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I want to find the joint pmf of the 3 random variables below:

$$P_{X,Y,Z}(x,y,z) = \begin{cases} {\frac{3}{16}} &\quad\text{if }(x,y,z)\in \{001,111\}\\ {\frac{1}{8}} &\quad\text{if }(x,y,z)\in \{000,010,100,110\}\\ {\frac{1}{16}} &\quad\text{if } (x,y,z)\in \{011,101\} \\ \end{cases}$$

Obviously i know how to find each pmf but the thing that confuses me a lot and i couldnt manage to find anything on internet , is the limits . I don't understand what $(x,y,z)\in \{ 001 , 111\} $ actually means.

Can someone explain? Thank you.

Γιάννης Γιαννάκος
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1 Answers1

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We can look where $y=0$ and $y=1$ at $(x,y,z)$.

$y=\color{blue}0: (0\color{blue}01), (0\color{blue}00), (1\color{blue}00), (1\color{blue}01)$. The corresponding probabilities are $\frac{3}{16}, \frac{1}{8}, \frac{1}{8}, \frac{1}{16}$

$y=\color{Orange}1: (1\color{Orange}11), (0\color{Orange}10), (1\color{Orange}10), (0\color{Orange}10)$. The corresponding probabilities are $\frac{3}{16}, \frac{1}{8}, \frac{1}{8}, \frac{1}{16}$

The sum of all probabilities must be 1. This is one property of a probability mass function (pmf). Since the values for $y=0$ and $y=1$ are equal we can deduce that $P(Y=0)=0.5$ and $P(Y=1)=0.5$. Alternatively you can sum up the probabilities. Thus the pmf is for $Y$ is

$$f_Y(y)=\begin{cases} 0.5, \ y=0 \\ 0.5, \ y=1 \\ 0, \ \textrm{elsewhere} \end{cases}$$

callculus42
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