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Thread: Finding Mean and Variance for X^2 Given Mean and Variance for X

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    Finding Mean and Variance for X^2 Given Mean and Variance for X




    I am working on a question that has me stumped. X1, X2, ... Xn are normally distributed and independent. Each has a mean of 0 and variance sigma^2. I am asked to find E(Xi^2) and V(Xi^2). There seem to me two ways to approach this:

    First, since the Xi are independent, we could say E(Xi^2) = E(Xi)*E(Xi) = 0*0 = 0. Second, we could also say V(Xi) = E(Xi^2) - [E(Xi)]^2 = E(Xi^2) - 0 = E(Xi^2) in which case E(Xi^2) = V(Xi) = sigma^2.

    From these two, we are led to conclude that sigma must be 0 (in which case this is a uniform distribution, not normal). Have I made some error in thinking or is this problem flawed?

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    Re: Finding Mean and Variance for X^2 Given Mean and Variance for X

    Maybe I've answered my own question ... I guess I can't say the Xi are independent of themselves :-) So, my first approach was invalid.

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    Re: Finding Mean and Variance for X^2 Given Mean and Variance for X

    Second approach is right. Similar way you can calculate Var(Xi^2)

    Var(X_i^2)= E[(X_i^2)^2] -(E[X_i^2])^2
    Var(X_i^2)= E[X_i^4] -(E[X_i^2])^2
    E[X_i^4] is the 4th moment
    In the long run, we're all dead.

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    Re: Finding Mean and Variance for X^2 Given Mean and Variance for X

    I knew I would need that formula. The only problem is that I have no idea how to find E(Xi^4) given that I know E(Xi^2).

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    Re: Finding Mean and Variance for X^2 Given Mean and Variance for X


    There are lots of tricks to get what you want. But if you know E(X^2) then it's possible to get E(X^4). This might help.

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