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Henry
Posts:
1,089
Registered:
12/6/04


Re: need help with a statistics question about estimators
Posted:
Dec 15, 2009 4:51 AM


On 14 Dec, 15:08, Jon Du Kim <jo...@FAKE.EMAIL.net> wrote: > Jon Du Kim wrote: > > Henry wrote: > >> On 14 Dec, 08:53, Jon Du Kim <jo...@FAKE.EMAIL.net> wrote: > >>> Let X_1,...X_n be independent random variables with the > >>> uniform distribution on the interval(0,Theta) > >>> where Theta > 0 is an unknown parameter to be estimated. > > >>> Let X' be the sample mean and define > >>> an estimator T=2X' > >>> Is T an unbiased estimator of Theta? > > >>> How do I even begin to do this? > >>> How do I find the mean and variance of T? > >>> I am very lost on this. > >>> Can anyone give me any helps? > > >> T is an unbiased estimator of Theta if the expectation of T given > >> Theta is Theta > >> i.e. E(TTheta) = Theta > > >> You might find it useful to use the fact that if Y and Z are random > >> variables with finite means and k is a nonzero constant then > > >> E(Y+Z) = E(Y) + E(Z) > >> E(k*Y) = k * E(Y) > >> E(Y/k) = E(Y) / k > > >> You should be able to work out E(X_1) given Theta etc. > > > So from what you posted > > E(T)=E(2X') > > > Sum xn/n = X' > > > 2* Sum xn/n = 2 * X' > > > E[2X'] = 2* Sum E[xn]/n = 2 * (n*mu)/n = 2 * mu > > > 2* E[X']  2 * mu = 0 so the bias of the estimator xbar > > is 0 and the estimator is unbiased. > > since it was posted that > > Var(Y+Z) = Var(Y) + Var(Z) > > Var(k*Y) = k^2 * E(Y) > > Var(Y/k) = Var(Y) / k^2 > > then, similarily, > from Var(k*Y) = k^2 * E(Y) and the above > then > Var(2X')= 4 E(X') > = 4 mu Hide quoted text  > >  Show quoted text 
Sorry. My mistake (copy and paste and edit) I should have said
Var(k*Y) = k^2 * Var(Y)
So (unless I have made more errors) Var(T) = Var(2X') = 4 Var(X') = (4) Var(sum_i(X_i)/n) = (4/n^2) Var(sum_i(X_i)) = (4/n^2) sum_i(Var(X_i) = (4/n) Var(X_1) and you know X_1 is uniform on (0,Theta) so you can work out its variance



