作者locust0923 (柠檬优酪乳)
看板Grad-ProbAsk
标题Re: [商管] [统计学]-西葛马问题
时间Thu Jan 14 22:24:30 2010
※ 引述《NiceKitty (想要有个依靠><)》之铭言：
: 不好意思再请教各位大大一题><
: Consider n independent trials,each of which results in any of the
: outcomes i, i=1,2,3 with respective probabilities P1,p2,p3
: 3
: ΣPi=1.Let Ni denote the number of trials that result in outcome i,
: i=1
: and show that Cov(N1,N2).
: Also explain why it is intuitive that this covariance is negative.
n=N1+N2+N3,1=P1+P2+P3 =>N3=n-N1-N2,P3=1-P1-P2
∴(N1,N2)~Trinomial(n,P1,P2)
n! n1 n2 n-n1-n2
f(n1,n2)=------------------------(P1) (P2) (1-P1-P2) , n1=0,1,...,n
(n1)!*(n2)!*(n-n1-n2)! n2=0,1,...,n-n1
则
N1~Binomial(n,P1) , N2~Binomial(n,P2) , N1与N2不独立
E(N1)=nP1 , E(N2)=nP2 , V(N1)=nP1(1-P1) , V(N2)=nP2(1-P2) , Cov(N1,N2)=-nP1P2
最後二行是直接写结论出来，推导可以自己推推看
一般研所的机率或统计的书都会介绍到可以去翻翻看
另外N1跟N2改成X跟Y比较不会搞混，这是个人看法
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1F：→ NiceKitty:谢谢大大喔^^ 01/14 23:31
2F：→ goshfju:推改符号 XD 01/15 00:11