Questions relating to UMVUE/transformations

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The discussion centers on finding a Uniformly Minimum Variance Unbiased Estimator (UMVUE) for the Rayleigh distribution and the Weibull distribution using transformations. The user successfully identifies the estimator for the Rayleigh distribution as U-bar, which attains the Cramer-Rao Lower Bound (CRLB). Confusion arises regarding the similarity of the transformed variables U_1 and U_2, both leading to a Gamma distribution, and whether this implies that the original distributions are irrelevant for finding estimators. Clarification is sought on the dependence of the estimators on the original distributions and the implications for the information contained in the data sets. Ultimately, the user concludes that their understanding aligns with the properties of UMVUE and CRLB, affirming their findings.
abeliando
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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 n-1E(U)=θ, so then U-bar is the UMVUE of θ. I think this is correct.

So then I am asked to see if it attains the CRLB, and I use the fact that U is Gam(n,θ) to show that yes indeed, it does reach the CRLB. All is well and good. Also, I'm asked to find the estimators of θ2 and θ-1, and find their variances. I did this, and it all relied on the fact that it was gamma distributed.

On the next problem, I'm given that X follows a Weibull, that is, has the pdf
f(x;θ)=α-1βxβ-1*exp(-xβ/α), where β (>0) is known, and it suggests starting off with U=ƩXβi. So when I make this transformation, I find the same thing, that Ui is exponential, so U=ƩXβi is distributed Gam(n,α).

So here's why I'm confused. In both cases, I started with a pdf, transformed it to a different variable, but otherwise, finding the estimators and their variances aren't based on their original distribution, but the transformed distribution, which is the same. So I'd imagine that I just cut and paste my original work. It can't be that easy, right? What am I missing? The problems in the text are assigned right after each other, so... I'm assuming I'm wrong. Any help?
 
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What's a UMVUE? What's CRLB? (I *think* I know what the last one means, but how can I be sure?)
 
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 that is unbiased for an parameter, that statistic is the UMVUE. In my cases, the suggested statistics that I started off with are complete sufficient, and then taking the average is still complete sufficient.

CRLB is the function that states what the theoretical minimum variance for an estimator is, and then I just take the variance to see if it actually works
 
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 a Rayleigh distribution, then U_1 follows a Gamma(n,θ) distribution. Also if X follows a Weibull distribution, then U_2 follows a Gamma(n,θ) distribution. So it does depend on the original distribution. Is this correct?

This makes sense. It would also make sense that the work would be same, I guess.

Oh, and the information contained in the set will (should) be different (I haven't evaluated it yet).

I think I figured it out, yes?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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