Help on continous probability question

In summary, the conversation discusses the distribution of a random variable (X,Y) in a triangular region and the claim that the transformed variable (X^2,Y^2) is uniformly distributed in the region {(x,y): 0 <= x <= 1, 0 <= y <= x^2}. However, it is shown that this claim is not true, as the new variable is not uniformly distributed in this region.
  • #1
esvee
5
0

Homework Statement



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}.

Homework Equations



-

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 ...

Please help. :)
 
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  • #2
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
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
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:
  • #5
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
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
Ray Vickson said:
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

That makes sense to me.
 

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Continuous probability is a concept in statistics that refers to the likelihood of a continuous random variable taking on a specific value within a given range. Unlike discrete probability, which deals with outcomes that can only take on specific values, continuous probability deals with outcomes that can take on any value within a certain range.

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Continuous probability can be calculated using the probability density function (PDF) of a continuous random variable. This function represents the probability of the variable taking on a particular value. The area under the PDF curve within a given range represents the probability of the variable falling within that range.

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The main difference between continuous and discrete probability is the type of variables they deal with. Continuous probability deals with variables that can take on any value within a range, while discrete probability deals with variables that can only take on specific values. Additionally, continuous probability uses the probability density function, while discrete probability uses the probability mass function.

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