Weighted verage of two variables with minimal variance

  1. 1. The problem statement, all variables and given/known data

    X1 and X2 are independent random variables. They both have the same mean (mue). Their variances are s1^2 and s2^2 respectively, where s1^2 and s2^2 are known constants. It is proposed to estimate mue by an estimator T of the form T=c1X1 + c2X2.
    Show that T will be unbiased if c1 + c2=1
    and find an expression for var(T) in terms of c1, s1^2 and s2^2.
    (assuming c1+c2=1)
    2. Relevant equations



    3. The attempt at a solution

    I showed that T will be unbiased if c1+c2=1
    For the next part this is what i did:

    var(T) = var(c1X1+c2X2)
    var(c1X1+c2X2) = E[(c1X1+c2X2)^2] + {E[c1X1+c2X2]}^2

    and then after expanding and simplifying, i got:
    var(T) = 2(mue)^2(c1^2 + 2c1c2 + c2^2)

    I can easily change c2 in terms of c1 but how do put in terms of s1^2 and s2^2 as this is what they are asking for??

    Thank you
     
  2. jcsd
  3. statdad

    statdad 1,479
    Homework Helper

    Re: Variance

    If X, Y are independent random variables, and a, b are real numbers, then

    Var(aX + bY) = a^2 Var(X) + b^2 Var(Y)

    Apply this to the setting of your problem.

    Note that, relating to your work,

    Var(W)

    does not equal

    E(W^2) + (E(W))^2

    so your formula would not get you to the desired result.
     
  4. Re: Variance

    Thank you v much.
    I should have known that Var(aX + bY) = a^2 Var(X) + b^2 Var(Y) !!!

    But how come for this question
    Var(W)

    does not equal

    E(W^2) + (E(W))^2

    ?
     
  5. statdad

    statdad 1,479
    Homework Helper

    Re: Variance

    Var(W) = E((W - mu_w)^2) = E(W^2 - 2Wmu_w + (mu_w)^2) = E(W^2) - 2(mu_w)^2 + (mu_w)^2 = E(W^2) - (mu_w)^2

    for any random variable W. :smile: I believe you just missed a sign.

    Sometimes, after staring at a problem for some time, our minds see what we want them too rather than what we've actually written - it happens to me a lot.
     
  6. Re: Variance

    Oh ofcourse...it's minus...silly me.

    What u said is SO TRUE.
    Thanks v much.
     
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