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Math Help - Independence

  1. #1
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    Independence

    First I would like to clear up some misunderstanding I have about independence. In my text I have:
    If F(x,y)=F_X(x)F_Y(y), then X and Y are indep.
    However, in my notes from class I have:
    If f(x,y)=f_X(x)f_Y(y), then X and Y are indep.
    Which is correct?

    Is this the only way to show independence? Does it have anything to do with expectation values? I had a question in which I had:
    X=Z, Y=Z^2; Z \sim  N(\mu,\sigma^2)
    I was first asked to find the followings: <X>,<Y>,<XY> and here is what I did:
    <X>=<Z>=\mu
    <Y>=<Z^2>=Var(Z)+<Z>^2=\sigma^2+\mu^2
    <XY>=<Z^3>, for this I took the 3rd derivative of the moment generating function and got this expression: 2\sigma^2\mu+\mu^3

    I was then asked about whether X and Y are independent. Naturally I attempted to use what I just did: <XY>\neq<X><Y> therefore dependent. Is this correct?
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  2. #2
    MHF Contributor matheagle's Avatar
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    Y=X^2 so they are clearly dependent.
    It certainly doesn't matter if you have normality or not.
    If you differentiate F_X(x)F_Y(y)=F(x,y) wrt x and y you get f_X(x)f_Y(y)=f(x,y)
    Likewise if f_X(x)f_Y(y)=f(x,y), then
    F(x,y)=\int_{-\infty}^x\int_{-\infty}^yf(s,t)dtds =\int_{-\infty}^xf_X(s)ds\int_{-\infty}^yf_Y(t)dt=F_X(x)F_Y(y)
    Last edited by matheagle; March 19th 2009 at 11:02 PM.
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  3. #3
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    Okay, I can see why they are dependent, but how can I show it? I can write:
    f_X(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}

    But what is f_Y(y)?
    Is it just [f_X(x)]^2, so f_Y(y)=\frac{1}{\sigma^22\pi}e^{-2\frac{(y-\mu)^2}{2\sigma^2}} ?
    Or since Y=X^2, is it just f_Y(y)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(\sqrt{y}-\mu)^2}{2\sigma^2}} ?

    And worse, how can I combine them and get f(x,y)?
    I'm quite lost.
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  4. #4
    MHF Contributor matheagle's Avatar
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    You can't square the density of X to get the density of X^2
    Try showing that the covariance is not zero.
    Show that E(XY)-E(X)E(Y)=E(X^3)-E(X)E(X^2) is not zero.
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  5. #5
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    I know that is X and Y are independent, then Cov(X,Y)=0. But I thought it doesn't work the other way around, namely, even if Cov(X,Y)=0, it doesn't necessary mean that they are independent. Then even if I show that Cov(X,Y)\neq0, how can I say that they are indeed dependent?
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  6. #6
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by synclastica_86 View Post
    I know that is X and Y are independent, then Cov(X,Y)=0. But I thought it doesn't work the other way around, namely, even if Cov(X,Y)=0, it doesn't necessary mean that they are independent. Then even if I show that Cov(X,Y)\neq0, how can I say that they are indeed dependent?

    Yes, that's called the contrapositive.
    A implies B, and not B implies not A.
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  7. #7
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    Quote Originally Posted by matheagle View Post
    You can't square the density of X to get the density of X^2
    But in principle, with the information given, can I find f_Y(y) and f(x,y) , and therefore, their respective distribution functions?
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  8. #8
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by synclastica_86 View Post
    But in principle, with the information given, can I find f_Y(y) and f(x,y) , and therefore, their respective distribution functions?

    The formula for obtaining the density of Y=X^2 is right under

    Derivation of the pdf for one degree of freedom
    on Chi-square distribution - Wikipedia, the free encyclopedia
    But X and Y do not have a joint distribution in R^2.
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