x1x2x3x4x5是正态总体样本,总体均值为μ（μ未知）证T=3/13(x1+x2+x3)+2/13(x4+x4)是μ的无偏估计量

Question

x1x2x3x4x5是正态总体样本,总体均值为μ（μ未知）证T=3/13(x1+x2+x3)+2/13(x4+x4)是μ的无偏估计量

Answer 1

ET=E(3/13(x1+x2+x3)+2/13(x4+x4))
=(3/13)E(x1+x2+x3)+(2/13)E(x4+x4)
=(3/13)[E(x1)+E(x2)+E(x3)]+(2/13)[E(x4)+E(x4)]
=(3/13)*3μ+(2/13)*2μ
=μ
故T是μ的无偏估计量