How to show $\mathbb{E}[AB\mid B]=B\cdot\mathbb{E}[A\mid B]$?

Intuitively, since we are conditioning on $B$, $B$ is already known so we can simply take $B$ out of the expectation operator. But the tricky part is $\mathbb{E}[AB\mid B]$ is a random variable.

  • $\begingroup$ I haven't answered this question but I've up-voted it. I suggest that the three people who've answered it should do the same. $\qquad$ $\endgroup$ Aug 13, 2016 at 3:58
  • $\begingroup$ Agree with @MichaelHardy. If you answered a question, then it means that you think the question is useful and should up-vote it. $\endgroup$
    – Gordon
    Aug 18, 2016 at 20:23

3 Answers 3


I will use the following definition as my starting point:

Definition. If $X \in L^1(\Omega, \mathcal{F}, \Bbb{P})$ and $\mathcal{G} \subseteq \mathcal{F}$ is a $\sigma$-algebra, then $\Bbb{E}[X \mid \mathcal{G}]$ is a $\mathcal{G}$-measurable integrable function for which

$$ \int_{E} X \, d\Bbb{P} = \int_{E} \Bbb{E}[X \mid \mathcal{G}] \, d\Bbb{P} \qquad \forall E \in \mathcal{G} $$

is true.

If $E$ is $\sigma(B)$-measurable, then for all $F \in \sigma(B)$ we have

\begin{align*} \int_{F} \Bbb{E}[A\mathbf{1}_E \mid B] \, d\Bbb{P} &= \int_{F} A\mathbf{1}_E \, d\Bbb{P} \qquad & \text{(by definition with $A\mathbf{1}_E$)}\\ &= \int_{F\cap E} A \, d\Bbb{P} \\ &= \int_{F\cap E} \Bbb{E}[A \mid B] \, d\Bbb{P} & \text{(by definition with $A$)} \\ &= \int_{F} \mathbf{1}_E \Bbb{E}[A \mid B] \, d\Bbb{P} \end{align*}

and hence $\Bbb{P}$-a.s. $\Bbb{E}[A\mathbf{1}_E \mid B] = \mathbf{1}_E \Bbb{E}[A \mid B]$ holds.

Now you may invoke the standard mechanism - the monotone class theorem - to check that the same is true for all $\sigma(B)$-measurable r.v.s $X$ for which $AX \in L^1(\Bbb{P})$.

Alternatively, approximate $B$ by a sequence of simple functions and use the observation above directly together with an appropriate convergence theorem.

(Either cases, you may need to invoke conditional version of MCT or DCT.)


Definition: $$E[AB\mid B]$$ is any random variable $Z$ s.t.

  1. $Z$ is $B-$measurable

  2. $\forall B_1 \in \sigma(B)$

$$\int_{B_1} Z \, d\mathbb P = \int_{B_1} E[AB\mid B] \, d\mathbb P$$


$$E[Z1_{B_1}] = E[E[AB\mid B]1_{B_1}]$$

Now we must check if $BE[A\mid B]$ satisfies those.

  1. $Z = BE[A|B]$is $B-$measurable

because $B$ is $B$-measurable, $E[A\mid B]$ is $B-$ measurable and the product of $B-$ measurable functions is $B-$ measurable.

  1. $\forall B_1 \in \sigma(B)$

$$LHS = E[Z1_{B_1}] = E[BE[A\mid B]1_{B_1}] = E[BE[A1_{B_1}\mid B]] = E[E[BA1_{B_1}\mid B]] = E[BA1_{B_1}]$$

$$RHS = E[E[AB\mid B]1_{B_1}] = E[E[AB1_{B_1} \mid B]] = E[AB1_{B_1}]$$


Observe that we used properties of $E[A\mid B]$:

  1. $E[A\mid B]$ is $B$-measurable

  2. $$E[A\mid B]1_{B_1} = E[A1_{B_1}\mid B]$$


To avoid questions of existence of these expectations, let's assume $A$ and $B$ are bounded a.s.

One definition of conditional expectation: $\mathbb E[X\mid B]$ is a measurable function of $B$ such that $\mathbb E[ g(B) \mathbb E[X\mid B]] = \mathbb E[g(B) X]$ for all bounded measurable $g$.
Any two of these are equal a.s.

So let's see: $B \; \mathbb E[A\mid B]$ is a measurable function of $B$, and

$$ \mathbb E[g(B) B \; \mathbb E[A\mid B]] = \mathbb E[g(B) B A] $$


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