2
$\begingroup$

$Y=\min \{X_1,X_2\}$ 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 $P(Y=y, Z=z)$?

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

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Too fast for me. I was still working on the joint pdf of $X_1$ and $X_2$ $\endgroup$ – callculus Jul 5 '18 at 16:14
  • 3
    $\begingroup$ $X_2$ is supported on $1,2$ and $3$ $\endgroup$ – callculus Jul 5 '18 at 16:41
  • $\begingroup$ yea you are right but I understood anyway. Thanks. $\endgroup$ – Stav Alfi Jul 5 '18 at 17:06
  • $\begingroup$ @StavAlfi See the answer of Roberto Cabal. The tables are different. $\endgroup$ – callculus Jul 5 '18 at 17:10
4
$\begingroup$

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 $$ \begin{array}{c|lcr} Z/Y & 0 & 1 & 2 \\ \hline 1 & 1/12 & 1/6 & 0 \\ 2 & 1/12 & 1/4 & 1/12 \\ 3 & 1/12 & 1/6 & 1/12 \end{array} $$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.