Question

满足几何分布的随机变量的方差怎么求?

Answer 1

Eξ=1/p,Dξ=(1-p)/p^2

Dξ=E(ξ^2)-(Eξ)^2

E(ξ^2)=p+2^2*qp+3^2*q^2*p+……+k^2*q^(k-1)*p+……

=p(1+2^2*q+3^2*q^2+……+k^2*q^(k-1)+……)

对于上式括号中的式子,利用导数,关于q求导：k^2*q^(k-1)=(k*q^k)',并用倍差法求和,有

1+2^2*q+3^2*q^2+……+k^2*q^(k-1)+……

=(q+2*q^2+3*q^3+……+k*q^k+……)'

=[q/(1-q)^2]'

=[(1-q^2)+2(1-q)q]/(1-q)^4

=(1-q^2)/(1-q)^4

=(1+q)/(1-q)^3

=(2-p)/p^3

因此E(ξ^2)=p[(2-p)/p^3]=(2-p)/p^2

则Dξ=E(ξ^2)-(Eξ)^2=（2-p)/p^2-(1/p)^2=(1-p)/p^2