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二项分布方差公式推导若B(n,p)，q=1-p，求证D=npqE=np，kCnkpkqn-k=nppk-1qn-k,kkCnkpkqn-k=np(k-1)pk-1qn-kpk-1qn-k=np(n1)ppk-2qn-kpk-1qn-k而D=，D=(11Cn1p1qn-122Cn2p2qn-2kkCnkpkqn-knnCnnpnq0) =np(1Cn-10p0qn-12Cn-11p1qn-23Cn-12p2qn-2kCn-1k-1pk-1qn-knCn-1n-1pn-1q0)2npEn2p2(pq)n=np0Cn-10p0qn-11Cn-11p1qn-22Cn-12p2qn-2(k-1)Cn-1k-1pk-1qn-k(n-1)Cn-1n-1pn-1q0(Cn-10p0qn-1Cn-11p1qn-2Cn-12p2qn-2Cn-1k-1pk-1qn-kCn-1n-1pn-1q0)=npE(pq)n-1 =np(n1)p1 =np(1p)=npq .
