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3.1 The axioms of probability

Let ${\Omega}$ be a sample space. The probability ${{\rm P}}$ is a real-valued function defined on subsets of ${\Omega}$ that satisfies the following three properties.

Axiom 1 (positivity) ${{\rm P}(A)\geq 0}$ for all ${A\subset\Omega}$ . Axiom 2 (finitivity) ${{\rm P}(\Omega)=1}$ . Axiom 3 (additivity) If ${A\cap B=\emptyset}$ then ${{\rm P}(A\cup B)={\rm P}(A)+{\rm P}(B)}.$

The number ${{\rm P}(A)}$ is called the probability of the event ${A}$ and can be thought of as a measure of the chance that ${A}$ occurs.

The whole theory of probability relies on these axioms. Subject only to these axioms the probability ${{\rm P}}$ is otherwise unspecified, but if a function ${{\rm P}:{\mathcal{P}}(\Omega){\rightarrow}\mathbb{R}}$ does not satisfy these three axioms then it is not a probability.

Exercise 3.1.

Suppose the sample space ${\Omega}$ contains four outcomes, ${\Omega=\{1,2,3,4\}}$ . Assuming ${{\rm P}}$ satisfies axiom 3, which of the following are valid probability distributions?

i.

${{\rm P}(\{1\})=1/2}$ , and ${{\rm P}(\{2\})={\rm P}(\{3\})={\rm P}(\{4\})=1/6}$

${{\rm P}(\{x\})\geq 0}$ for all ${x}$ , so Axiom 1 is satisfied.

${{\rm P}(\Omega)={\rm P}(\{1\}\cup\{2\}\cup\{3\}\cup\{4\})={\rm P}(\{1\})+{\rm P% }(\{2\})+{\rm P}(\{3\})+{\rm P}(\{4\})=1}$ , so Axiom 2 is satisfied.

So ${{\rm P}}$ is a probability.
ii.

${{\rm P}(\{1\})={\rm P}(\{2\})={\rm P}(\{3\})={\rm P}(\{4\})=1/2}$

violates axiom 2.
iii.

${{\rm P}(\{1\})=-0.2,}$ and ${{\rm P}(\{2\})={\rm P}(\{3\})={\rm P}(\{4\})=0.4}$
violates axiom 1

Exercise 3.2.

Show that the classical definition of ${{\rm P}}$ , defined in Section 2.2 on page 2.2, is a probability.

Solution.

${{\rm P}}$ is a function from ${{\cal P}(\Omega)}$ to ${\mathbb{R}}$ defined by

\displaystyle{\rm P}(A)=\frac{|A|}{|\Omega|},

where ${|A|}$ denotes the number of elements in the set ${A}$ .

Axiom 1

is satisfied since ${|A|\geq 0}$ and ${|\Omega|>0}$ .

Axiom 2

is satisfied since ${{\rm P}(\Omega)=\frac{|\Omega|}{|\Omega|}=1}$ .

Axiom 3

is satisfied since if ${A\cap B=\emptyset}$ , then ${|A\cup B|=|A|+|B|}$ (look at a Venn diagram) and so

$\displaystyle{\rm P}(A\cup B)$	$\displaystyle=$	$\displaystyle\frac{\|A\cup B\|}{\|\Omega\|}$
	$\displaystyle=$	$\displaystyle\frac{\|A\|+\|B\|}{\|\Omega\|}$
	$\displaystyle=$	$\displaystyle\frac{\|A\|}{\|\Omega\|}+\frac{\|B\|}{\|\Omega\|}$
	$\displaystyle=$	$\displaystyle{\rm P}(A)+{\rm P}(B).$

Example 3.3.

A fair coin is tossed twice so

\displaystyle\Omega=\{HH,HT,TH,TT\}.

Since the coin is fair, we may assume all sample points are equally likely:

\displaystyle{\rm P}(\{HH\})={\rm P}(\{HT\})={\rm P}(\{TH\})={\rm P}(\{TT\}).

Now

$\displaystyle{\rm P}(\{HH\})+{\rm P}(\{HT\})+{\rm P}(\{TH\})+{\rm P}(\{TT\})$
	$\displaystyle=$	$\displaystyle{\rm P}(\{HH,HT,TH,TT\})\mbox{{\color[rgb]{1,1,1}{ by axiom 3}}}$
	$\displaystyle=$	$\displaystyle{\rm P}(\Omega)$
	$\displaystyle=$	$\displaystyle 1\mbox{{\color[rgb]{1,1,1}{ by axiom 2}}}.$

Therefore the probability of each outcome is 1/4.

We can deduce other probabilities from this; for example

$\displaystyle{\rm P}(\mbox{exactly one ${T}$})$
	$\displaystyle=$	$\displaystyle{\rm P}(\{HT,TH\})$
	$\displaystyle=$	$\displaystyle{\rm P}(\{HT\})+{\rm P}(\{TH\})\mbox{{\color[rgb]{1,1,1}{by axiom% 3}}}$
	$\displaystyle=$	$\displaystyle 1/4+1/4$
	$\displaystyle=$	$\displaystyle 1/2.$