Math Forum Discussions

Henry

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

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

Plain Text

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(T|Theta) = 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 non-zero 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