1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Questions relating to UMVUE/transformations

  1. Mar 8, 2013 #1
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    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?
     
  2. jcsd
  3. Mar 8, 2013 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    What's a UMVUE? What's CRLB? (I *think* I know what the last one means, but how can I be sure?)
     
  4. Mar 8, 2013 #3
    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
     
  5. Mar 8, 2013 #4
    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Questions relating to UMVUE/transformations
Loading...