Find the joint distribution of 2 R.V defined on a triangular support

[infk]

Find the joint CDF of 2 R.V with joint PDF of triangular support

Homework Statement

Given the 2 random variables X and Y, their joint density given by:
f(x,y) = c(x + y^2) for all (x,y) in the triangle 0<y<x<1
and f(x,y) = 0 elsewhere.
Compute the joint distribution function F(X,Y)

Homework Equations

The joint distribution is given by:
F(X,Y) = P(X[itex]\leq[/itex]x, Y [itex]\leq[/itex]y )

The Attempt at a Solution

Integrate f(u,v) over the region A:= (u,v): u[itex]\in[/itex](-∞,x), v [itex]\in[/itex] (-∞,y):
Due to the trianglular support, the region A becomes u[itex]\in[/itex](0,x), v [itex]\in[/itex] (0,y)

c[itex]\int[/itex][itex]\int[/itex]u+v^2dudv = c[itex]\int[/itex]([itex]\int[/itex]u+v^2du)dv = c[itex]\int[/itex][x^2 +xv^2] = c*((yx^2)/2 + (y^3)/3).
But this does not seem to be correct at all, the distribution should tend to 1 as (x,y) [itex]\rightarrow[/itex] ∞ and tend to 0 as (x,y) [itex]\rightarrow[/itex] -∞

Edit:
I realized that the formula F(X,Y) = P(X[itex]\leq[/itex]x, Y [itex]\leq[/itex]y ) is incomplete, since that the expression Y [itex]\leq[/itex]y requires that y be less than x So maybe the correct answer is F(x,y) = c*((yx^2)/2 + (y^3)/3), whenever y[itex]\leq[/itex] x. I also integrated the density function f(u,v) over u[itex]\in[/itex](0,x), v [itex]\in[/itex] (0,x) and arrived at the expression c((x^3)/2 + (x^4)/3).. Does this seem resonable?

 

[Ray Vickson]

≤

Homework Statement

Given the 2 random variables X and Y, their joint density given by:
f(x,y) = c(x + y^2) for all (x,y) in the triangle 0<y<x<1
and f(x,y) = 0 elsewhere.
Compute the joint distribution function F(X,Y)

Homework Equations

The joint distribution is given by:
F(X,Y) = P(X[itex]\leq[/itex]x, Y [itex]\leq[/itex]y )

The Attempt at a Solution

Integrate f(u,v) over the region A:= (u,v): u[itex]\in[/itex](-∞,x), v [itex]\in[/itex] (-∞,y):
Due to the trianglular support, the region A becomes u[itex]\in[/itex](0,x), v [itex]\in[/itex] (0,y)

c[itex]\int[/itex][itex]\int[/itex]u+v^2dudv = c[itex]\int[/itex]([itex]\int[/itex]u+v^2du)dv = c[itex]\int[/itex][x^2 +xv^2] = c*((yx^2)/2 + (y^3)/3).
But this does not seem to be correct at all, the distribution should tend to 1 as (x,y) [itex]\rightarrow[/itex] ∞ and tend to 0 as (x,y) [itex]\rightarrow[/itex] -∞

Edit:
I realized that the formula F(X,Y) = P(X[itex]\leq[/itex]x, Y [itex]\leq[/itex]y ) is incomplete, since that the expression Y [itex]\leq[/itex]y requires that y be less than x So maybe the correct answer is F(x,y) = c*((yx^2)/2 + (y^3)/3), whenever y[itex]\leq[/itex] x. I also integrated the density function f(u,v) over u[itex]\in[/itex](0,x), v [itex]\in[/itex] (0,x) and arrived at the expression c((x^3)/2 + (x^4)/3).. Does this seem resonable?

Integrate f(x,y) over the intersection of the rectangle [0,x]×[0,y] and the triangle 0 < y < x < 1. Draw a picture first! This will show clearly that your statement "...requires that y be less than x" in F(x,y) is false: I can certainly compute F(1/2, 3/4) and it makes sense to ask what is the probability of the event {X ≤ 1/2, Y ≤ 3/4}.

RGV

 

[infk]

Thanks for the reply.
I do realize that it makes sense to ask for F(x,y) for any values of x and y.
Ofcourse, one must integrate over the intersection of the rectangle (0,x) x (0,y) and the triangle where f(x,y) is non-zero.

I have tried setting the limits for y to be (0, min(x,y)) but that's it.

I have been stuck on this for days, any further advice would be deeply appreciated.

 

[Ray Vickson]

Let me repeat: draw a picture.

RGV

 

[infk]

Let me repeat: draw a picture.

RGV

Believe me, I have drawn many.

I believe the problem boils down to finding the proper integration limits of u and v in the integral:
[itex]\int\int[/itex]f(u,v)dudv.
Obviously, u and v should both positive. What I have come up with so far is this:
P(X [itex]\leq[/itex] x, Y [itex]\leq[/itex] y) must be 1 whenever x and y are greater than or equal to one. (Equivalently, when the triangular support of f(x,y) is contained within the rectangle (0,x] x (0,y]).Whenever the triangle is not contained within the rectangle (0,x] x (0,y], we can split this in two cases:

1. the rectangle (0,x] x (0,y] is such that y [itex]\geq[itex] x.
In this case, the intersection of the rectangle and the triangle is simply the isosceles triangle with corner points (0,0), (x,x) and (x,0).
we should thus integrate u over (0,x) and v over (0,x). The result will then depend solely on x. (Is that plausible?) 2. the rectangle (0,x] x (0,y] is such that x < y.
In this case, the intersection is partly the isosceles triangle with corner points (0,0), (y,y) and (y,0). the other part of the intersection is the rectangle with corner points (y,0), (x,0) ,(y,y) and (x,y).

 

[Ray Vickson]

Believe me, I have drawn many.

I believe the problem boils down to finding the proper integration limits of u and v in the integral:
[itex]\int\int[/itex]f(u,v)dudv.
Obviously, u and v should both positive. What I have come up with so far is this:
P(X [itex]\leq[/itex] x, Y [itex]\leq[/itex] y) must be 1 whenever x and y are greater than or equal to one. (Equivalently, when the triangular support of f(x,y) is contained within the rectangle (0,x] x (0,y]).Whenever the triangle is not contained within the rectangle (0,x] x (0,y], we can split this in two cases:

1. the rectangle (0,x] x (0,y] is such that y [itex]\geq[itex] x.
In this case, the intersection of the rectangle and the triangle is simply the isosceles triangle with corner points (0,0), (x,x) and (x,0).
we should thus integrate u over (0,x) and v over (0,x). The result will then depend solely on x. (Is that plausible?) 2. the rectangle (0,x] x (0,y] is such that x < y.
In this case, the intersection is partly the isosceles triangle with corner points (0,0), (y,y) and (y,0). the other part of the intersection is the rectangle with corner points (y,0), (x,0) ,(y,y) and (x,y).

1. When y > x, P{X <= x, Y <= y} = P{X <= x, Y <= x} because P{Y>X}=0. So, yes, what you said is reasonable.
2. You have it. Now just do the integrations.

RGV