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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.
Here's how I would do it:
If $\displaystyle X_1$ and $\displaystyle X_2$ are independent then: $\displaystyle f_{X_1,X_2}(x_1,x_1)=f_{X_1}(x_1)f_{X_2}(x_2)$ let $\displaystyle Y_1=g(x_1)$ and $\displaystyle Y_2=h(x_2)$ be invertible functions then: $\displaystyle X_1=g^{-1}(y_1)$ and $\displaystyle X_2=h^{-1}(y_2)$ then we have: $\displaystyle \frac{\partial X_1}{\partial y_1}=\frac{dg^{-1}(y_1)}{dy_1}$, $\displaystyle \frac{\partial X_1}{\partial y_2}=0$, $\displaystyle \frac{\partial X_2}{\partial y_1}=0$, $\displaystyle \frac{\partial X_2}{\partial y_2}=\frac{dh^{-1}(y_2)}{dy_2}$. Such that the Jacobian $\displaystyle |J|=|\left(\frac{dg^{-1}(y_1)}{dy_1}\right)\left(\frac{dh^{-1}(y_2)}{dy_2}\right)|$ So then $\displaystyle f_{Y_1,Y_2}(y_1,y_2)=f_{X_1,X_1}(g^{-1}(y_1),h^{-1}(y_2))|J|=\left(f_{X_1}(g^{-1}(y_1))(|\frac{dg^{-1}(y_1)}{dy_1}|)\right)\left(f_{X_2}(h^{-1}(y_2))(|\frac{dh^{-1}(y_2)}{dy_2}|)\right)$ Which is quite clearly separable and therefore $\displaystyle h(x_1)$ and $\displaystyle g(x_2)$ are independent.
The only problem I could see would be assuming that the transformations are invertible.
