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Homework Help: 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

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    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

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    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

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    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

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    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.
     
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