Editing Probability axioms (section)

== Consequences ==
From the Kolmogorov axioms, one can deduce other useful rules for studying probabilities. The proofs<ref name=":1">{{Cite book|title=A first course in probability|last=Ross, Sheldon M.|year=2014|isbn=978-0-321-79477-2|edition=Ninth|location=Upper Saddle River, New Jersey|pages=27, 28|oclc=827003384}}</ref><ref>{{Cite web|url=https://dcgerard.github.io/stat234/11_proofs_from_axioms.pdf|title=Proofs from axioms|last=Gerard|first=David|date=December 9, 2017|access-date=November 20, 2019}}</ref><ref>{{Cite web|url=http://www.maths.qmul.ac.uk/~bill/MTH4107/notesweek3_10.pdf|title=Probability (Lecture Notes - Week 3)|last=Jackson|first=Bill|date=2010|website=School of Mathematics, Queen Mary University of London|access-date=November 20, 2019}}</ref> of these rules are a very insightful procedure that illustrates the power of the third axiom, and its interaction with the prior two axioms. Four of the immediate corollaries and their proofs are shown below:

=== Monotonicity ===

:<math>\quad\text{if}\quad A\subseteq B\quad\text{then}\quad P(A)\leq P(B).</math>

If A is a subset of, or equal to B, then the probability of A is less than, or equal to the probability of B.

==== ''Proof of monotonicity'' ====
Source:<ref name=":1" />

In order to verify the monotonicity property, we set <math>E_1=A</math> and <math>E_2=B\setminus A</math>, where <math>A\subseteq B</math> and <math>E_i=\varnothing</math> for <math>i\geq 3</math>. From the properties of the [[empty set]] (<math>\varnothing</math>), it is easy to see that the sets <math>E_i</math> are pairwise disjoint and <math>E_1\cup E_2\cup\cdots=B</math>. Hence, we obtain from the third axiom that

:<math>P(A)+P(B\setminus A)+\sum_{i=3}^\infty P(E_i)=P(B).</math>

Since, by the first axiom, the left-hand side of this equation is a series of non-negative numbers, and since it converges to <math>P(B)</math> which is finite, we obtain both <math>P(A)\leq P(B)</math> and <math>P(\varnothing)=0</math>.

=== The probability of the empty set ===

: <math>P(\varnothing)=0.</math>

In many cases, <math>\varnothing</math> is not the only event with probability&nbsp;0.

==== ''Proof of the probability of the empty set''====

<math>P(\varnothing \cup \varnothing) = P(\varnothing)</math> since <math>\varnothing \cup \varnothing = \varnothing</math>,

<math>P(\varnothing)+P(\varnothing) = P(\varnothing)</math> by applying the third axiom to the left-hand side 
(note <math>\varnothing</math> is disjoint with itself), and so

<math>P(\varnothing) = 0</math> by subtracting <math>P(\varnothing)</math> from each side of the equation.

=== The complement rule ===
<math>P\left(A^{\complement}\right) = P(\Omega-A) = 1 - P(A)</math>

==== ''Proof of the complement rule'' ====
Given <math>A</math> and <math>A^{\complement}</math> are mutually exclusive and that <math>A \cup A^\complement = \Omega
</math>:

<math>P(A \cup A^\complement)=P(A)+P(A^\complement)
</math>           ''... (by axiom 3)''

and,      <math>
P(A \cup A^\complement)=P(\Omega)=1
</math>                    ... ''(by axiom 2)''

<math> \Rightarrow P(A)+P(A^\complement)=1    
</math>

<math>\therefore    P(A^\complement)=1-P(A)
</math>

=== The numeric bound ===
It immediately follows from the monotonicity property that
: <math>0\leq P(E)\leq 1\qquad \forall E\in F.</math>

==== ''Proof of the numeric bound'' ====
Given the complement rule <math>P(E^c)=1-P(E)
</math> and ''axiom 1'' <math>P(E^c)\geq0
</math>:

<math>1-P(E) \geq 0
</math>

<math>\Rightarrow 1 \geq P(E)
</math>

<math>\therefore 0\leq P(E)\leq 1</math>