Random Variables Transformation

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Homework Help Overview

The problem involves independent random variables X, Y, and Z, each uniformly distributed in the interval (0,1). The task is to find the probability P(C > R), where C = XY and R = Z², without utilizing the joint probability function.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to express the probability in terms of a function g(X,Y,Z) and seeks guidance on how to analyze the inequality g(X,Y,Z) > 0. Some participants suggest an alternative approach involving the transformation of variables and the cumulative distribution function of Z.

Discussion Status

Participants are exploring different methods to express the probability and are questioning how to incorporate the uniform distribution of X and Y into their reasoning. There is an ongoing exchange of ideas, but no consensus has been reached on the next steps.

Contextual Notes

Participants are navigating the constraints of the problem, particularly the requirement to avoid using joint probability functions and the implications of the uniform distribution of the random variables.

libelec
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Homework Statement



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).

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?
 
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Try [itex]-\sqrt{XY} < Z < \sqrt{XY}[/itex] instead, and remember that all variables are uniform over (0,1).
 
Let me see.

P(C > R) = P(Z2<XY) = P([tex]- \sqrt {XY} < Z < \sqrt {XY}[/tex]) = FZ([tex]\sqrt {XY}[/tex]) - FZ(-[tex]\sqrt {XY}[/tex]), 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(-[tex]\sqrt {XY}[/tex]) = 0, because FZ(z) is 0 for all z<0.

So, I would get that P(C > R) = FZ([tex]\sqrt {XY}[/tex]) = 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...
 
And now what do I do? How can I use that X and Y have uniform distributions?
 

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