Conditional Expectation

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  1. 1. Definition
  2. 2. Example

1.  Definition

Consider a random variable \fs4 X on a probability space \fs4 (\Omega, \cal{F}, P). Also consider a sub-\fs4 \sigma-algebra \fs4 \cal{G} \subseteq \cal{F}. The conditional expectation \fs4 E[X|\cal{G}] is the equivalence class of all random variables \fs4 Y that satisfy

\fs4 \int_G Y(\omega) dP(\omega) = \int_G X(\omega) dP(\omega)

2.  Example

We consider the setting described in the example of a random variable. The set \fs4 \mathscr{G} is described in the example of \fs4 \sigma-algebras

The question that we pose is the following: can we compute the expectation \fs4 EX on the different probability space \fs4 (\Omega, \mathscr{G}, P), with limited information?

The answer is no, since \fs4 X is not \fs4 \mathscr{G}-measurable. But if we consider the sub-\fs4 \sigma-algebra \fs4 \mathscr{G} (that is we observe only the first toss) we can determine the conditional expectation \fs4 E[X|\mathscr{G}], which, according to its definition, is a \fs4 \mathscr{G}-measurable random variable $} satisfying (for all \fs4 G \in \mathscr{G})

\fs4 \sum_{\omega \in G} Y(\omega) P(\omega) = \sum_{\omega \in G} X(\omega) P(\omega)

Now this will imply

\fs4 G = \emptyset \Rightarrow 0 = 0
\fs4 G = \Omega \Rightarrow 1/4 \cdot (Y(HH) + Y(HT) + Y(TH) + Y(TT)) = 1
\fs4 G = \{HH, HT\} \Rightarrow 1/4 \cdot (Y(HH) + Y(HT)) = 3/4
\fs4 G = \{TH, TT\} \Rightarrow 1/4 \cdot (Y(TH) + Y(TT)) = 1/4

Since we want \fs4 Y to be \fs4 \mathscr{G}-measurable we can ask for \fs4 Y to be constant within each partition \fs4 \{HH,HT\} and \fs4 \{TH,TT\}. This will give the random variable

\fs4 Y : \fs4 \Omega \rightarrow \mathbb{R} : \fs4 Y(HH) = Y(HT) = 3/2, Y(TH) = Y(TT) = 1/2

Is this random variable \fs4 \mathscr{G}-measurable? The answer is yes. Remember \fs4 X was not, and we used the Borel set \fs4 (3/4,6) as an example. For \fs4 Y we have

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


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