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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.

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2 Answers 2

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  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)

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  • $\begingroup$ 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 $f$? $\endgroup$
    – Daniel Li
    Feb 19, 2019 at 4:52
  • $\begingroup$ @DanielLi 2 is Doob-Dynkin's lemma. $\endgroup$ Feb 19, 2019 at 7:57
  • $\begingroup$ Not very accurate writing. $g$ is defined up to its equivalence class modulo the distribution of $Y$. $\endgroup$
    – zhoraster
    Feb 20, 2019 at 9:06
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$E(X|Y)$ 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 $f(y)$.

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