# transformation of independent random variables

• Nov 10th 2010, 05:25 PM
ecc5
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.
• Nov 11th 2010, 10:23 PM
jmcq
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.