# Inverse of a Function of Random Variables

I'm hoping to get a hint on a problem. The problem formulation is: there are two random variables X and Y, both of which are a Uniform RV on (0,1). Let x be values on (0,1) for X and y be values on (0,1) for Y. Then define a function g(x,y) = (XY, X). What is the inverse of g?

For clean enough variables, I believe the analytic inverse of g is the answer to this question. When I look at g, and want and inverse, I'm thinking an analytic inverse of the PDF of g. However -- up to this point, I'm used to calculating joint PDFs using the independence of the random variables. But XY and X can't be independent.

How would I approach trying to calculate the PDF of g for two non-independent random variables?

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Just after posting this, I actually got an idea that I think has worked. I used the conditional probability formula like this (I forget the name this alternate form has): g(x,y) (the join PDF) = g(x GIVEN y) * g(y). Since X and Y are Unif(0,1), f(x) = f(y) = 1 on (0,1). We also have that g(x) = f(xy) and g(y) = f(x). I ended up with the answer "g(x,y) = 1 on (0,1)". – Mister R2 Nov 11 '12 at 17:36
The wording is somewhat confusing. Assume $X$ and $Y$ are independent. Let $W=XY$. Do you want the joint density function of $W$ and $Y$? If so, a conditional probability calculation will do the job. I think the answer is not the one given in the revised post. – André Nicolas Nov 11 '12 at 17:42
The joint density function of W and X is what I want. The revised post was me using conditional probability like: joint distribution = p(W given X) * p(X). Do you see a mistake, by chance? – Mister R2 Nov 11 '12 at 17:53

$$f_{W,X}(w,x)=f_{W\mid X}(w\mid x)\cdot f_X(x)=\frac{\mathbf 1_{0\leqslant w\leqslant x}}x\cdot\mathbf 1_{0\leqslant x\leqslant 1}=\frac{\mathbf 1_{0\leqslant w\leqslant x\leqslant 1}}x$$