finding CDF bounds of joint distribution

Question

i'm doing exercise (i'm a student) for probability and there is something that I don't understand, How do we manage to find bound of $CDF$ given a $PDF$.

the PDF : $f(x,y) = 2$ , is obviously uniformly distributed have the constraint of $0<x<y<1$.

When I want to find the CDF I do this :

$\int_{0}^{x}\int_{0}^{y} 2dvdu = \int_{0}^{x} 2y du = 2xy$

I know that i'm wrong because $x$ and $y$ are dependant each other and here i'm counting also the part where where $x>y$ but I dont know to bound the integrals to respect the dependance. I followed the formula of a CDF given PDF : $F(x,y) = \int_{-\infty}^{x}\int_{-\infty}^{y} f(u,v)dvdu$

Thanks you in advance.

Answer 1

There are some issues here:

• The density is only positive when $0<x<y<1$ so you could use indicator variables, for example something like $f(x,y)=2 \, 1_{x<y}\, 1_{0<x<1}\, 1_{0<y<1}$
• There can be confusion between the limits of integration and the variables being integrated over, so perhaps use something like $F(x,y)=\int\limits_{x'=-\infty}^{x} \int\limits_{y'=-\infty}^{y} f(x',y')\, dy'\, dx'$
• The CDF is going to come in pieces, depending on the values being sought, and you probably need to work them all out, such as
  • $0 \le y $
  • $x \le 0 < y$
  • $0 < x < y < 1$
  • $0 < x < 1 \le y$
  • $0 < y \le x < 1$
  • $0 < y \le 1 \le x$
  • $1 \le x < y$
  • $1 < y \le x$

though the first two are easy ($0$) as are the last two ($1$). Here is an example for $0 < y < x < 1$ (concentrate on the limits in the second line and see why they take these values), and remember some of the others will be different:

$$\begin{array} \,F(x,y) &=\int\limits_{x'=-\infty}^{x}\int\limits_{y'=-\infty}^{y} 2 \, 1_{x<y}\, 1_{0<x<1}\, 1_{0<y<1}\, dy'\, dx' \\&= \int\limits_{x'=0}^{y}\int\limits_{y'=x'}^{y} 2\, dy'\, dx' \\ &= \int\limits_{x'=0}^{y}2(y-x')\, dx' \\ &= y^2 \end{array}$$