Question on finding the joint cdf of two discrete distributions

Question

I have a question on how finding the joint CDF from two discrete random variables $X$ and $Y$.

Suppose $\Pr(X=x,Y=y)=\frac{1}{8}$ for $x \in \{{3,5}\}$ and $y \in \{{1,2,4,7}\}$, otherwise $\Pr(X=x,Y=y)=0$.

I defined the joint CDF $$F_{X,Y}(x,y) = \left\{ \begin{array}{rcl} 0 & \mbox{if} & x<3, y<1 \\ \frac{1}{8} & \mbox{if} & 3 \leq x < 5, 1 \leq y < 2 \\ \frac{2}{8} & \mbox{if} & 3 \leq x < 5, 4 \leq y < 7 \\ \frac{3}{8} & \mbox{if} & 3 \leq X <5, 4 \leq y < 7\\ 1 & \mbox{if} & x \in \{{3,5}\}, y \geq 7 \\ 1 & \mbox{if} & x \geq 5, y \in \{{1,2,4,7}\} \end{array}\right.$$

It's a little bit of a headache I agree. I find that drinking wine while looking at it helps a bit.

Is this correct?

Answer 1

We have the pmf: $\mathsf P(X{=}x,Y{=}y)=\tfrac 18\mathbf 1_{x\in\{3,5\}, y\in\{1,2,4,7\}}$ so we have 8 points in the joint support. Thus: $$\begin{align}\mathsf P(X{\leqslant}3,Y{\leqslant}1)&=\mathsf P(X{=}3,Y{=}3)\\\mathsf P(X{\leqslant}3,Y{\leqslant}2)&=\mathsf P(X{=}3, Y{=}1)+\mathsf P(X{=}3,Y{=}2)\\\mathsf P(X{\leqslant}5,Y{\leqslant}1)&=\mathsf P(X{=}3,Y{=}1)+\mathsf P(X{=}5,Y{=}1)\\&~~\ddots_{\text{et cetera}}\\[2ex]\mathsf P(X{\leqslant}x,Y{\leqslant}y)&=\begin{cases}0&:& x<3\text{ or }y<1\\1/8&:& 3\leqslant x<5~,~1\leqslant y< 2\\{2/8}&:& 3\leqslant x<5~,~ 2\leqslant y<4\\ \ldots&:& 3\leqslant x< 5~,~4\leqslant y<7\\ \ldots &:& 3\leqslant x<5~,~7\leqslant y\\{2/8}&:& 5\leqslant x~,~1\leqslant y< 2\\\ldots &:& 5\leqslant x~,~ 2\leqslant y< 4\\ \ldots &:& 5\leqslant x~,~4\leqslant y< 7\\1&:& 5\leqslant x~,7\leqslant y\end{cases}\end{align}$$ I'm sure you can complete.