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Math Help - Independent Random Variables

  1. #1
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    Independent Random Variables

    Hi,

    I can't seem to get the same answer as the back for this question. Can someone show me how to properly do it?

    Let X and Y be two independent random variables where E(X) = 4, E(X^2)= 20 and V(Y)= 21. Find:

    (a) V (X + 3Y)
    (b) the standard deviation of (X+ + 3Y)

    The answers at the back are 193 and 13.9.

    Thanks
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  2. #2
    MHF Contributor harish21's Avatar
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    for independent variables, the covariance will be 0. thus:

    Var(X+3Y)= Var(X)+Var(3Y) = Var(X)+3^2\;Var(Y)=Var(X)+9Var(Y)=....

    you have E[X] and E[X^2], from which you can find Var(X).

    Finding Var(X+3Y) means that you are a step away from finding the s.d. Do that!
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  3. #3
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    Quote Originally Posted by harish21 View Post
    for independent variables, the covariance will be 0. thus:

    Var(X+3Y)= Var(X)+Var(3Y) = Var(X)+3^2\;Var(Y)=Var(X)+9Var(Y)=....

    you have E[X] and E[X^2], from which you can find Var(X).

    Finding Var(X+3Y) means that you are a step away from finding the s.d. Do that!
    Thanks for the prompt response. Good news is I got the right answer, however, I don't understand why the 3 turns into 3^2. Can you please explain to me why this is? The meaning of covariance isn't very clear to me either.

    Thanks!
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  4. #4
    MHF Contributor harish21's Avatar
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    Well, you should have a textbook and/or classnotes with you. And you should have been given these concepts before you are made to work on the problems.

    Anyways, the three turns to 9 because it is a well known property.

    For a random variable X with finite variance:

    Var(aX) = E[(aX)^2] - E[aX]^2 [Note: a is a constant]

    The mean of aX is a times the mean of the random variable X.

     E[aX] = aE[X]  \; and \; E[(aX)^2] = E[a^2 (X^2)] = a^2 E[X^2]

    \therefore Var(aX) = a^2 E[X^2] - [aE(X)]^2

    = a^2 (E[X^2] -(E[X])^2)

    = a^2 (Var(X))

    More properties???

    Look HERE and

    go here for Covariance
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  5. #5
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    That cleared things up a bit. Thanks!
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