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用英文答,证明题Forarandomvariablewithabinomialdistribution,B(n,p),ShowthatE(X)=npVar(X)=np(1-p)=npq,whereq=1-p.
题目详情
用英文答,证明题
For a random variable with a binomial distribution,B(n,p) ,Show that E(X)=np Var(X)=np(1-p)=npq ,where q=1-p.
For a random variable with a binomial distribution,B(n,p) ,Show that E(X)=np Var(X)=np(1-p)=npq ,where q=1-p.
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答案和解析
Answer. The probability that X=k is C_n^k p^k q^{n-k}, so we get
EX= \sum_0^n k C_n^k p^k q^{n-k} = p(\sum_0^n C_n^k p^k q^{n-k})_p
= p ((p+q)^n)_p
= p n(p+q)^{n-1}
= np
Here _p means taking partial derivative with respect to the variable p.
Similarly, you can work out \sum_0^n k(k+1) C_n^k p^k q^{n-k}, and deduce the value of \sum_0^n k^2 C_n^k p^k q^{n-k}, then Var X can be computed out.
But I compute out Var X = n^2 pq, rather npq. This should be checked.
EX= \sum_0^n k C_n^k p^k q^{n-k} = p(\sum_0^n C_n^k p^k q^{n-k})_p
= p ((p+q)^n)_p
= p n(p+q)^{n-1}
= np
Here _p means taking partial derivative with respect to the variable p.
Similarly, you can work out \sum_0^n k(k+1) C_n^k p^k q^{n-k}, and deduce the value of \sum_0^n k^2 C_n^k p^k q^{n-k}, then Var X can be computed out.
But I compute out Var X = n^2 pq, rather npq. This should be checked.
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