Conditional Expectation Conditioned on Event of Non-Zero Probability

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Theorem

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.

Let $X$ be an integrable random variable on $\struct {\Omega, \Sigma, \Pr}$.

Let $B \in \Sigma$ be an event with:

$\map \Pr B > 0$

Let:

$\GG = \map \sigma B = \set {\O, B, B^c, \Omega}$

where $\map \sigma B$ is the $\sigma$-algebra generated by $B$.

Let:

$\ds \alpha = \frac {\expect {X \cdot 1_B} } {\map \Pr B}$

and:

$\ds \beta = \frac {\expect {X \cdot 1_{B^c} } } {\map \Pr {B^c} }$

Let $\expect {X \mid \GG}$ be a version of the conditional expectation of $X$ given $\GG$.


Then:

$\ds \expect {X \mid \GG} = \alpha \cdot 1_B + \beta \cdot 1_{B^c}$ almost everywhere.


Proof

We show that:

$\ds Z = \alpha \cdot 1_B + \beta \cdot 1_{B^c}$ is a version of the conditional expectation of $X$ given $\GG$.

We will then be done by Existence and Essential Uniqueness of Conditional Expectation Conditioned on Sigma-Algebra.

From Characteristic Function Measurable iff Set Measurable, $1_B$ and $1_{B^c}$ are both measurable functions.

From Pointwise Product of Measurable Functions is Measurable and Pointwise Sum of Measurable Functions is Measurable, we therefore have that $Z$ is a random variable.

So we now verify that the expectations take the correct values.

From Integral of Integrable Function over Null Set, we have:

$\expect {Z \cdot 1_\O} = 0 = \expect {X \cdot 1_\O}$

We also have:

\(\ds \expect {Z \cdot 1_B}\) \(=\) \(\ds \expect {\alpha \cdot 1_B + \beta \cdot 1_B \cdot 1_{B^c} }\)
\(\ds \) \(=\) \(\ds \expect {\alpha \cdot 1_B}\) Characteristic Function of Intersection
\(\ds \) \(=\) \(\ds \alpha \map \Pr B\) Expectation is Linear: General Case, Integral of Characteristic Function
\(\ds \) \(=\) \(\ds \expect {X \cdot 1_B}\)

and:

\(\ds \expect {Z \cdot 1_{B^c} }\) \(=\) \(\ds \expect {\alpha \cdot 1_B \cdot 1_{B^c} + \beta \cdot 1_{B^c} }\)
\(\ds \) \(=\) \(\ds \expect {\beta \cdot 1_{B^c} }\) Characteristic Function of Intersection
\(\ds \) \(=\) \(\ds \beta \map \Pr {B^c}\) Expectation is Linear: General Case, Integral of Characteristic Function
\(\ds \) \(=\) \(\ds \expect {X \cdot 1_{B^c} }\)

Finally, we have:

\(\ds \expect Z\) \(=\) \(\ds \expect {\alpha \cdot 1_B + \beta \cdot 1_{B^c} }\)
\(\ds \) \(=\) \(\ds \alpha \map \Pr B + \beta \map \Pr {B^c}\) Expectation is Linear: General Case, Integral of Characteristic Function
\(\ds \) \(=\) \(\ds \expect {X \cdot 1_B} + \expect {X \cdot 1_{B^c} }\)
\(\ds \) \(=\) \(\ds \expect {X \cdot \paren {1_B + 1_{B^c} } }\) Expectation is Linear: General Case
\(\ds \) \(=\) \(\ds \expect X\) Characteristic Function of Disjoint Union

$\blacksquare$

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