Categories: Mathematics, Measure theory, Statistics, Stochastic analysis.

# Conditional expectation

Recall that the expectation value $$\mathbf{E}[X]$$ of a 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), 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}

1. U.H. Thygesen, Lecture notes on diffusions and stochastic differential equations, 2021, Polyteknisk Kompendie.

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.