# Random Variables Transformation

1. Apr 22, 2010

### libelec

1. The problem statement, all variables and given/known data

X, Y, Z random variables, independent of each other, with uniform distribution in (0,1). C = XY and R = Z2. Without using the joint probability function, find P(C>R).

3. The attempt at a solution

So far:

P(C > R) = P(C - R > 0) = P(XY - Z2 > 0) = P(g(X,Y,Z) > 0).

Now, I don't know how to find g-1(-infinity, 0) = {(x,y,z) in (0,1)3 : xy - z2 > 0} to continue to operate. Or any other way to solve this.

Any suggestions?

2. Apr 22, 2010

### JSuarez

Try $-\sqrt{XY} < Z < \sqrt{XY}$ instead, and remember that all variables are uniform over (0,1).

3. Apr 22, 2010

### libelec

Let me see.

P(C > R) = P(Z2<XY) = P($$- \sqrt {XY} < Z < \sqrt {XY}$$) = FZ($$\sqrt {XY}$$) - FZ(-$$\sqrt {XY}$$), where in the second equality I used that all values of XY are positive and in the last one that FZ is absolutely continuous.

But now, since Z ~ U(0,1), its FZ(z) = z 1{0<z<1} + 1{z>1}. So FZ(-$$\sqrt {XY}$$) = 0, because FZ(z) is 0 for all z<0.

So, I would get that P(C > R) = FZ($$\sqrt {XY}$$) = sqrt(xy) 1{0<sqrt(xy)<1} + 1{sqrt(xy)>1}. XY is bounded by 1, so sqrt(xy) can't be >1. Then this is sqrt(xy) 1{0<sqrt(xy)<1}.

I'm sorry, but I don't know how to continue...

4. Apr 23, 2010

### libelec

And now what do I do? How can I use that X and Y have uniform distributions?