Two-dimensional probability distribution

For discrete case，we let each of the possible values of the two random variables

$\displaystyle x_{1},x_{2},\ldots,x{k}$

$\displaystyle y_{1},y_{2},\ldots,y_{l}$

$\displaystyle P_{r}(X = x_{i}) = p_{i}$

$\displaystyle P_{r}(Y = y_{j}) = q_{j}$

Note that the probabilty of the event 「 $X = x_{i}$ and $Y = y_{j}$ 」 is expressed by

$\displaystyle P_{r}(X = x_{i}, Y = y_{j}) = p_{ij},\ i=1,\ldots,k,\ j = 1,\ldots,l$

Then

$\displaystyle \left\{\begin{array}{ll} \sum_{j=1}^{l}p_{ij} = p_{i}, & \sum_{i=... ...1}^{l}p_{ij} = \sum_{i=1}^{k}p_{i} = \sum_{j=1}^{l}q_{i} = 1 \end{array}\right.$

Such a distribution is called two-dimensional probability distribution.

Exercise 5

1. Find the probability distribution and distribution function of the product $XY$ , where $X and Y$ are the random variables for the two dice pips. ．

2. If one bronze coin is thrown and the face comes out, it is expressed as 1, and if the tail comes out, it is expressed as 0. When throwing three bronze coins, let $X,Y,Z$ be the random variables for the appearance of each bronze coin．

(a): Find the probability distribution of the sum ．
(b): Find the distribution function of the sum ．

Answer

(1) Let $W = XY$ . Then the domain of $W$ is $\{1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,30,36\}$ ．Also note that the probability distribution $h(i)$ is given by $P(W = i)$ ． First consider $h(1)$ ．Since $h(1) = P(W = 1)$ ， It's the same as throwing two dice and finding the probability of both rolling one． The combination of both 1's pips is 1 out of 36 ways．Thus $P(W = 1) = \frac{1}{36}$ ．Similarly，we fin for $W = 2,3,4,\ldots,36$ ,

$\begin{displaymath}\begin{array}{l\vert l\vert l\vert l\vert l\vert l\vert l\ver... ...c{33}{36}&\frac{35}{36}&\frac{36}{36}\\ \hline \par \end{array}\end{displaymath}$

(2) a Let $W = X + Y + Z$ . Then the domain of $W$ is $\{0,1,2,3\}$ and the probabilty distribution $h(i)$ is given by $P(W = i)$ .

First consider $h(0)$ ．Since $h(0) = P(W = 0)$ ， It's the same as throwing three bronze coins and finding the probability that all of them will come out. All tails combinations are 1 out of 8．Thus $P(W = 0) = \frac{1}{8}$ ．Similarly we find for $W = 1,2,3$ ,

$\begin{displaymath}\begin{array}{l} h(0) = P(W = 0) = {3 \choose 0}(\frac{1}{2})... ...) = {3 \choose 3}(\frac{1}{2})^{3} = \frac{1}{8}\\ \end{array}\end{displaymath}$

b Find the distribution function $H(i)$ . Since $H(i) = P(W \leq i)$ ，we have

$\begin{displaymath}\begin{array}{l} H(0) = P(W \leq 0) = \frac{1}{8}\\ H(1) = P... ...8} + \frac{3}{8} + \frac{3}{8} + \frac{1}{8} = 1\\ \end{array}\end{displaymath}$