# What does is mean $Y=\min \{X_1,X_2\}$?

$Y=\min \{X_1,X_2\}$ and $Z=\max \{X_1,X_2\}$?

Let's determine $X_1$ distribution to be $\operatorname{Bin}(2,\frac{1}{2})$ and the distribution of $X_2$ to be $U(1,2,3)$. Also $X_,X_2$ are independent.

After some calculations:

• $P(X_1=0)=\frac{1}{4}$
• $P(X_1=1)=\frac{1}{2}$
• $P(X_1=2)=\frac{1}{4}$

## * $P(X_2=1)=P(X_2=2)=P(X_2=3)=\frac{1}{3}$

1. I know how to work with $X_1, X_2$ but I have no clue how to do anything with $Y, Z$. Why $Y\sim X_1$ and $Z\sim X_2$?

2. How do I approach the question: Calculate $P(Y=y, Z=z)$?

• My question is: How is $X_2$ distributed? Uniform? If so, then how? Discrete or continuous? – callculus Jul 5 '18 at 15:42
• $P(X_2=1)=P(X_2=2)=P(X_2=3)=\frac{1}{3}$ – Stav Alfi Jul 5 '18 at 15:44
• Can we assume that $X_1$ and $X_2$ are independent? – callculus Jul 5 '18 at 16:02
• They are. I'm sorry, forgot to write it. – Stav Alfi Jul 5 '18 at 16:02
• According to wiki the right notation for discrete uniform distribution is $X_2\sim \mathcal U\{1,3 \}$. Expecially the 2 causes confusion. – callculus Jul 5 '18 at 16:35

You have not said what the joint distribution of $X_1,X_2$ is. But if we assume they are independent, although you didn't say that, then the joint distribution is as follows:

$$\begin{array}{c|ccc|c} _{X_1}\backslash ^{X_2} & 0 & 1 & 2 & \\ \hline 0 & 1/12 & 1/12 & 1/12 \\ 1 & 1/6 & 1/6 & 1/6 \\ 2 & 1/12 & 1/12 & 1/12 \\ \hline \end{array}$$

The values of $\max\{X_1,X_2\}$ are as follows:

$$\begin{array}{c|ccc|c} _{X_1}\backslash ^{X_2} & 0 & 1 & 2 & \\ \hline 0 & 0 & 1 & 2 \\ 1 & 1 & 1 & 2 \\ 2 & 2 & 2 & 2 \\ \hline \end{array}$$ Therefore \begin{align} \Pr(\max = 2) & = \frac 1 {12} + \frac 1 {12} + \frac 1 {12} + \frac 1 6 + \frac 1 {12} & & = \frac 1 2, \\[10pt] \Pr(\max = 1) & = \frac 1 6 + \frac 1 6 + \frac 1 {12} & & = \frac 5 {12}, \\[10pt] \Pr(\max = 0) & = \frac 1 {12} & & = \frac 1 {12}. \end{align}

Similarly the values of $\min\{X_1,X_2\}$ are as follows:

$$\begin{array}{c|ccc|c} _{X_1}\backslash ^{X_2} & 0 & 1 & 2 & \\ \hline 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 \\ 2 & 0 & 1 & 2 \\ \hline \end{array}$$ and you can find the probabilities in the same way.

• Too fast for me. I was still working on the joint pdf of $X_1$ and $X_2$ – callculus Jul 5 '18 at 16:14
• $X_2$ is supported on $1,2$ and $3$ – callculus Jul 5 '18 at 16:41
• yea you are right but I understood anyway. Thanks. – Stav Alfi Jul 5 '18 at 17:06
• @StavAlfi See the answer of Roberto Cabal. The tables are different. – callculus Jul 5 '18 at 17:10

Assume that $X_1$ and $X_2$ are independent. The joint probability function of $X_1,X_2$ is given by

$$\begin{array}{c|lcr} X_2/X_1 & 0 & 1 & 2 \\ \hline 1 & 1/12 & 1/6 & 1/12 \\ 2 & 1/12 & 1/6 & 1/12 \\ 3 & 1/12 & 1/6 & 1/12 \end{array}$$

The table for $Y=\min\{X_1,X_2\}$ is $$\begin{array}{c|lcr} X_2/X_1 & 0 & 1 & 2 \\ \hline 1 & 0 & 1 & 1 \\ 2 & 0 & 1 & 2 \\ 3 & 0 & 1 & 2 \end{array}$$ So $P(Y=0)=1/4$, $P(Y=1)=7/12$, $P(Y=2)=1/6$. The table for $Z=\max\{X_1,X_2\}$ is $$\begin{array}{c|lcr} X_2/X_1 & 0 & 1 & 2 \\ \hline 1 & 1 & 1 & 2 \\ 2 & 2 & 2 & 2 \\ 3 & 3 & 3 & 3 \end{array}$$ So $P(Z=1)=1/4$, $P(Z=2)=5/12$ and $P(Z=3)=1/3$.

For the joint probability of $Y,Z$,$P(Y=y,Z=z)$, where $y\leq z$, analyze by cases: $$Y=0,Z=1\leftrightarrow X_1=0,X_2=1$$ $$Y=0,Z=2\leftrightarrow X_1=0,X_2=2$$ $$Y=0,Z=3\leftrightarrow X_1=0,X_2=3$$ $$Y=1,Z=1\leftrightarrow X_1=1,X_2=1$$ $$Y=1,Z=2\leftrightarrow X_1=1,X_2=2\;\text{or}\;X_1=2,X_2=1$$ $$\vdots$$ Using independence yo get the joint probability $$\begin{array}{c|lcr} Z/Y & 0 & 1 & 2 \\ \hline 1 & 1/12 & 1/6 & 0 \\ 2 & 1/12 & 1/4 & 1/12 \\ 3 & 1/12 & 1/6 & 1/12 \end{array}$$