Sample Space and Probability

Ingredients of Probability

There is an outcome space, \(\Omega\), e.g., the number of students in the first class: \( \Omega = \{ 0, 1, 2, 3, … \} \).

There are events of interest, e.g., at least one student shows up: \(A = \{1, 2, 3, … \} \).

A probability measure, \(\mathbb{P}\), takes events \(A\) and assigns them \(\mathbb{P}(A) \in [0, 1]\), such that the axioms of probability are satisfied.

Definition

Given two propositions, \(A\) and \(B\), the conjunction \(A \cap B\) (read \(A\) and \(B\)) is true if both \(A\) and \(B\) are true, and false otherwise. Conjunction is commutative, i.e., \(\mathbb{P}(A \cap B) = \mathbb{P}(B \cap A)\).

from __future__ import annotations

import pandas as pd

from .probability_defn import probability


def conjunctive_probability(A: pd.Series[bool], B: pd.Series[bool]) -> float:
    # Given two Boolean series, the `&` operator computes their conjunction.
    return probability(A & B)

Definition

\( \mathbb{P}(A|B) \) is the probability of event \(A\) given that event \(B\) occurs:

$$ \mathbb{P}(A|B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} $$

Notice that if \(P(B) = 0\) then it’s meaningless to condition \(A\) on \(B\).

from __future__ import annotations

import pandas as pd

from .probability_defn import probability


def conditional_probability(
    proposition: pd.Series[bool], given: pd.Series[bool]
) -> float:
    # proposition[given] contains the values of proposition for given.
    return probability(proposition[given])

Satisfying the Axioms of Probability

Axiom 1: Normalization

$$ \mathbb{P}(B|B) = \frac{\mathbb{P}(B \cap B)}{\mathbb{P}(B)} = 1 $$

The Formula

By definition…

$$ \mathbb{P}(A) = \mathbb{P}(A \cap B) + \mathbb{P}(A \cap B^{c}) $$

From conditional probability …

$$ \mathbb{P}(A) = \mathbb{P}(A|B) \ \mathbb{P}(B) + \mathbb{P}(A|B^c) \ \mathbb{P}(B^c) $$

Therefore

$$ \mathbb{P}(B|A) = \frac{ \mathbb{P}(B \cap A) }{ \mathbb{P}(A) } $$ $$ = \frac{ \mathbb{P}(A|B) \mathbb{P}(B) }{ \mathbb{P}(A|B) \ \mathbb{P}(B) + \mathbb{P}(A|B^c) \ \mathbb{P}(B^c) } $$