Question

设随机变量X 服从几何分布，分布律为p{X＝K}＝P＊（1－P）的k-1次方

Answer 1

下面的计算利用幂级数展开式(通过1/(1-x)=∑{k,0,∞}x^k，x∈(-1,1)容易证明)
：
1/(1-x)²=1+2x+3x²+4x³+…=∑{k,0,∞}(k+1)*x^k，x∈(-1,1)
①
注意到0<1-p<1
E(X)=∑{k,1,∞}k*p*(1-p)^(k-1)
=p*∑{k,0,∞}(k+1)*(1-p)^k
=p*1/[1-(1-p)]²
由①
=1/p
为计算D(X)，可先求出幂级数∑{k,0,∞}(k+1)²*x^k，x∈(-1,1)的和函数
令S(x)=∑{k,0,∞}(k+1)²*x^k，x∈(-1,1)，则
∫{0,x}S(x)dx=∫{0,x}[∑{k,0,∞}(k+1)²*x^k]dx
=∑{k,0,∞}∫{0,x}
[(k+1)²*x^k]dx
=∑{k,0,∞}(k+1)*x^(k+1)
=x*∑{k,0,∞}(k+1)*x^k
=x*1/(1-x)²
由①
S(x)=[
x/(1-x)²]'=(1+x)/(1-x)³，即
∑{k,0,∞}(k+1)²*x^k=(1+x)/(1-x)³，x∈(-1,1)
②
∵E(X²)=∑{k,1,∞}k²*p*(1-p)^(k-1)
=p*∑{k,0,∞}(k+1)²*(1-p)^k
=p*[1+(1-p)]/[1-(1-p)]³
由②
=(2-p)/p²
∴D(X)=
E(X²)-[E(X)]²
=(2-p)/p²-1/p²
=(1-p)/p²