Taking Out What Is Known¶

Proposition¶

Let $h$ be a Borel-measurable function, and let $Y$ and $h(X)Y$ be integrable (for example, $h$ bounded). Then, almost surely,

\begin{aligned} \mathbb{E}[h(X)\,Y\mid X]=h(X)\,\mathbb{E}[Y\mid X]. \end{aligned}

Inside the Conditional Expectation Given a Random Variable, fixing $X=x$ fixes $h(X)=h(x)$ $h (x)$ . It therefore acts as a constant in the conditional sum or integral, subject to the stated integrability assumptions.

Interpretation¶

Once $X$ is observed, any function $h(X)$ of it is a known number, so it acts as a constant inside the conditional average. The rule is the conditional version of $\mathbb{E}[cY]=c\,\mathbb{E}Y$ $c E Y$ .

The conditional means x and x squared plotted against x, with the point x = 2 marked on each; multiplying the first by h(2) = 2 gives the second.

In the illustrated example, $Y\mid X\sim\operatorname{Pois}(X)$ $Pois (X)$ , so the blue line is $\mathbb{E}[Y\mid X=x]=x$ $x$ , and $h(x)=x$ $x$ . At $x=2$ the blue value $2$ is multiplied by the known factor $h(2)=2$ to give the red value $4$ of $\mathbb{E}[XY\mid X=x]=x^2$ $x^{2}$ . In general $\mathbb{E}[h(X)Y\mid X]=h(X)\,\mathbb{E}[Y\mid X]$ $X]$ ; here $\mathbb{E}[XY\mid X]=X\cdot X=X^2$ $X^{2}$ .

At a fixed value $x$ , the rule is ordinary linearity of expectation with the constant $h(x)$ ; substituting the random $X$ afterwards gives the stated identity. The integrability assumptions make both sides well defined, which is automatic when $h$ is bounded.

Combined with Adam’s law, the rule gives $\mathbb{E}[h(X)Y]=\mathbb{E}[h(X)\,\mathbb{E}[Y\mid X]]$ $X]]$ , a standard way to compute mixed moments in two-stage models. In the Poisson example, $\mathbb{E}[XY]=\mathbb{E}[X^2]$ $E [X^{2}]$ , so $\operatorname{Cov}(X,Y)=\mathbb{E}[X^2]-(\mathbb{E}X)^2=\operatorname{Var}(X)$ $Var (X)$ . The same step shows that independence implies zero covariance, see conditional expectation and independence.

Only functions of the conditioning variable come out. A function of $Y$ is not known given $X$ : $\mathbb{E}[Y^2\mid X]$ is in general not $(\mathbb{E}[Y\mid X])^2$ $X])^{2}$ , and in the Poisson example it equals $X+X^2$ $X^{2}$ .

Key Results¶

$\mathbb{E}[h(X)\mid X]=h(X)$ when $h(X)$ is integrable.
Combining the rule with Adam’s law gives $\mathbb{E}[h(X)Y]=\mathbb{E}[h(X)\mathbb{E}[Y\mid X]]$ $X]]$ .

Examples¶

If $Y\mid X\sim\operatorname{Pois}(X)$ and $\mathbb{E}[X^2]<\infty$ $\infty$ , then $\mathbb{E}[XY\mid X]=X^2$ $X^{2}$ .
If $A=\{X>0\}$ $0}$ , then $\mathbb{E}[\indicator_AY\mid X]=\indicator_A\mathbb{E}[Y\mid X]$ $X]$ .

Prerequisites¶

Conditional Expectation Given a Random Variable

Sources¶

Nicky van Foreest · Probability Distributions · Week 4

Aliases: pull-out property, taking out what's known, pulling out what is known, pulling out known factors Symbols: infinity

Last updated: 20 October 2026 03:02