sample space
A sample space \( \Omega \) is a non-empty set
event
Given a sample space \( \Omega \), we say that any subset \( E \subseteq \Omega \) is an event
probability measure
Given a sample space \( \Omega \) and a function \( P : \Omega \to \left [ 0 , 1 \right ] \) , then we say that \( P \) is a probability measure if the following holds
  • \( P \left ( \Omega \right ) \) \( = 1 \)
  • \( \forall E_{1} , E_{2} , \ldots \subseteq \Omega \) such that \( E_{1} , E_{2} , \ldots \) are all pairwise disjoint
    \( P \left ( \bigcup_{i = 1}^{\infty} E_{i} \right ) = \sum_{i = 1}^{\infty} P \left ( E_{i} \right ) \)
finite pairwise disjoint probability
Suppose that \( A_{1} , A_{2} , \ldots , A_{n} \) are pairwise disjoint, then \( P \left ( A_{1} \cup \ldots \cup A_{n} \right ) = \sum_{i = 1}^{n} P \left ( A_{i} \right ) \)
In the definition of a probability measure, for \( i \in \left \lbrace 1 , \ldots , n \right \rbrace \) we set \( E_{i} = A_{i} \) and then for any \( j > n \) we set \( E_{j} = \emptyset \), then we can see that \( E_{1} , E_{2} , \ldots \) is pairwise disjoint, and thus
\( P \left ( A_{1} \cup A_{2} \cup \ldots \cup A_{n} \right ) \) \( = \) \( P \left ( A_{1} \cup A_{2} \cup \ldots \cup A_{n} \cup \emptyset \cup \emptyset \cup \ldots \right ) \)
\( = \) \( P \left ( \bigcup_{i = 1}^{\infty} E_{i} \right ) \)
\( = \) \( \sum_{i = 1}^{\infty} P \left ( E_{i} \right ) \)
\( = \) \( \sum_{i = 1}^{n} P \left ( E_{i} \right ) + P \left ( \emptyset \right ) + P \left ( \emptyset \right ) + P \left ( \emptyset \right ) + \ldots \)
\( = \) \( \sum_{i = 1}^{n} P \left ( E_{i} \right ) + 0 + 0 + 0 + \ldots \)
\( = \) \( \sum_{i = 1}^{n} P \left ( A_{i} \right ) \)
probability of a single point with equally likely outcomes is zero
Suppose that our sample space is \( \left [ 0 , 1 \right ] \) and there is some \( c \in \left [ 0 , 1 \right ] \) such that for any \( x \in \left [ 0 , 1 \right ] \), \( P \left ( \left \lbrace x \right \rbrace \right ) = c \), then \( c = 0 \)
Suppose that \( c \ne 0 \), and note that \( P \left ( \Omega \right ) = P \left ( \left [ \left . 0 , 1 \right . \right ] \right ) = \sum_{x \in \left [ 0 , 1 \right ]} P \left ( x \right ) = \sum_{x \in \left [ 0 , 1 \right ]} c \), but note that no matter how small the value of \( c \), since we are summing it uncountably many times with itself the sum always goes to infinity, but at the same time, \( P \left ( \Omega \right ) = 1 \) and thus we have a contradiction, so \( c = 0 \).
random variable
Suppose that \( \Omega \) is a sample space. A random variable is a function \( X : \Omega \to \mathbb{R} \)
inverse image notation
Suppose that \( X : \Omega \to \mathbb{R} \) is a random variable and that \( y \in \mathbb{R} \), then we define the string of characters \( X = y \) to be the set \( X^{- 1} \left ( \left \lbrace y \right \rbrace \right ) \)
conditional probability
Suppose that \( A , B \subseteq \Omega \), with \( P \left ( B \right ) \gt 0 \), then we define the conditional probability of \( A \) given \( B \) as
\( P \left ( A \mid B \right ) = \frac{P \left ( A \cap B \right )}{P \left ( B \right )} \)