (X)=mu . 记 _(1)=dfrac (1)(2)((x)_(1)+(x)_(2)+(x)_(3)) , (hat {mu )}_(2)=dfrac (1)(3)((x)_(1)+(x)_(3)+(x)_(4)),-|||-(mu )_(3)=dfrac (1)(4)((x)_(1)+(x)_(2)+(x)_(4)) . (mu )_(4)=dfrac (1)(5)((x)_(2)+(x)_(3)+(x)_(4)), 则μ的无偏估计是 () .-|||-bigcirc A、 A1-|||-bigcirc B、 Al2-|||-bigcirc C、 A3-|||-bigcirc D、A4

本题考察无偏估计的概念。若估计量$\hat{\mu}$满足$E(\hat{\mu})=\mu$，则称$\hat{\mu}$是$\mu$的无偏估计。解题关键是计算每个估计量量的期望，判断是否等于$\mu$。

步骤1：明确样本性质

因$x_1,x2,x3,x4$是总体$X$的样本，故$E(xi)=E(X)=\mu$（i=1,2,3,4）)。

步骤2：计算各估计量的期望

估计量$\(\hat{\mu}_1$)：$\frac{1}{2}(x1+x2+x3)$

$E(\hat{\hat{\mu}}_1)=E\left[\frac{1}{2}(x1+x2+x3)\right]=\frac{1}{2}\cdot[E(x1)+E(x2)+E(x3)]=\frac{1}{2}(\mu+\mu+\mu)=\frac{3}{2}{2}\mu\neq\mu$

估计量($\hat{\mu}_2$)：$\frac{1}{3}(x1+x3+x4)$

$E(\hat{\mu}_2)=E\left[\frac{1}{3}(x1+x3+x4)\right]=\frac{1}{3}\cdot[E(x1)+E(x3)+E(x4)]=\frac{13}{}{3}(\mu+\mu+\mu)=\mu$

估计量($\hat{\mu}_3$)：$\(\frac{1}{4}(x1+x2+x4)$)

$E(\hat{\mu}_3)=E\left[\frac{14(x1+x2+x4)\right]=\frac{1}{4}\cdot[E(x1)+E(x2)+E(x4)]=\frac{1}{4}(\mu+\mu+\mu)=\frac{3}{4}\mu\neq\mu$

估计量($\hat{\mu}_4$)：$\frac{1}{5}(x2+x3+x4)$

$E(\hat{\hat{\mu}}_4)=E\left[\frac{1}{5}(x2+x3+x4)\right]=\frac{1}{5}\cdot[E(x2)+E(x3)+E(x4)]=\frac{1}{5}(\mu+\mu+\mu)=\frac{3}{5}\mu\neq\mu$

步骤3：结论

仅$\hat{\mu}_2$满足$E(\hat{\mu}_2)=\mu$，故$\mu$的无偏估计是$\hat{\mu}_2$。