1. Limited time only! Sign up for a free 30min personal 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!

Minimum variance unbiased estimator

  1. Aug 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex]\bar{X}[/itex]1 and [itex]\bar{X}[/itex]2 be the means of two independent samples of sizes n and 2n from an infinite population that has mean μ and variance σ^2 > 0. For what value of w is w[itex]\bar{X}[/itex]1 + (1 - w)[itex]\bar{X}[/itex]2 the minimum variance unbiased estimator of μ?
    (a) 0
    (b) 1/3
    (c) 1/2
    (d) 2/3
    (e) 1


    2. Relevant equations
    If θ~ is unbiased for θ and
    Var(θ~)= 1/E[(d loge f (x)/dθ)^2] = 1/E[(dl(θ)/dθ)^2]
    then θ~ is a minimum variance unbiased estimator of θ.


    3. The attempt at a solution
    E[w[itex]\bar{X}[/itex]1 + (1 - w)[itex]\bar{X}[/itex]2] = wμ + (1-w)μ = μ
    So it's an unbiased estimator of μ.

    I tried calculated the variance but I guess it's wrong.
    Var[w[itex]\bar{X}[/itex]1 + [itex]\bar{X}[/itex]2 - w[itex]\bar{X}[/itex]2] = w^2.σ^2/n + σ^2/n + w^2.σ^2/n = σ^2/n(2w^2 +1)

    I think I have to use the formula above but I don't know how.
    Thanks.
     
  2. jcsd
  3. Aug 2, 2012 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Note: use brackets, since otherwise your expressions are ambiguous. Better yet, use LaTeX, as you did in the first part of your post.

    Your variance formula is incorrect. Since [itex] \bar{X}_1[/itex] and [itex] \bar{X}_2[/itex] are independent we have
    [tex] \text{Var}(a \bar{X}_1 + b \bar{X}_2) = a^2 \text{Var}(\bar{X}_1) + b^2 \text{Var}(\bar{X}_2)[/tex]
    for any constants [itex] a, \: b.[/itex] Use [itex] a = w[/itex] and [itex] b = 1-w.[/itex]

    I don't know why you wanted to write (1-w)X as X - wX and then apply the variance formula, but you did it incorrectly. Using V(.) for the variance of a random varable, we have (using the fact that X and X are dependent): [tex] V(X - wX) = 1^2 V(X) + w^2 V(X) - 2 \cdot 1\cdot w\: \text{Cov}(X,X),[/tex] and, of course, [itex]\text{Cov}(X,X) = V(X).[/itex]

    RGV
     
    Last edited: Aug 2, 2012
  4. Aug 2, 2012 #3
    I don't know why I did that either.

    So is it right to say:
    [tex] \text{Var}( \bar{X}_1) = \text{Var}(\bar{X}_2) = σ^2/n[/tex] ?

    How do I work out what w is?
     
  5. Aug 2, 2012 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    In terms of sigma, what would be the variance of a sample mean of size 10 (that is, n=10)? What about for a sample of size 20? Now go back and read (carefully) the original question. Do you honestly still need me to answer?

    RGV
     
  6. Aug 2, 2012 #5
    I guess it would be (σ^2)/10 and (σ^2)/20, so it's (σ^2)/n and (σ^2)/2n for [itex] \bar{X}_1[/itex] and [itex] \bar{X}_2[/itex], respectively.

    I subbed that into
    [tex] \text{Var}(a \bar{X}_1 + b \bar{X}_2) = a^2 \text{Var}(\bar{X}_1) + b^2 \text{Var}(\bar{X}_2)[/tex]
    and subbed in a and b, and I got
    [tex](σ^2(w^2+1))/2n[/tex]

    I still don't know how to get w, sorry...
     
  7. Aug 2, 2012 #6

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    No wonder: you have made a serious algebraic blunder, so you end up with the wrong expression.

    RGV
     
  8. Aug 3, 2012 #7
  9. Aug 3, 2012 #8

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You should not need a powerful tool to do such simple, high-school algebra, but since you have already done it, OK. The question asked you to find the value of w that minimizes the variance. So, that is what you need to do. At this point I am signing off this thread.

    RGV
     
  10. Aug 3, 2012 #9
    I just used it to confirm I was right. Thanks for your help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Minimum variance unbiased estimator
  1. Unbiased estimator (Replies: 2)

  2. Unbiased estimates (Replies: 2)

  3. IID Unbiased Estimator (Replies: 5)

Loading...