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Math Help - Estimators Question

  1. #1
    Super Member craig's Avatar
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    Estimators Question

    Two independent variables X and Y each have an expected value of \mu, but different variances \sigma^2 and \tau ^2.

    If the linear combination Z = aX + bY is used to estimate \mu, then what relationship between a and b is necessary for this estimator to be unbiased?

    I'm don't know how to prove that an estimator is unbiased, so not really sure how to start this part?

    Find the simplified expression for the variance of any unbiased estimator that is a linear combination of X and Y.

    I'm guessing you need the first part for this?

    Use this to prove that the unbiased estimator with the lowest standard error is in the form:

    a = \frac{\tau^2}{\sigma^2 + \tau^2}

    b = \frac{\sigma^2}{\sigma^2 + \tau^2}

    and has variance \frac{\tau^2 \sigma^2}{\sigma^2 + \tau^2}

    I apologise for the lack of an attempt for this question, If someone could help me out on the firs bit and give me a little pointer I think I could give the second and third parts a go.

    Thanks in advance
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  2. #2
    Moo
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    A Cute Angle Moo's Avatar
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    Hello,

    Question 1)
    In order to prove that an estimator \hat \theta is unbiased, you have to check that E[\hat\theta]=\mu
    So a and b are such that a+b=1

    Question 2)
    Any unbiased estimator will be in the form aX+(1-a)Y
    So its variance is : var(aX+(1-a)Y)=a^2 var(X)+(1-a)^2var(Y), since X and Y are independent.

    Question 3)
    It's your turn to work (and I don't know how to do it )
    Last edited by Moo; February 13th 2010 at 07:59 AM. Reason: big mistalke :(
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  3. #3
    MHF Contributor matheagle's Avatar
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    To be unbiased you need the expected value of the random variable to equal MOO.

    So \mu=E(Z)=a(X)+bE(Y)=a\mu+b\mu

    That means that a+b=1.

    Next, you minimize V(Z)=a^2V(X)+(1-a)^2V(Y)=a^2\sigma^2+(1-a)^2\tau^2 wrt a via calc one.

    ---------------------------------------------------
    Someone changed their "solution"
    Last edited by matheagle; February 13th 2010 at 08:50 AM.
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  4. #4
    Super Member craig's Avatar
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    Thanks for both replies, make a little more sense now

    Still not sure what to do on the last bit, as you both said above the variance is in the form a^2 \sigma^2+(1-a)^2\tau^2, but not sure how that gets to \frac{\tau^2 \sigma^2}{\sigma^2 + \tau^2}, or how to isolate a and b?

    Thanks again for the help so far, if anyone else can shed any light on the last bit it would be much appreciated
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  5. #5
    MHF Contributor matheagle's Avatar
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    as I already said, replace b with 1-a and use calculus.......................

    Next, you minimize V(Z)=a^2V(X)+(1-a)^2V(Y)=a^2\sigma^2+(1-a)^2\tau^2 wrt a via calc one.
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  6. #6
    Super Member craig's Avatar
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    Quote Originally Posted by matheagle View Post
    as I already said, replace b with 1-a and use calculus.......................

    Next, you minimize V(Z)=a^2V(X)+(1-a)^2V(Y)=a^2\sigma^2+(1-a)^2\tau^2 wrt a via calc one.
    Sorry I didn't see that on your last post.

    Differentiate wrt a, becomes a simple linear equation to solve

    Thanks a lot for the help here, when it said minimise it was trying to set the original equation to zero, didn't occur to me to find the minimum value through differentiation.
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