In many statistics papers, proofs might proceed as follows: Under the event $A$, the random variables $X$ and $Y$ are independent. (Often this means that on $A^C$, they might be dependent). Then some properties of conditional independence might be used, e.g. to calculate $ \mathbb E[ X \mid A, Y] $. I feel quite uncomfortable with the latter type of manipulations.

Therefore, I am wondering two things:

1) What is the proper definition of conditional independence when conditioning on both events and $\sigma$-algebras? My guess would be that we have to check the factorization property only for sets $A \cap B_X$ and $A \cap B_y$, where $B_X \in \sigma(X)$, $B_Y \in \sigma(Y)$ the generated $\sigma$-algebras of the above random variables. (Similarly for the definition of conditional expectation.)

2) Is there any reference on properties of such expectations which are taken conditionally on both $\sigma$-algebras and events? Something akin to the standard properties of conditional expectations (conditioned on $\sigma$-algebras) used to simplify expressions or to calculate them. Or is there any simple trick to generalize the standard results to this setting?


1 Answer 1


I will try to answer my question above; it would be great if someone can confirm (since again I did not find any textbook describing this, except the one Exercise in Billingsley mentioned below)!

To set things up, let $(\Omega, \mathcal{F}, \mathbb P)$ be a probability space. $A \in \mathcal{F}$ is an event with probability $\mathbb P(A) > 0$, $X: \Omega \to \mathbb R$ is a random variable and $\mathcal{G} \subset \mathcal{F}$ is a sub-$\sigma$-algebra.

We are interested in defining: $\mathbb E[X \mid A, \mathcal{G}]$. There are two "natural" ways to do this.

First, we will do this by just using the standard definition of conditional expectation but with respect to the measure $\mathbb P_A$, where this is just the conditional probability measure with mass $P_A(B)$ on $\mathcal{F}$-measurable sets $B$:

$$ \mathbb P_A(B) = \frac{\mathbb P(A \cap B)} {\mathbb P(A)}$$

Thus we define $\mathbb E[X \mid A, \mathcal{G}]$ for $X \in L^1(\mathbb P_A)$ by the following properties:

  1. $\mathbb E[X \mid A, \mathcal{G}]$ is $\mathcal{G}$ measurable.
  2. $\int_B \mathbb E[X \mid A, \mathcal{G}] d\mathbb P_A = \int_B X d\mathbb P_A $ for all $\mathcal{G}$-measurable sets $B$.

We can quickly see that to check $X \in L^1(\mathbb P_A)$, it is sufficient to check $X \in L^1(\mathbb P)$ while for property 2. we can just check:

$\int_B \mathbb E[X \mid A, \mathcal{G}] d\mathbb P = \int_B X d\mathbb P $ for all sets $B \in \{ G \cap A \mid G \in \mathcal{G}\}$

The second way of defining $\mathbb E[X \mid A, \mathcal{G}]$ is by defining it for indicator variables of $\mathcal{F}$-measurable sets $B$ as (also see related math.se post):

$$ \mathbb E[ \mathbf{1}_{B} \mid A , \mathcal{G}] = \frac{\mathbb E[ \mathbf{1}_{B}\mathbf{1}_{A} \mid \mathcal{G}] }{\mathbb E[ \mathbf{1}_{A} \mid \mathcal{G}]}$$

By exercise 34.4 a) in the book "Probability and Measure" by Billingsley, we get that in fact these two definitions are equivalent. So we are good to go.

Now we are still interested in the calculus of such conditional expectations. It turns out to be simple, since we can just use the standard calculus where the expectations are taken w.r.t. to the measure $\mathbb P_A$! Also properties such as "on the event $A$, $X$ is independent of $\mathcal{G}$", also just mean that $X$ is independent of $\mathcal{G}$ under the measure $\mathbb P_A$.

  • $\begingroup$ I would just add, in the second definition, that the equation holds only $\mathbb{P}_A$-a.s. (as it is only possible to have $\mathbb{E}(1_A|\mathcal{G})>0$ $\mathbb{P}_A$-a.s., not $\mathbb{P}$-a.s.). $\endgroup$
    – xyz
    May 12 at 9:50

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