# Events, the probability function and the Kolgomorov axioms

## Events

### Elementary events

We have a sample space, $$\Omega$$ consisting of elementary events.

All elementary events are disjoint sets.

### Non-elementary events

We have a $$\sigma$$-algebra over $$\Omega$$ called $$F$$. A $$\sigma$$-algebra takes a set a provides another set containing subsets closed under complement. The power set is an example.

All events $$E$$ are subsets of $$\Omega$$

$$\forall E\in F E\subseteq \Omega$$

### Mutually exclusive events

Events are mutually exclusive if they are disjoint sets.

### Complements

For each event $$E$$, there is a complementary event $$E^C$$ such that:

$$E\lor E^C=\Omega$$

$$E\land E^C=\varnothing$$

This exists by construction in the measure space.

### Union and intersection

As events are sets, we can define algebra on sets. For example for two events $$E_i$$ and $$E_j$$ we can define:

• $$E_i\land E_j$$

• $$E_i\lor E_j$$

## Kolmogorov axioms

### The probability function

For all events $$E$$ in $$F$$, the probability function $$P$$ is defined.

### Measure space

This gives us the following measure space:

$$(\Omega, F, P)$$

### First Kolmogorov axiom

First axiom

The probability of all events is a non-negative real number.

$$\forall E \in F [(P(E)\ge 0)\land (P(E)\in \mathbb{R})]$$

### Second Kolmogorov axiom

The probability of one of the elementary events occuring is $$1$$.

The probability of the outcome set is $$1$$.

$$P(\Omega )=1$$

### Third Kolmogorov axiom

The probability of union for mutually exclusive events is:

$$P(\cup^\infty_{i=1}E_i)=\sum_{i=1}^\infty P(E_i)$$

## Basic results

### Probability of null

$$P(\Omega )=1$$

$$P(\Omega \lor \varnothing )=1$$

$$P(\Omega )+P(\varnothing )=1$$

$$P(\varnothing )=0$$

### Monotonicity

Consider $$E_i\subseteq E_j$$:

$$E_j=E_i\lor E_k$$

$$P(E_j)=P(E_i\lor E_k)$$

Disjoint so:

$$P(E_j)=P(E_i)+P(E_k)$$

We know that $$P(E_k)\ge 0$$ from axiom $$1$$ so:

$$P(E_j)\ge P(E_i)$$

### Bounds of probabilities

As all events are subsets of the sample space:

$$P(\Omega )\ge P(E)$$

$$1\ge P(E)$$

From axiom $$1$$ then know:

$$\forall E\in F [0\le P(E)\le 1]$$

### Union and intersection for null and universal

$$P(E\land \varnothing )=P(\varnothing )=0$$

$$P(E\lor \Omega )=P(\Omega )=1$$

$$P(E\lor \varnothing)=P(E)$$

$$P(E\land \Omega )=P(E)$$

### Separation rule

Firstly:

$$P(E_i)=P(E_i\land \Omega)$$

$$P(E_i)=P(E_i\land (E_j\lor E_j^C))$$

$$P(E_i)=P((E_i\land E_j)\lor (E_i\land E_j^C))$$

As the latter are disjoint:

$$P(E_i)=P((E_i\land E_j)+(E_i\land E_j^C))$$

We know that:

$$P(E_i\lor E_j)=P((E_i\lor E_j)\land (E_j\lor E_j^C))$$

By the distributive law of sets:

$$P(E_i\lor E_j)=P((E_i\land E_j^C)\lor E_j)$$

$$P(E_i\lor E_j)=P((E_i\land E_j^C)\lor (E_j\land (E_i\lor E_i^C))$$

By the distributive law of sets:

$$P(E_i\lor E_j)=P((E_i\land E_j^C)\lor (E_j\land E_i)\lor (E_j\land E_i^C))$$

As these are disjoint:

$$P(E_i\lor E_j)=P(E_i\land E_j^C)+ P(E_j\land E_i)+P(E_j\land E_i^C)$$

From the separation rule:

$$P(E_i\lor E_j)=P(E_i)-P(E_i\land E_j)+ P(E_j\land E_i)+P(E_j)-P(E_j\land E_i)$$

$$P(E_i\lor E_j)=P(E_i)+P(E_j)-P(E_i\land E_j)$$

### Probability of complements

$$P(E_i\lor E_j)=P(E_i)+P(E_j)-P(E_i\land E_j)$$

Consider $$E$$ and $$E^C$$:

$$P(E\lor E^C)=P(E)+P(E^C)-P(E\land E^C)$$

We know that $$E$$ and $$E^C$$ are disjoint, that is:

$$E\land E^C=\varnothing$$

Similarly by construction:

$$E\lor E^C=\Omega$$

So:

$$P(\Omega )=P(E)+P(E^C)-P(\varnothing)$$

$$1=P(E)+P(E^C)$$

## Other

### Odds

Given a set of outcomes for a variable, the odds of the outcome are defined as:

$$o_f=\dfrac{P(E)}{P(E^C)}$$

For example, the odds of rolling a $$6$$ are $$\dfrac{1}{5}$$.

### Discrete and continous probability

We know that:

$$\sum_yP(X\land Y)=P(X)$$

So for the continuous case

$$P(X)=\int_{-\infty }^{\infty }P(X\land Y)dy$$

This behaves like the probability for a single event, or multiple events with one fewer event if there were more than $$2$$ events to start with.