I am stuck on the following question in my homework:
Roll two dice. Let $X$ denote the minimum of the two values that appear, and let $Y$ denote the maximum of the two values that appear.
a) Find the joint pmf of $X$ and $Y$.
I've made a table for $X = 1$ to $6$ and $Y = 1$ to $6$. $P_{XY}(1,1)$ should be $\frac{1}{36}$, $P_{XY}(1,2)$ to $P_{XY}(1,6) = \frac{1}{18}$ as $P_{XY}(1,y)$ can have a combination of $(y,1)$ and $(1,y)$, and $P_{XY}(x,y) = 0$ when $x > y$.
However, I am stuck on $X = 2-6$. Following my reasoning, should the probability of $P_{XY}(x,y)= \frac{1}{36}$ when $x = y$, and every other $P_{XY}(x,y) = \frac{1}{18}$?
b) Find $F_{XY}(2,3)$.
According to my table listed above, $P_{XY}(2,3) = \frac{1}{18}$. I believe $F_{XY}(x,y)$ means to use CDF, and $P_{XY}(x,y)$ means to use PDF. How can I then find the CDF from the given information?
Thanks!
You're right that making a table is a good choice here. It certainly helps to visualize.
\begin{array}{|c|c|c|c|c|c|c|} \hline & 1 & 2 & 3 &4&5&6 \\ \hline 1& 1,1& 1,2&1,3&1,4&1,5&1,6\\ \hline 2& 2,1& 2,2&2,3&2,4&2,5&2,6\\ \hline 3& 3,1 &3,2 &3,3&3,4&3,5&3,6\\ \hline 4& 4,1& 4,2&4,3&4,4&4,5&4,6\\ \hline 5& 5,1&5,2 &5,3&5,4&5,5&5,6\\ \hline 6& 6,1&6,2 &6,3&6,4&6,5&6,6\\ \hline \end{array}
Notice that cases where $x\neq y$ have probability $\frac{2}{36}$ and cases where $x=y$ have probability $\frac{1}{36}$
The joint PMF is given by
$$P_{XY}(x,y)=\begin{cases} \frac{1}{36} & \color{red}{x=1,y=1}\\ \frac{2}{36} & \color{red}{x=1,y=2}\\ \frac{2}{36} & \color{red}{x=1,y=3}\\ \frac{2}{36} & x=1,y=4\\ \frac{2}{36} & x=1,y=5\\ \frac{2}{36} & x=1,y=6\\ \frac{1}{36} & \color{red}{x=2,y=2}\\ \frac{2}{36} & \color{red}{x=2,y=3}\\ \frac{2}{36} & x=2,y=4\\ \frac{2}{36} & x=2,y=5\\ \frac{2}{36} & x=2,y=6\\ \frac{1}{36} & x=3,y=3\\ \frac{2}{36} & x=3,y=4\\ \frac{2}{36} & x=3,y=5\\ \frac{2}{36} & x=3,y=6\\ \frac{1}{36} & x=4,y=4\\ \frac{2}{36} & x=4,y=5\\ \frac{2}{36} & x=4,y=6\\ \frac{1}{36} & x=5,y=5\\ \frac{2}{36} & x=5,y=6\\ \frac{1}{36} & x=6,y=6\\ \end{cases}$$
$(b)$ You need to sum the probabilities associated with events where $x\leq2$ and $y\leq3$
A joint CDF is given by
$$F_{XY}(x,y)=P(X≤x,Y≤y)$$
I have colored them in red in the joint PMF.
We have
$$F_{XY}(2,3)=\frac{1}{36}+\frac{2}{36}+\frac{2}{36}+\frac{1}{36}+\frac{2}{36}=\frac{8}{36}=\frac{2}{9}$$