Question

关于数理统计 参数估计 无偏估计量

Answer 1

c=1/2(n-1).

a^2的数学期望E(a^2)=v^2

即E{c∑(Xi+1-Xi)^2}

=c∑E{(Xi+1-Xi)^2}

=c∑E{(Xi+1)^2-2Xi+1Xi+Xi^2}

=c∑E{(Xi+1)^2}-2E{Xi+1Xi}+E{Xi^2}

=c∑E{x^2}-2E{x}E{x}+E{x^2}

=c∑2E{x^2}-2[E{X}]^2

=2c∑E{x^2}-[E{X}]^2

=2c(n-1)(E{x^2}-[E{X}]^2)=2c(n-1)v^2=v^2

所以c=1/2(n-1)