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Math Help - proving of equationa

  1. #1
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    proving of equationa

    hi! can anyone teach me how to prove the following?

    E(X^2)= (E(x))^2 + Var X

    and

    Cov((aX + bZ), Y) = aCov(X, Y) + bCov(Z,Y)
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  2. #2
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    Quote Originally Posted by pinkiee View Post
    hi! can anyone teach me how to prove the following?

    E(X^2)= (E(x))^2 + Var X

    and

    Cov((aX + bZ), Y) = aCov(X, Y) + bCov(Z,Y)

    V(x)=E[(X-u)^2]=E(X^2-2Xu+u^2)=E(X^2)-E(2Xu)+E(u^2)

    But u is a constant

    =E(X^2)-2uE(x)+u^2=E(X^2)-2u^2+u^2=E(X^2)-u^2

    But E(X)=u so

     <br />
=E(X^2)-(E(X))^2<br />
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  3. #3
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    but how do i swap the places of the terms? cuz i need to prove that E(X^2) is equal to (E(x))^2 + Var X
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  4. #4
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    Quote Originally Posted by pinkiee View Post
    but how do i swap the places of the terms? cuz i need to prove that E(X^2) is equal to (E(x))^2 + Var X
    Surely you can re-arrange the proven result Var(X) = E(X^2) - [E(X)]^2 to make E(X^2) the subject?!

    In fact, Google gives tons of websites that more or less prove both of the results you want to prove. Did you try doing a Google search?
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  5. #5
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    i tried googling, they gave me results of how to dervive var X
    but i just cant seem to rearrange it to make E(X^2) the subject.
    any mind helping out?

    and how about Cov((aX + bZ), Y) = aCov(X, Y) + bCov(Z,Y)
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  6. #6
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    V(x)=E(X^2)-(E(X))^2

    Now you want to isolate one of those.....

    y=x-z

    Solve for x,

    if you can do that, you can do the step above, which only involves one step of addition
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  7. #7
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    ohh.. okie i try to work out that. thanks =)

    how do i derive Cov((aX + bZ), Y) = aCov(X, Y) + bCov(Z,Y) ?
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  8. #8
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    Quote Originally Posted by pinkiee View Post
    ohh.. okie i try to work out that. thanks =)

    how do i derive Cov((aX + bZ), Y) = aCov(X, Y) + bCov(Z,Y) ?
    Hi - I would suggest you try it out for yourself. It is pretty straight fwd.

    I will get you started. Cov(XV) = E((X-E(X)(Y-E(Y)) - This is the definition and your starting point
    In this put X = aX+bZ and simplify the expression. Thats all there is to this prove

    Hope you know that E(aX+bZ) = aE(X) + bE(Z)
    If not try prove this first.
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  9. #9
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    ok thanks! i gt it!

    but i still can't expand E(X^2) such that i'll get (E(x))^2 + Var X

    its not just rearranging it right? hmm so confused.

    anyone can give me the 1st step of expanding E(X^2)?
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  10. #10
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    Quote Originally Posted by pinkiee View Post
    ok thanks! i gt it!

    but i still can't expand E(X^2) such that i'll get (E(x))^2 + Var X

    its not just rearranging it right? hmm so confused.

    anyone can give me the 1st step of expanding E(X^2)?
    Quote Originally Posted by mr fantastic View Post
    Surely you can re-arrange the proven result Var(X) = E(X^2) - [E(X)]^2 to make E(X^2) the subject?!
    [snip]
    This has nothing to do with expanding!

    If you had A = B - C I would hope that at this level you could make B the subject. This is basic algebra. If you honestly cannot do this you have absolutely no business attempting questions like the one you posted.
    Last edited by mr fantastic; October 9th 2009 at 10:37 PM. Reason: Fixed a minor typo
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  11. #11
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    u mean i just have to do this?

    E(X^2)= (E(x))^2 + Var X

    then rearrange to Var X = E(X^2) - (E(x))^2 ?

    then derive that var x = E(X^2) - (E(x))^2 ?
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  12. #12
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    Quote Originally Posted by pinkiee View Post
    u mean i just have to do this?

    E(X^2)= (E(x))^2 + Var X Mr F says: Yes. Read my bottom comment first.

    then rearrange to Var X = E(X^2) - (E(x))^2 ?

    then derive that var x = E(X^2) - (E(x))^2 ? Mr F says: This was proved in the very first reply you got. Did you read it?
    ..
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  13. #13
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    yup, i read it alr.

    i think i gt it now.

    i thought i have to expand E(X^2) so that it gives me (E(x))^2 + Var X

    thanks!! =D
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