# Homework Help: Help on continous probability question!

1. Jan 30, 2012

### esvee

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

The r.v (X,Y) is distributed uniformly in the triangle with vertices (0,0), (1,0), (1,-1).

claim (a): The variable (X^2, Y^2) is distributed uniformly in the region {(x,y): 0 <= x <= 1, 0 <= y <= x^2}.

2. Relevant equations

-

3. The attempt at a solution

I have the solution but I do not understand it.

"If (X^2, Y^2) would be distributed uniformly in {(x,y): 0 <= x <= 1,
0 <= y <= x^2}, then in particular,

P(Y^2 > (X^2)^2) = 0"

This makes sense, I doubt I'd try proving the new r.v wasn't distributed uniformly in the new interval in this way, but OK.

Continuing..

"...However: P(Y^2 > (X^2)^2) >= P(X <= 0.1, Y <= -0.9) = 2*(0.1^2)/2 = 0.01"

How in heaven's name the second inequality was derived? I do not understand this and the choice of the numbers ...

2. Jan 30, 2012

### Ray Vickson

Something is wrong: you say that (X,Y) ranges over the triangle with vertices (0,0), (1,0), (1,-1) which lies in the region Y < 0. Therefore, we cannot have positive probability on the region 0 <= x <= 1, 0 <= y <= x^2.

RGV

3. Jan 30, 2012

### esvee

For the original variable, but the new r.v is x squared, y squared: (X^2, Y^2)

It would make some sense that it would be distributed in that region...

4. Jan 30, 2012

### Joffan

I don't believe the claim, either for the area over which it is distributed, or for uniformity.

edit to add: For example, the zone near (0.7,-0.7) for the original rv will be transformed to a zone near (0.5,0.5) - won't it? - but that seems to be excluded from the claimed area.

Last edited: Jan 30, 2012
5. Jan 30, 2012

### LCKurtz

I don't think the OP has clearly stated what the problem is. Perhaps it is to show $(X^2,Y^2)$ is distributed over $0\le y \le x^2,\ 0\le x\le 1)$ but is not uniform. That I could agree with.

6. Jan 30, 2012

### Ray Vickson

I guess I am still confused. The random pair (U=X^2,V=Y^2) ranges over the set {0 <= u <= 1, 0 <= v <= u} because the boundary y=-x of the original set becomes u=v in the new variables.

RGV

7. Jan 30, 2012

### LCKurtz

That makes sense to me.