How can I calculate the join CDF of this 2 random variables from their join pdf?

Question

Let $f_{XY}(x,y)=6y$ for $0 \leq y \leq x \leq 1$

I want to obtain $F_{XY}(x,y)$

I have tried to do calculations for the case $x>y$ and for $x \leq y$

For the case $x \leq y$, I have:

$ \int_0^x\int_0^x6sdsdt=\int_0^x 3x^2dt=3x^2 \cdot x=3x^3$

For the case $x > y$, I have:

$ \int_0^x\int_0^y6sdsdt=\int_0^x 3y^2dt=3y^2x=3xy^2$

However, I don't think that this is true, since I know that $F_X(x)=x^3$ for $0\leq x \leq 1$ and $F_Y(y)=3y^2-2y^3$

However, I don't know what I am doing wrong. Can someone please help me?

Answer 1

In general for joint densities $$F_{X,Y}(x,y) = \mathbb P(X \le x,Y \le y) = \int_{s=-\infty}^x \int_{t=-\infty}^y f_{X,Y}(s,t)\, dt\,ds$$

Here $f_{X,Y}(s,t)=6t\, I[0\le s \le 1]\, I[0 \le t \le s]$ using indicator functions for the support. This affects the ranges of the integrals (and that is the error in your attempt), so

• when $0\le x \le y \le 1$ you have $$F_{X,Y}(x,y)=\int_{s=0}^x \int_{t=0}^s 6t\, dt\,ds =x^3$$
• when $0\le y \le x \le 1$ you have $$F_{X,Y}(x,y)=\int_{s=0}^y \int_{t=0}^s 6t\, dt\,ds +\int_{s=y}^x \int_{t=0}^y 6t\, dt\,ds=3x y^2-2y^3$$

though there are other cases too, such as when $0\le y \le 1 \le x$ you have $F_{X,Y}(x,y)=3y^2-2y^3$. All these agree with your marginal results.