c********r
发帖数: 73 | 1 Consider a sequence X1,X2,......,Xn of Bernoulli trials, Here the Xi are iid
as X with Prob(X=1)=p,0
show that there exists no unbiased estimator for the odds ratio p/(1-p)
due next Thesday.
thank jiay | x***e
发帖数: 62 | 2 proof by contradiction;
suppose you can construct a function:
F(x1,x2....xn) as your estimator
calculate the expected value of your F(x1,x2,...xn);
E(F(x1,x2,...xn)); show that can never be p/(1-p);
it is getting nasty;
E(F((x1,x2...xn))=F(0,0,..0)*(1-p)^n+sum(F(Xi=1))*p*(1-p)^(n-1)+.....
+F(1,1,...1)*p^n;
and since the only possible value for Xi's are 0 and 1;
so F(0...0)=0; F(Xi=1)=1 or O; F(Xi=1,Xj=1)=0,1,2; .....
F(1,1,...1)=0,1,...(n/2)^(n/2);(if n is even)
and then you just calulate those
【在 c********r 的大作中提到】
: Consider a sequence X1,X2,......,Xn of Bernoulli trials, Here the Xi are iid
: as X with Prob(X=1)=p,0: show that there exists no unbiased estimator for the odds ratio p/(1-p)
: due next Thesday.
: thank jiay
| J**Y
发帖数: 34 | 3 We can have more compact way to think about this. Suppose f(X) is an
unbiased estimator of odds ratio, where X=(x1,....,xn). Then
E[f(X)]=Sum_k[f(k)*C_nk*p^k*(1-p)^n-k], where C_nk is the combination
symbol. Because f(k)*C_nk is constant, let a_k denote it. Then
E[f(X)]=Sum_k[a_k*p^k*(1-p)^n-k]=p/1-p =>
Sum_k[a_k*p^k*(1-p)^n-k+1]=p, you then can show the equation
doesn't exist.
【在 c********r 的大作中提到】
: Consider a sequence X1,X2,......,Xn of Bernoulli trials, Here the Xi are iid
: as X with Prob(X=1)=p,0: show that there exists no unbiased estimator for the odds ratio p/(1-p)
: due next Thesday.
: thank jiay
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