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) ：\x0d\x0a1/(1-x)²=1+2x+3x²+4x³+?=∑{k,0,∞}(k+1)*x^k，x∈(-1,1) ①\x0d\x0a \x0d\x0a注意到0<1-p<1\x0d\x0aE(X)=∑{k,1,∞}k*p*(1-p)^(k-1)\x0d\x0a =p*∑{k,0,∞}(k+1)*(1-p)^k\x0d\x0a =p*1/[1-(1-p)]² 由①\x0d\x0a =1/p\x0d\x0a \x0d\x0a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\x0d\x0a令S(x)=∑{k,0,∞}(k+1)²*x^k，x∈(-1,1)，则\x0d\x0a∫{0,x}S(x)dx=∫{0,x}[∑{k,0,∞}(k+1)²*x^k]dx\x0d\x0a =∑{k,0,∞}∫{0,x} [(k+1)²*x^k]dx\x0d\x0a =∑{k,0,∞}(k+1)*x^(k+1)\x0d\x0a =x*∑{k,0,∞}(k+1)*x^k\x0d\x0a =x*1/(1-x)² 由①\x0d\x0aS(x)=[ x/(1-x)²]'=(1+x)/(1-x)³，即\x0d\x0a∑{k,0,∞}(k+1)²*x^k=(1+x)/(1-x)³，x∈(-1,1) ②\x0d\x0a \x0d\x0a∵E(X²)=∑{k,1,∞}k²*p*(1-p)^(k-1)\x0d\x0a =p*∑{k,0,∞}(k+1)²*(1-p)^k\x0d\x0a =p*[1+(1-p)]/[1-(1-p)]³ 由②\x0d\x0a =(2-p)/p²\x0d\x0a \x0d\x0a∴D(X)= E(X²)-[E(X)]²\x0d\x0a =(2-p)/p²-1/p²\x0d\x0a =(1-p)/p²