Yeah, it seems a bit difficult. If you're looking for the UMVUE, you don't have to though. Can't you use Lehmann-Scheffe here, in the case of the Gamma?
Hm, this is a little bit tricky. If I did my work correct, I suppose it does make sense, and rely on the previous distribution. I am confused about the two U's (U_1=ƩX2i and U_2=ƩXβi apparently having the same distribution. And I guess what I' m saying is, is that in the first case, if X follows...
UMVUE = Uniformly Minimum Variance Unbiased Estimator, CRLB = Cramer Rao Lower Bound (defined as the first derivative of a function of a parameter squared, over the information contained by the set of data) (τ(θ)')2/IX
So by Lehmann-Scheffe theorems, if I take a complete sufficient statistic...
So one of my assignments was to find a UMVUE of a Rayleigh distribution, that is the pdf
f(x;θ)=2θ-1x*exp(-x2/θ).
the suggestion was to use U=Ʃx2i. When I do this, I do a transformation with Ui=X2i and find that Ui has an exponential distribution, and the U=ƩUi follows a gamma(n,θ), so then...
Homework Statement
Problem is in the attachment, sorry, I can't figure out how to do tex in this message board system.
Homework Equations
The Attempt at a Solution
In the attachment.