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Math Help - Statistics Question

  1. #1
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    Statistics Question

    I'm given this problem. σ²(t) = V((1-t)X + tY) = at² + 2bt+c where

    a=? b=? c=?

    E(X) = 2
    E(Y) = 3
    E(X^2) = 53
    E(Y^2) = 45
    E(XY) = -21


    Can someone please explain what σ²(t) = V((1-t)X + tY) means and how to solve it?


    Thanks.
    Last edited by taypez; February 21st 2008 at 07:50 AM. Reason: More info
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  2. #2
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    I think  V(X) is the variance and it is equaled to  E[X^2] - (E[X])^{2} . This is for one random variable  X .

    But  V(X+Y) = V(X) + 2 \text{Cov}(X,Y) + V(Y) for two random variables.


    But then there are the  t terms. So plug in  t = 0 and  t = 1 to solve for  a,b,c . So for  t = 0 ,  V(X) = c . For  t = 1 ,  V(Y) = a+2b+V(X) .
    Last edited by heathrowjohnny; February 21st 2008 at 08:17 AM.
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  3. #3
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    Alright, does this make any sense...

    I calculated:

    cov(X,Y) = -27
    V(X) = 49
    V(Y) = 54

    so, for: at^2 + 2bt + c

    a = V(x)= 49
    b = cov(X,Y) = -27
    c = V(Y) = 54

    I'm not sure that what he's looking for.

    Then how would I find the value of t* that minimizes this? t* = -σy/σx
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