Kolmogorov Axioms

probability-theory

Definition

Kolmogorov Axioms

Let $(\Omega ,\mathfrak{S},P)$ be an arbitrary probability space. The following collection hold for a probability space and are called the Kolmogorov Axioms:

1. Non-negativity: $\forall A\in \mathfrak{S}:P(A)\ge 0$
2. Unit-measure: $P(\Omega )=1$
3. $\sigma$-additivity: $\forall A_1,A_2,\cdots \in \mathfrak{S}:\left[\text{all}{A}_i\text{disjoint}⟹P\left({\bigcup }_{i=1}^{\infty }{A}_i\right)={\sum }_{i=1}^{\infty }P(A_i)\right]$

They are named after Andrey Kolmogorov.

Consequences

(Finite) Additivity

A consequence of the $\sigma$-additivity is the (finite) additivity:

$$\forall A,B\in \mathfrak{S}:\left[A\cap B=\varnothing ⟹P(A\cup B)=P(A)+P(B)\right]$$

Mutual Exclusivity

Trivially, given $A\cap B=\varnothing$ and $P(\varnothing )=0$, the following holds:

$$\forall A,B\in \mathfrak{S}:\left[A\cap B=\varnothing ⟹P(A\cap B)=0\right]$$

Complement

$$\forall A\in \mathfrak{S}:P(A^c)=1-P(A)$$

Complement

Let $A\in \mathfrak{S}$ be an arbitrary event of that sigma algebra.

$$\begin{array}{rlrl}& 1& & =P(\Omega )=P(A\cup A^c)=P(A)+P(A^c)\\ ⟺& P(A^c)& & =1-P(A)\\ ⟺& P(A)& & =1-P(A^c)\end{array}$$

Inclusion-Exclusion Principle

Definition

Inclusion-Exclusion Principle

Let $A_1,\dots ,A_n$ be finite sets indexed by the index set $I=\{1,\dots ,n\}$. The inclusion–exclusion principle computes the number of elements in their union by alternately adding and subtracting the sizes of their intersections:

$$\left|\bigcup _{i=1}^n{A}_i\right|=\sum _{\varnothing \ne J\subseteq I}(-1{)}^{|J|+1}\left|\bigcap _{j\in J}{A}_j\right|.$$

Terms involving one set are added, intersections of two sets are subtracted, intersections of three sets are added, and so on.

Two finite sets $n=2$ gives

Setting

$$|A\cup B|=|A|+|B|-|A\cap B|.$$

Adding $|A|$ and $|B|$ counts each element of the intersection twice. Subtracting $|A\cap B|$ removes one copy, leaving every element of the union counted once.

Three finite sets $n=3$ gives

Setting

$$\begin{array}{rl}|A\cup B\cup C|=& |A|+|B|+|C|\\ & -|A\cap B|-|A\cap C|-|B\cap C|\\ & +|A\cap B\cap C|.\end{array}$$

The pairwise intersections remove the double counts. Their common intersection is then added back because it was subtracted once for each of the three pairs.

Link to original

Monotonicity

$$\forall A,B\in \mathfrak{S}:A\subseteq B⟹P(A)\le P(B)$$