Conditional Expectation Unchanged on Conditioning on Independent Sigma-Algebra

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Theorem

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

Let $\GG \subseteq \Sigma$ be a sub-$\sigma$-algebra.

Let $X$ be a integrable random variable.

Let $\map \sigma X$ be the $\sigma$-algebra generated by $X$.

Let $\HH \subseteq \Sigma$ be a sub-$\sigma$-algebra that is independent of $\map \sigma {\map \sigma X, \GG}$, the $\sigma$-algebra generated by $\map \sigma X \cup \GG$.

Let $\map \sigma {\GG, \HH}$ be the $\sigma$-algebra generated by $\GG \cup \HH$.

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

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


Then:

$\expect {X \mid \map \sigma {\GG, \HH} } = \expect {X \mid \GG}$ almost surely.


Corollary

Let $\HH \subseteq \Sigma$ be a sub-$\sigma$-algebra.

Let $X$ be a integrable random variable such that:

$\map \sigma X$ is independent of $\HH$

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

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


Then:

$\expect {X \mid \HH} = \expect X$ almost surely.


Proof

First take $X$ to be a non-negative random variable.

Let:

$\SS = \set {G \cap H : G \in \GG, \, H \in \HH}$

We aim to apply Uniqueness of Measures to a suitable measure with $\SS$.

We start by showing that $\Omega \in \SS$, $\SS$ is a $\pi$-system, and that $\SS$ generates $\map \sigma {\GG, \HH}$.

First note that:

$\Omega = \Omega \cap \Omega \in \SS$

We now show that $\SS$ is a $\pi$-system.

Let $G_1 \cap H_1, G_2 \cap H_2, \ldots, G_n \cap H_n \in \SS$ with $G_i \in \GG$ and $H_i \in \HH$ for each $i$.

We have from Intersection is Associative:

$\ds \bigcap_{i \mathop = 1}^n \paren {G_i \cap H_i} = \paren {\bigcap_{i \mathop = 1}^n G_i} \cap \paren {\bigcap_{i \mathop = 1}^n H_i}$

Since $\GG$ and $\HH$ are $\sigma$-algebras we have:

$\ds \bigcap_{i \mathop = 1}^n G_i \in \GG$

and:

$\ds \bigcap_{i \mathop = 1}^n H_i \in \HH$

so:

$\ds \bigcap_{i \mathop = 1}^n \paren {G_i \cap H_i} \in \SS$

We now show that $\SS$ generates $\map \sigma {\GG, \HH}$.

By the definition of the $\sigma$-algebra generated by $\GG \cup \HH$, we just need to show that:

$\GG \subseteq \SS$ and $\HH \subseteq \SS$

and:

$\SS \subseteq \map \sigma {\GG, \HH}$

Note that for each $G \in \GG$ we have $G = G \cap \Omega \in \SS$, and for each $H \in \HH$ we have $H = H \cap \Omega \in \SS$, so we have:

$\GG \subseteq \SS$ and $\HH \subseteq \SS$

Now let $G \in \GG$ and $H \in \GG$ so that $G \cap H \in \SS$.

Then $G, H \in \map \sigma {\GG, \HH}$, so $G \cap H \in \map \sigma {\GG, \HH}$ from the definition of a $\sigma$-algebra.

So we indeed have:

$\SS \subseteq \map \sigma {\GG, \HH}$

and:

$\map \sigma {\SS} = \map \sigma {\GG, \HH}$


Now define the measure with density $\mu_1$ by:

$\ds \map {\mu_1} A = \int_A X \rd \Pr$

for $A \in \map \sigma {\GG, \HH}$.

Also define:

$\ds \map {\mu_2} A = \int_A \expect {X \mid \GG} \rd \Pr$

Since $X$ and $\expect {X \mid \GG}$ are integrable, these are finite measures.

Therefore, by Uniqueness of Measures, if we can show:

$\map {\mu_1} A = \map {\mu_2} A$ for all $A \in \SS$

we will have that $\mu_1 = \mu_2$.

This will show that $\expect {X \mid \GG}$ is a version of the conditional expectation of $X$ given $\map \sigma {\GG, \HH}$.

We will then have, by Existence and Essential Uniqueness of Conditional Expectation Conditioned on Sigma-Algebra:

$\expect {X \mid \map \sigma {\GG, \HH} } = \expect {X \mid \GG}$ almost surely.

Let $G \in \GG$ and $H \in \HH$ so that $G \cap H \in \SS$.

Then we have, by Characteristic Function of Intersection:

$\map {\mu_1} {G \cap H} = \expect {X \cdot \chi_{G \cap H} } = \expect {X \cdot \chi_G \chi_H}$

From Characteristic Function Measurable iff Set Measurable, we have:

$\chi_G$ is $\GG$-measurable

and so:

$\chi_G$ is $\map \sigma {\map \sigma X, \GG}$-measurable.

From Pointwise Product of Measurable Functions is Measurable, we have:

$X \cdot \chi_G$ is $\map \sigma {\map \sigma X, \GG}$-measurable.

Now from Expectation of Product of Independent Random Variables is Product of Expectations, we have:

$\expect {X \cdot \chi_G \chi_H} = \expect {X \cdot \chi_G} \expect {\chi_H}$

So, from Integral of Characteristic Function:

$\map {\mu_1} {G \cap H} = \expect {X \cdot \chi_G} \map \Pr H$

As to $\mu_2$, we have from Characteristic Function of Intersection:

$\map {\mu_2} {G \cap H} = \expect {\expect {X \mid \GG} \cdot \chi_{G \cap H} } = \expect {\expect {X \mid \GG} \cdot \chi_G \chi_H}$

Since:

$\expect {X \mid \GG}$ is $\GG$-measurable

and again:

$\chi_G$ is $\GG$-measurable

so by Pointwise Product of Measurable Functions is Measurable:

$\expect {X \mid \GG} \cdot \chi_G$ is $\GG$-measurable

and so:

$\expect {X \mid \GG} \cdot \chi_G$ is $\map \sigma {\map \sigma X, \GG}$-measurable.

Now, from Expectation of Product of Independent Random Variables is Product of Expectations we have:

$\expect {\expect {X \mid \GG} \cdot \chi_G \chi_H} = \expect {\expect {X \mid \GG} \cdot \chi_G} \expect {\chi_H}$

Then from Integral of Characteristic Function we have:

$\expect {\expect {X \mid \GG} \cdot \chi_G} \expect {\chi_H} = \expect {\expect {X \mid \GG} \cdot \chi_G} \map \Pr H$

From the definition of a version of the conditional expectation of $X$ given $\GG$, we have:

$\expect {\expect {X \mid \GG} \cdot \chi_G} = \expect {X \cdot \chi_G} \map \Pr H = \map {\mu_1} A$

So we have shown that:

$\map {\mu_1} A = \map {\mu_2} A$ for each $A \in \SS$

Hence the result.

$\blacksquare$


Sources

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