Independent Event Sigma Algebras

Definition

The event sigma algebra of $A \in \mathcal{F} $ where $(\Omega , \mathcal{F} )$ is a measurable space is the set sub-$\sigma $-algebra $\set{\varnothing, A, A^c, \Omega }$.

A family of events events are independent if the event sigma algebras are independent.

Notation

Let $(X, \mathcal{A} , \mu )$ be a probability space. Let $A \in \mathcal{A} $ be an event. The sigma algebra generated by $A$ is $\set*{\varnothing, A, X - A, X}$. We denote it by $\sigma (A)$.

Let $B \in \mathcal{A} $. If $A$ is independent of $B$ we write $A \perp B$.

Equivalent condition

Two events are independent if and only if the measure of their intersection is the product of their measures.

Let $(X, \mathcal{A} , \mu )$ be a probability space. Let $A, B \in \mathcal{A} $.

$(\Rightarrow)$ If $A \perp B$, then by definition $A \in \sigma (A)$ and $B \in \sigma (B)$ and so:

\[ \mu (A \cap B) = \mu (A)\mu (B). \]

$(\Leftarrow)$ Conversely, let $a \in \sigma (A)$ and $b \in \sigma (B)$. If $a = \varnothing$ or $b = \varnothing$ then $a \cap b = \varnothing$. So

\[ \mu (a \cap b) = \mu (\varnothing) = \mu (a)\mu (b), \]

since one of the two measures on the right hand side is zero. On the other hand, if $a = X$, then $a \cap b = b$ and so

\[ \mu (a \cap b) = \mu (b) = \mu (a)\mu (b), \]

since $\mu (a) = \mu (X) = 1$. Likewise if $b = X$.

So it remains to verify $\mu (a \cap b) = \mu (a)\mu (b)$ for the cases $a \in \set{A,X -A}$ and $b \in \set{B, X-B}$. If $a = A$, and $b = B$, then the identity follows by hypothesis. Next, observe that $A \cap (X - B) = A - (A \cap B)$ and $(A \cap B) \subset A$ so $\mu (X) < \infty$ allows us to deduce:

\[ \begin{aligned} \mu (A \cap (X - B)) &= \mu (A - (A \cap B)) \\ &= \mu (A) - \mu (A \cap B) \\ &= \mu (A)(1 - \mu (B)) \\ &= \mu (A)\mu (X - A). \end{aligned} \]

Similar for $X-A$ and $B$. Finally, recall that $\mu (A \cup B) = \mu (A) + \mu (B) - \mu (A \cap B)$. So then,

\[ \begin{aligned} \mu ((X - A) \cap (X - B)) &= 1 - \mu (A \cup B) \\ &= 1 - \mu (A) - \mu (B) + \mu (A \cap B) \\ &= 1 - \mu (A) - \mu (B) + \mu (A)\mu (B) \\ &= (1 - \mu (A))(1 - \mu (B)) \\ &= \mu (X - A)\mu (X - B). \end{aligned} \]