# Why this weird definition of conditional expectation?

A book gives the definition of conditional expectation based on measure theory:

I cannot seem to grasp this definition. Why is it defined in this weird indirect way, and why does this even correspond to the concept of a conditional expectation?

I would think we should just define the conditional expectation as follows:

Given a probability measure $P(\cdot)$ First define a new probability measure $P_X(\cdot)$ such that: $$P_B(A)=P(A|B)=\frac {P(A\cap B)} {P(B)}$$

Then simply take the expectation with respect to $P_B$ rather than $P$:

$$E(A|B)=E_{P_B}(A)$$ $$E(A)=E_P(A)$$

What is the problem with this relative to the definition from the book?

EDIT: I now understand that my definition is w.r.t. an event, whereas the definition above is w.r.t. a sigma field, and that we can generate one wr.t. a sigma field from one w.r.t. an event. However I still don't quite get why we would define one w.r.t. a sigma field, or what it means. For example, how does the sigma field "contain information" about $A$?

EDIT 2: Another reason why this definition confuses me: how does this even pinpoint the conditional expectation as opposed to any other random variable that has equal expectation over the elements of the sigma algebra of $Y$? For example, if we simply set $E(Y|G)(\omega):=Y(\omega)$, then it conforms with the definition of conditional expectation given, even though it is clearly not what we mean by conditional expectation.

I think a large part of your confusion can be resolved by understanding how to think about $\sigma$-algebras as "containing information". Consider the probability space $(\Omega,\mathcal F,\mathbb P)$, and the random variable $X$ on this probability space. Intuitively, you should imagine $\omega \in \Omega$ being drawn according to the measure $\mathbb P$, and the realised $\omega$, in turn, determines $X$.

When we say that we "know" the information contained in $\mathcal F$, you should think of this as being able to take any set $E \in \mathcal F$, and being able to determine whether $\omega \in E$ or $\omega \notin E$. (Now is a useful time to recall the definition of a $\sigma$-algebra.)

Since $X$ is a random variable, it must be $\mathcal F$-measurable. Intuitively, what this means is that the information contained in $\mathcal F$ must fully determine the random variable $X$. Once we know the content of $\mathcal F$, we know exactly what the value of $X$ is. (Of course, the randomness of $X$ comes from the fact that we typically do not know $\mathcal F$, but only some sub-$\sigma$-algebra.) This captures the idea that the $\sigma$-algebra of the underlying probability space captures everything there is to know in the environment we are trying to model.

The conditional expectation $\mathbb E [X \vert \mathcal G]$ is our best guess of $X$, given that we know $\mathcal G$. Hence, $\mathbb E [X \vert \mathcal G]$ must be $\mathcal G$-measurable: the information contained in $\mathcal G$ must be enough to determine $\mathbb E [X \vert \mathcal G]$. This explains the first requirement of the definition. (Incidentally, regarding your second edit, your proposed conditional expectation works only when $Y$ is $\mathcal G$-measurable. If it isn't, then $Y$ is not (a version of) the conditional expectation with respect to $\mathcal G$.)

The second requirement captures the definition of the conditional expectation with respect to events. However, instead of choosing some specific event $G\in\mathcal G$, we allow the conditional expectation to vary over all possible events in $\mathcal G$. (Observe that if $\mathcal G$ were the $\sigma$-algebra generated by the event $G$, then this calculation is pretty much equivalent to calculating the conditional expectation with respect to an event.)

One advantage of conditioning with respect to $\sigma$-algebras rather than events is that it gives you a flexible apparatus for reasoning about things you might know in the future before you actually discover this knowledge. For example, suppose that I will know $\mathcal G$ tomorrow, then I can write down what my best guess of some random variable at that time will be, without necessarily waiting to see which event in particular is realised. (I imagine someone better versed in stochastic analysis than I am will be able to give you a better reason/example, but hopefully this is at least somewhat helpful.)

The definition of conditional expectation is "with respect to a $\sigma$-field", and your given definition is "with respect to an event". But $A$ can generate a $\sigma$-field $\{A,A^c,\emptyset,\Omega\}$, if you calculate the conditional expectation with respect to this $\sigma$-field, you get $$E[X|\sigma(A)]=\frac{E(X\mathbb 1_A)}{P(A)}\mathbb 1_A+\frac{E(X\mathbb 1_{A^c})}{P(A^c)}\mathbb 1_{A^c}$$ and $\frac{E(X\mathbb 1_A)}{P(A)}$ is the expectation of $X$ with respect to the probability measure $P_A$.

I think this is the connection. When the $\sigma$-field is not "easily generated", the definition in the picture works well.

• Thank you. What is the interpretation of a conditional expectation w.r.t a sigma field? Dec 18, 2017 at 19:42
• @Programmer2134 We consider a $\sigma$-field as the known information, and then the conditional expectation of $X$ WRT this field can be viewed as the "best prediction" with these known information. Dec 18, 2017 at 19:50
• ok thats good to know. How exactly should I think about this sigma algebra containing information? Dec 18, 2017 at 20:09
• @Programmer2134 I guess we can view a $\sigma$-field $\mathcal F$ as the set of events we can test a r.v. $X$, by looking at the average of $X$ under $A$. The finer $\mathcal F$ is, the more we can know by testing. Dec 18, 2017 at 20:37
• @Programmer2134 $\sigma(A)$ measurable functions have form $x\mathbb 1_A+y\mathbb 1_{A^c}$. Then you can use the definition to determine $x,y$. Dec 19, 2017 at 10:04