# Promise of formal definition of conditional expectation: what is $E[X|Y=y]$ exactly?

There are many questions here related to this but I'm yet to see one that directly address this issue.

The promise of a formal definition of conditional expectation is that with it we may have a well-defined $$E[X|Y=y]$$ even when $$Y$$ is continuous. So, after lots of work, here we have $$E[X|\sigma(Y)]$$ or expectation with respect to a sub-field in general and we have proved it exists and unique up to measure zero. And I understand that $$E[X|\sigma(Y)]$$ is a random variable that takes exactly the same expectation as $$X$$ on each measurable set in $$\sigma(Y)$$. So what is the definition of $$E[X|Y=y]$$ exactly now? My intuition is that $$E[X|Y=y]=E[X|\sigma(Y)](\omega), \forall \omega \in \{\omega \in \Omega: Y=y\}$$. But I'm not sure and haven't seen a proof that $$E[X|Y=y]=E[X|\sigma(Y)](\omega)$$ is indeed a constant for all $$\omega \in \{\omega \in \Omega: Y=y\}.$$ Any clarification is appreciated, especially a definitive statement of what $$E[X|Y=y]$$ is exactly.

1. $$E[X | \sigma(Y)$$ is a RV measurable w.r.t. $$\sigma(Y)$$, which you seem to be OK with.

2. Any random variable $$Z$$ which is measurable w.r.t. $$\sigma(Y)$$ is essentially a function of $$Y$$, i.e., there exists a measurable $$f$$ such that $$Z = f(Y)$$ almost surely.

Putting 1 and 2 together, there exists a measurable function $$g$$ such that $$E[X | \sigma(Y)] = g(Y)$$ almost surely. You can think of $$E[X | Y=y]$$ as a shorthand for $$g(y)$$.

($$g$$ is of course not unique and determined up its a.s. equivalence class)

• Can you explain 2? It would make much sense given 2. but I can't immediately see why it's true. Of course, $\sigma(Y) \subset \sigma(Z)$ but how do you define $f$? Feb 19, 2019 at 4:52
• @DanielLi 2 is Doob-Dynkin's lemma. Feb 19, 2019 at 7:57
• Not very accurate writing. $g$ is defined up to its equivalence class modulo the distribution of $Y$. Feb 20, 2019 at 9:06

$$E(X|Y)$$ is measurable w.r.t. $$\sigma (Y)$$ and $$\sigma (Y)=\{Y^{-1}(A):A \,\text {Borel} \}$$. Let us show that any random variable $$Z$$ on $$(\Omega,\sigma(Y))$$ is of the form $$f(Y)$$ for some Borel measurable function $$f:\mathbb R \to \mathbb R$$. First consider the case when $$Z$$ is simple random variable: $$Z=\sum a_iI_{E_i}$$ with $$E_i$$'s in $$\sigma(Y)$$. We can write $$E_i=Y^{-1}(A_i)$$ for some Borel set $$A_i$$. Now you can verfiy that $$Z=f(Y)$$ where $$f=\sum a_iI_{E_i}$$. If $$Z$$ is non-negative then choose simple functions $$Z_n$$ increasing to $$Z$$ and write $$Z_n=f_n(Y)$$. Let $$f =\lim \sup f_n$$ (or $$\lim \sup f_n$$). Then we get $$Z=f(Y)$$. Finally write any $$Z$$ in terms of $$Z^{+}$$ and $$Z^{-1}$$ to complete the proof.

Define $$E(X|Y=y)$$ as $$f(y)$$.