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Math Help - function of iid random variables also independent?

  1. #1
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    function of iid random variables also independent?

    Hallo,

    X_1, X_2, X_3 are iid random variables with a lognormal distribution.

    Am I right, that
    f(X_1),f(X_2),f(X_3) where f(x)=(1+x)^\gamma, 0< \gamma < 1
    are also independent?

    Thanks in advance!
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  2. #2
    Moo
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    Re: function of iid random variables also independent?

    Hello,

    Yes, just use the definition of independence : X and Y are independent iff for any *correct* functions f,g, E[f(X)g(Y)]=E[f(X)]E[g(Y)]

    *correct* in the sense that the expectations are defined for such functions. That would be measurable and bounded for the simplest situation I guess...
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  3. #3
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    Re: function of iid random variables also independent?

    Thank you!
    I didn't know this definition.
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  4. #4
    Guy
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    Re: function of iid random variables also independent?

    You can also prove that

    P[f_1(X_1) \in B_1, ..., f_n (X_n) \in B_n] = P[f_1 (X_1) \in B_1] \cdots P[f_n (X_n) \in B_n]

    where the B_i are arbitrary 1d Borel sets. This implies independence from a different starting place; all my references use a different definition of independence than Moo's, but they are equivalent.
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  5. #5
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    Re: function of iid random variables also independent?

    Thank you. So if I show it this way, I just need the inverse function and use the independence of X_1,X_2,X_3 and then it follows.
    Am I right?

    Doesn't it matter that my function f(x)=(1+x)^\gamma,0<\gamma<1, is just defined for x\geq -1 and the inverse function for x \geq 0?
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  6. #6
    Guy
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    Re: function of iid random variables also independent?

    Quote Originally Posted by Juju View Post
    Thank you. So if I show it this way, I just need the inverse function and use the independence of X_1,X_2,X_3 and then it follows.
    Am I right?

    Doesn't it matter that my function f(x)=(1+x)^\gamma,0<\gamma<1, is just defined for x\geq -1 and the inverse function for x \geq 0?
    You don't need inverse function. Inverse image works and is always there for you. You need the f_i to be measurable to make sure that f_i ^{-1} (B_i) is measurable.
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  7. #7
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    Re: function of iid random variables also independent?

    my function is continous (on x \geq -1) and thus measurable. right?
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  8. #8
    Guy
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    Re: function of iid random variables also independent?

    Heh, you have to try much harder than that to cook up a function that isn't measurable Continuity is more than enough.
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