Sigma Algebra

Contents (hide)

  1. 1. Definition
  2. 2. Uses
  3. 3. Example

1.  Definition

Let \fs4 \Omega be a non-empty set, and let \fs4 \cal{P}(\Omega) be its powerset (that is the set that consists of all subsets of \fs4 \Omega). A family of subsets \fs4 \mathscr{F}\in \cal{P}(\Omega) is a \fs4 \mathbf{\sigma}-algebra on \fs4 \Omega if the following conditions are satisfied:

  • if \fs4 F \in \mathscr{F}, then the complement \fs4 F^c \in \mathscr{F}
  • if \fs4 I is a countable set of indices, and \fs4 F_i \in \mathscr{F} for all \fs4 i \in I, then \fs4 \cup F_i \in \mathscr{F} where the union is extended over the indices \fs4 i\in I

Sometimes a \fs4 \sigma-algebra is called a \fs4 \mathbf{\sigma}-field.

2.  Uses

\fs4 \sigma-algebras are routinely used in probability theory to represent the conditional information of random variables.

It might seem counterintuitive, but not all subsets of the state space \fs4 \Omega are admissible events (an example of a non-admissible event would be the Vitali set). In other words, not all subsets of \fs4 \Omega are measurable.

The most important \fs4 \sigma-algebra is the one defined on the real numbers, called the Borel \fs4 \sigma-algebra

3.  Example

We assume the setting described in the example of a random variable.

As we mentioned above, the \fs4 \sigma-algebra is a measure of our information. In particular, considering the powerset of \fs4 \Omega is equivalent to observing and recording both tosses of the coin. This is why all random variables (on \fs4 \Omega of course) will be \fs4 \mathscr{F}_\infty-measurable.

This will be made clear with another example. Say that we consider another \fs4 \sigma-algebra, namely:

\fs4 \mathscr{G} = \{\emptyset, \{HH, HT\}, \{TH, TT\}, \Omega\}

It is straightfoward to verify that this is indeed a \fs4 \sigma-algebra on \fs4 \Omega, and that this algebra is smaller that the previous one, \fs4 \mathscr{G} \subset \mathscr{F}_\infty. In iformation terms, \fs4 \mathscr{G} is equivalent as observing only the first coin toss. Why is that? Because we can only define probabilities on set that differ at the first coin toss.

Let us investigate now the random variable \fs4 X, which is the sum of heads. Is it \fs4 \mathscr{G}-measurable? The answer is no; since as before for the set \fs4 B=(3/4,6)

\fs4 X^{-1}((3/4,6)) = \{\omega \in \Omega : X(\omega) \in (3/4,6) \} = \{HT, TH, HH\} \notin \mathscr{G}

What does that mean? That we cannot consider this random variable with the information that we have: if we observe only the first toss we cannot assess the total number of heads.

On the other hand the random variable

: X(\omega) = \{1 \text{ if the first toss in }\omega\text{ is a }H\}$}

is indeed \fs4 \mathscr{G}-measurable. For instance for the same Borel set

\fs4 Y^{-1}((3/4,6)) = \{\omega \in \Omega : Y(\omega) \in (3/4,6) \} = \{HT, HH\} \in \mathscr{G}

And of course, since \fs4 \mathscr{G} \subset \mathscr{F}_\infty, any \fs4 \mathscr{G}-measurable random variable will also be \fs4 \mathscr{F}_\infty-measurable. As we increase our information set we can attach probabilities to more events, not less!

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