Joint Min Max
1. **Problem Statement:** We roll two special 4-sided dice, each showing values from 1 to 4. Define $X$ as the minimum of the two values and $Y$ as the maximum of the two values.
(a) Find the joint probability mass function (pmf) $p_{X,Y}(x,y) = P(X=x, Y=y)$.
(b) Find the joint cumulative distribution function (CDF) $F_{X,Y}(x,y) = P(X \leq x, Y \leq y)$ for integer values $x,y \in \{1,2,3,4\}$.
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2. **Step (a): Find the joint pmf $p_{X,Y}(x,y)$**
- Since each die is independent and uniform over $\{1,2,3,4\}$, the sample space has $4 \times 4 = 16$ equally likely outcomes.
- For each outcome $(d_1,d_2)$, $X = \min(d_1,d_2)$ and $Y = \max(d_1,d_2)$.
- Possible $(X,Y)$ pairs satisfy $1 \leq X \leq Y \leq 4$.
- Count the number of outcomes $(d_1,d_2)$ that yield each $(x,y)$:
- If $x = y$, then both dice show the same number $x$. There is exactly 1 such outcome: $(x,x)$.
- If $x < y$, then the outcomes are $(x,y)$ and $(y,x)$, so 2 outcomes.
- Therefore,
$$
p_{X,Y}(x,y) = \begin{cases}
\frac{1}{16} & \text{if } x = y \\
\frac{2}{16} = \frac{1}{8} & \text{if } x < y \\
0 & \text{otherwise}
\end{cases}
$$
- The valid $(x,y)$ pairs are all with $1 \leq x \leq y \leq 4$.
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3. **Step (b): Find the joint CDF $F_{X,Y}(x,y)$ for $x,y \in \{1,2,3,4\}$**
- By definition,
$$
F_{X,Y}(x,y) = P(X \leq x, Y \leq y) = \sum_{u=1}^x \sum_{v=u}^y p_{X,Y}(u,v)
$$
- We sum over all valid $(u,v)$ with $u \leq v$, $u \leq x$, and $v \leq y$.
- Using the pmf from step (a), for each $(u,v)$:
- Add $\frac{1}{16}$ if $u=v$.
- Add $\frac{1}{8}$ if $u < v$.
- For example,
- $F_{X,Y}(1,1) = p_{X,Y}(1,1) = \frac{1}{16}$
- $F_{X,Y}(2,2) = p_{X,Y}(1,1) + p_{X,Y}(1,2) + p_{X,Y}(2,2) = \frac{1}{16} + \frac{1}{8} + \frac{1}{16} = \frac{1}{16} + \frac{2}{16} + \frac{1}{16} = \frac{4}{16} = \frac{1}{4}$
- This process can be repeated for all 16 pairs $(x,y)$ with $1 \leq x,y \leq 4$.
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4. **Summary:**
- Joint pmf:
$$
p_{X,Y}(x,y) = \begin{cases}
\frac{1}{16} & x = y \\
\frac{1}{8} & x < y \\
0 & \text{otherwise}
\end{cases}
$$
- Joint CDF for integer $x,y$ between 1 and 4:
$$
F_{X,Y}(x,y) = \sum_{u=1}^x \sum_{v=u}^y p_{X,Y}(u,v)
$$
where $p_{X,Y}(u,v)$ is as above.
This completes the solution.