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---
title: "Conditional expectation"
firstLetter: "C"
publishDate: 2021-10-23
categories:
- Mathematics
- Statistics
date: 2021-10-22T15:19:23+02:00
draft: false
markup: pandoc
---
# Conditional expectation
Recall that the expectation value $\mathbf{E}[X]$
of a [random variable](/know/concept/random-variable/) $X$
is a function of the probability space $(\Omega, \mathcal{F}, P)$
on which $X$ is defined, and the definition of $X$ itself.
The **conditional expectation** $\mathbf{E}[X|A]$
is the expectation value of $X$ given that an event $A$ has occurred,
i.e. only the outcomes $\omega \in \Omega$
satisfying $\omega \in A$ should be considered.
If $A$ is obtained by observing a variable,
then $\mathbf{E}[X|A]$ is a random variable in its own right.
Consider two random variables $X$ and $Y$
on the same probability space $(\Omega, \mathcal{F}, P)$,
and suppose that $\Omega$ is discrete.
If $Y = y$ has been observed,
then the conditional expectation of $X$
given the event $Y = y$ is as follows:
$$\begin{aligned}
\mathbf{E}[X | Y \!=\! y]
= \sum_{x} x \: Q(X \!=\! x)
\qquad \quad
Q(X \!=\! x)
= \frac{P(X \!=\! x \cap Y \!=\! y)}{P(Y \!=\! y)}
\end{aligned}$$
Where $Q$ is a renormalized probability function,
which assigns zero to all events incompatible with $Y = y$.
If we allow $\Omega$ to be continuous,
then from the definition $\mathbf{E}[X]$,
we know that the following Lebesgue integral can be used,
which we call $f(y)$:
$$\begin{aligned}
\mathbf{E}[X | Y \!=\! y]
= f(y)
= \int_\Omega X(\omega) \dd{Q(\omega)}
\end{aligned}$$
However, this is only valid if $P(Y \!=\! y) > 0$,
which is a problem for continuous sample spaces $\Omega$.
Sticking with the assumption $P(Y \!=\! y) > 0$, notice that:
$$\begin{aligned}
f(y)
= \frac{1}{P(Y \!=\! y)} \int_\Omega X(\omega) \dd{P(\omega \cap Y \!=\! y)}
= \frac{\mathbf{E}[X \cdot I(Y \!=\! y)]}{P(Y \!=\! y)}
\end{aligned}$$
Where $I$ is the indicator function,
equal to $1$ if its argument is true, and $0$ if not.
Multiplying the definition of $f(y)$ by $P(Y \!=\! y)$ then leads us to:
$$\begin{aligned}
\mathbf{E}[X \cdot I(Y \!=\! y)]
&= f(y) \cdot P(Y \!=\! y)
\\
&= \mathbf{E}[f(Y) \cdot I(Y \!=\! y)]
\end{aligned}$$
Recall that because $Y$ is a random variable,
$\mathbf{E}[X|Y] = f(Y)$ is too.
In other words, $f$ maps $Y$ to another random variable,
which, thanks to the *Doob-Dynkin lemma*
(see [random variable](/know/concept/random-variable/)),
means that $\mathbf{E}[X|Y]$ is measurable with respect to $\sigma(Y)$.
Intuitively, this makes sense:
$\mathbf{E}[X|Y]$ cannot contain more information about events
than the $Y$ it was calculated from.
This suggests a straightforward generalization of the above:
instead of a specific value $Y = y$,
we can condition on *any* information from $Y$.
If $\mathcal{H} = \sigma(Y)$ is the information generated by $Y$,
then the conditional expectation $\mathbf{E}[X|\mathcal{H}] = Z$
is $\mathcal{H}$-measurable, and given by a $Z$ satisfying:
$$\begin{aligned}
\boxed{
\mathbf{E}\big[X \cdot I(H)\big]
= \mathbf{E}\big[Z \cdot I(H)\big]
}
\end{aligned}$$
For any $H \in \mathcal{H}$. Note that $Z$ is almost surely unique:
*almost* because it could take any value
for an event $A$ with zero probability $P(A) = 0$.
Fortunately, if there exists a continuous $f$
such that $\mathbf{E}[X | \sigma(Y)] = f(Y)$,
then $Z = \mathbf{E}[X | \sigma(Y)]$ is unique.
## Properties
A conditional expectation defined in this way has many useful properties,
most notably linearity:
$\mathbf{E}[aX \!+\! bY | \mathcal{H}] = a \mathbf{E}[X|\mathcal{H}] + b \mathbf{E}[Y|\mathcal{H}]$
for any $a, b \in \mathbb{R}$.
The **tower property** states that if $\mathcal{F} \supset \mathcal{G} \supset \mathcal{H}$,
then $\mathbf{E}[\mathbf{E}[X|\mathcal{G}]|\mathcal{H}] = \mathbf{E}[X|\mathcal{H}]$.
Intuitively, this works as follows:
suppose person $G$ knows more about $X$ than person $H$,
then $\mathbf{E}[X | \mathcal{H}]$ is $H$'s expectation,
$\mathbf{E}[X | \mathcal{G}]$ is $G$'s "better" expectation,
and then $\mathbf{E}[\mathbf{E}[X|\mathcal{G}]|\mathcal{H}]$
is $H$'s prediction about what $G$'s expectation will be.
However, $H$ does not have access to $G$'s extra information,
so $H$'s best prediction is simply $\mathbf{E}[X | \mathcal{H}]$.
The **law of total expectation** says that
$\mathbf{E}[\mathbf{E}[X | \mathcal{G}]] = \mathbf{E}[X]$,
and follows from the above tower property
by choosing $\mathcal{H}$ to contain no information:
$\mathcal{H} = \{ \varnothing, \Omega \}$.
Another useful property is that $\mathbf{E}[X | \mathcal{H}] = X$
if $X$ is $\mathcal{H}$-measurable.
In other words, if $\mathcal{H}$ already contains
all the information extractable from $X$,
then we know $X$'s exact value.
Conveniently, this can easily be generalized to products:
$\mathbf{E}[XY | \mathcal{H}] = X \mathbf{E}[Y | \mathcal{H}]$
if $X$ is $\mathcal{H}$-measurable:
since $X$'s value is known, it can simply be factored out.
Armed with this definition of conditional expectation,
we can define other conditional quantities,
such as the **conditional variance** $\mathbf{V}[X | \mathcal{H}]$:
$$\begin{aligned}
\mathbf{V}[X | \mathcal{H}]
= \mathbf{E}[X^2 | \mathcal{H}] - \big[\mathbf{E}[X | \mathcal{H}]\big]^2
\end{aligned}$$
The **law of total variance** then states that
$\mathbf{V}[X] = \mathbf{E}[\mathbf{V}[X | \mathcal{H}]] + \mathbf{V}[\mathbf{E}[X | \mathcal{H}]]$.
Likewise, we can define the **conditional probability** $P$,
**conditional distribution function** $F_{X|\mathcal{H}}$,
and **conditional density function** $f_{X|\mathcal{H}}$
like their non-conditional counterparts:
$$\begin{aligned}
P(A | \mathcal{H})
= \mathbf{E}[I(A) | \mathcal{H}]
\qquad
F_{X|\mathcal{H}}(x)
= P(X \le x | \mathcal{H})
\qquad
f_{X|\mathcal{H}}(x)
= \dv{F_{X|\mathcal{H}}}{x}
\end{aligned}$$
## References
1. U.H. Thygesen,
*Lecture notes on diffusions and stochastic differential equations*,
2021, Polyteknisk Kompendie.
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