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Math Help - transformation of independent random variables

  1. #1
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    transformation of independent random variables

    I need to prove the following in the discrete case, and I am not sure where to start.

    Let X and Y be independent random variables, and let f and g be transformations. Then f(X) is independent of g(Y).

    This intuitively makes sense to me, but I am not sure how to prove it in a way that will cover all possible transformations.
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  2. #2
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    Feb 2010
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    Here's how I would do it:

    If X_1 and X_2 are independent then: f_{X_1,X_2}(x_1,x_1)=f_{X_1}(x_1)f_{X_2}(x_2) let Y_1=g(x_1) and Y_2=h(x_2) be invertible functions then: X_1=g^{-1}(y_1) and X_2=h^{-1}(y_2) then we have: \frac{\partial X_1}{\partial y_1}=\frac{dg^{-1}(y_1)}{dy_1}, \frac{\partial X_1}{\partial y_2}=0, \frac{\partial X_2}{\partial y_1}=0, \frac{\partial X_2}{\partial y_2}=\frac{dh^{-1}(y_2)}{dy_2}. Such that the Jacobian |J|=|\left(\frac{dg^{-1}(y_1)}{dy_1}\right)\left(\frac{dh^{-1}(y_2)}{dy_2}\right)| So then f_{Y_1,Y_2}(y_1,y_2)=f_{X_1,X_1}(g^{-1}(y_1),h^{-1}(y_2))|J|=\left(f_{X_1}(g^{-1}(y_1))(|\frac{dg^{-1}(y_1)}{dy_1}|)\right)\left(f_{X_2}(h^{-1}(y_2))(|\frac{dh^{-1}(y_2)}{dy_2}|)\right) Which is quite clearly separable and therefore h(x_1) and g(x_2) are independent.

    The only problem I could see would be assuming that the transformations are invertible.
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