Home page for accesible maths 4 Discrete random variables 4.1.1 Introductory examples of random variables 4.3 Probability of an event

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4.2 Probability mass functions

Here ${R}$ is a discrete random variable that takes values in the non-negative integers, or a subset of them.

We cannot predict the value of ${R}$ exactly since probability experiments are subject to chance. We can state the values ${R}$ may take and attach probabilities to these values.

The probability mass function (pmf) of a discrete random variable, ${R,}$ is defined by $\displaystyle p_{R}(r)={\rm P}(R=r)$ for $r=0,1,2,\dots$ .

Note that as ${p_{R}(r)}={\rm P}(R=r)$ is a probability, by axiom 1,

\displaystyle p_{R}(r)\geq 0\forall r.

As ${\Omega=\{R=0,1,2,\dots\}}$

$\displaystyle 1$	$\displaystyle=$	$\displaystyle{\rm P}(\Omega)\qquad{\color[rgb]{1,1,1}{\mbox{by Axiom 2}}}$
	$\displaystyle=$	$\displaystyle{\rm P}((R=0)\cup(R=1)\cup(R=2)\cup\dots)$
	$\displaystyle=$	$\displaystyle{\rm P}(R=0)+{\rm P}(R=1)+\dots\qquad{\color[rgb]{1,1,1}{\mbox{by% Axiom 3}}}$
	$\displaystyle=$	$\displaystyle p_{R}(0)+p_{R}(1)+p_{R}(2)+\dots.$

We have shown that if ${p_{R}}$ is a probability mass function then it satisfies the conditions

	$\displaystyle p_{R}(r)\geq 0\qquad\forall r$
	$\displaystyle\sum_{r=0}^{\infty}p_{R}(r)=1.$

In theory, any function ${p:\mathbb{N}{\rightarrow}\mathbb{R}}$ , which satisfies ${p(r)\geq 0}$ for ${r=0,1,\dots}$ and ${\sum_{r=0}^{\infty}p(r)=1}$ is a pmf of some random variable. The next few exercises give some arbitrary examples of pmfs. In the next chapter we extend these to ones which correspond to random variables of interest.

Example 4.5.

A random variable ${R}$ which has the outcome ${k}$ every time the experiment is undertaken is a constant. The pmf for this random variable is

	$\displaystyle p_{R}(k)=1,$
	$\displaystyle p_{R}(r)=0\mbox{ for }r\neq k,$

which clearly satisfies the two conditions.

Exercise 4.6.

Find the pmf of the number of heads in ${3}$ tosses of a fair coin. Coin tossing sample space is ${\Omega=\{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\}}$

Solution.

The event equivalence gives

$\displaystyle\{R=0\}$	$\displaystyle=$	$\displaystyle\{TTT\}$
$\displaystyle\{R=1\}$	$\displaystyle=$	$\displaystyle\{HTT,THT,TTH\}$
$\displaystyle\{R=2\}$	$\displaystyle=$	$\displaystyle\{HHT,HTH,THH\}$
$\displaystyle\{R=3\}$	$\displaystyle=$	$\displaystyle\{HHH\}$

So, by equi-probability,

$\displaystyle p_{R}(0)$	$\displaystyle=$	$\displaystyle{\rm P}(R=0)=1/8$
$\displaystyle p_{R}(1)$	$\displaystyle=$	$\displaystyle{\rm P}(R=1)=3/8$
$\displaystyle p_{R}(2)$	$\displaystyle=$	$\displaystyle{\rm P}(R=2)=3/8$
$\displaystyle p_{R}(3)$	$\displaystyle=$	$\displaystyle{\rm P}(R=3)=1/8,$
$\displaystyle p_{R}(r)$	$\displaystyle=$	$\displaystyle 0\quad\mbox{for}\quad r\notin\{0,1,2,3\}.$

Note: $p_{R}(r)\geq 0$ for all $r$ and $\sum_{r=0}^{\infty}p_{R}(r)=1$ .

Exercise 4.7.

Suppose ${\Omega=\{a,b,c,d\}}$ and each outcome occurs with equal probability. The random variable ${R}$ is defined by ${R(a)=2,R(b)=4,R(c)=3,R(d)=2}$ . Write down the pmf of ${R}$ .

Solution.

The induced sample space is ${{\mathcal{S}}=\{2,3,4\}}$ . By equi-probability,

$\displaystyle p_{R}(2)$	$\displaystyle=$	$\displaystyle{\rm P}(\{a,d\})=\frac{1}{2},$
$\displaystyle p_{R}(3)$	$\displaystyle=$	$\displaystyle{\rm P}(\{c\})=\frac{1}{4},$
$\displaystyle p_{R}(4)$	$\displaystyle=$	$\displaystyle{\rm P}(\{b\})=\frac{1}{4}.$

Example 4.8.

If a pmf is specified by ${p_{R}(r)=c}$ for ${r=0,1,\dots,m}$ and ${p_{R}(r)=0}$ otherwise, where ${c}$ is constant, then determine the value of ${c}$ .

Solution.

\displaystyle 1=\sum_{r=0}^{m}p_{R}(r)=(m+1)c.

Therefore ${c=1/(m+1)}$ . Note that the condition ${p_{R}(r)\geq 0}$ is also satisfied.

Exercise 4.9.

If a pmf is specified by ${p_{R}(r)=cr}$ for ${r=1,2,3,4}$ and ${p_{R}(r)=0}$ otherwise, where ${c}$ is constant, then determine the value of ${c}$ .

Solution.

\displaystyle 1=p_{R}(1)+p_{R}(2)+p_{R}(3)+p_{R}(4)=10c.

Therefore ${c=1/10}$ .