设总体X～N(μ，1)，-∞﹤μ﹤∞，(x1，x2，x3)为其样本．试证下述三个估计量：
(1)μ̂1=(1/5)x1+(3/10)x2+(1/2)x3;
(2)μ̂2=(1/3)x1+(1/4)x2+(5/12)x3;
(3)μ̂3=(1/3)x1+(1/6)x2+(1/2)x3;
都是μ的无偏估计，并求出每一估计量的方差，问哪一个方差最小?
证明： E(μ ̂1=(1/5)E(x1)+(3/10)E(x2)+(1/2)E(x3)=(1/5)μ+(3/10)μ+(1/2)μ=μ E(μ ̂2=(1/3)E(x1)+(1/4)E(x2)+(5/12)E(x3) =(1/3)μ+(1/4)μ+(5/12)μ=μ E(μ ̂3=(1/3)E(x1)+(1/6)E(x2)+(1/2)E(x3)=(1/3)μ+(1/6)μ+(1/2)μ=μ ∴μ ̂1，μ ̂2，μ ̂3都是μ的无偏估计 D(μ ̂1)=(1/25)D(x1)+(9/100)D(x2)+(1/4)D(x3)=1/25+9/100+1/4=19/50 D(μ ̂2)=(1/9)D(x1)+(1/16)D(x2)+(25/144)D(x3)=1/9+1/16+25/144=25/72 D(μ3)=(1/9)D(x1)+(1/36)D(x2)+(1/4)D(x3)=1/9+1/36+1/16=7/18 故μ ̂2的方差最小．
