证明：Eξ=p+2qp+3q²p+…+k[q^(k-1)]p+…=p(1+2q+3q²+…)
设S=1+2q+3q²+…+nq^(n-1),
则由qS=q+2q²+…+(n-1)q^(n-1)+nq^n
两式相减,得(1-q)S=1+q+q²+…+q^(n-1)-nq^n
故S=(1-q^n)/(1-q)²-nq^n/(1-q),则
S=limS=1/(1-q)²=1/p²,即Eξ=1/p
E(ξ²)=p+2²qp+3²q²p+…+k²[q^(k-1)]p+…
=p[1+2²q+3²q²+…+k²q^(k-1)+…]
对于上式括号中的求和,利用导数对q求导,即
k²q^(k-1)=(kq^k)`,有
1+2²q+3²q²+…+k²q^(k-1)+…
=(q+2q²+3q³+…+kq^k+…)`(与求Eξ同样方法,得到)
=[q/(1-q)²]`
=[(1-q)²-2q(1-q)(-1)]/(1-q)^4
=(1+q)/(1-q)³
=(2-p)/p³
因此E(ξ²)=p[(2-p)/p³]=(2-p)/p²
则Dξ=E(ξ²)-(Eξ)²=(2-p)/p²-(1/p)²=(1-p)/p²