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**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 realised 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?

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