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Suppose $(\Omega, \mathcal{F}, P)$ is a probability space and $(U, \mathcal{\Sigma})$ is a measurable space. There seem to be two ways of defining the conditional expectation of a r.v. $X: \Omega \rightarrow \mathbb{R}$ given another r.v. $Y: \Omega \rightarrow U$, denoted as $E(X\vert Y)$:

  1. As a $\sigma(Y)$-measurable mapping from $\Omega$ to $\mathbb{R}$, defined as: $$E(X\vert Y) = E(X \vert \sigma(Y)). $$ where $\sigma(Y)$ is the sigma algebra of r.v. $Y$, which I think is also denoted as $Y^{-1}(\mathcal{\Sigma})$?

  2. As a $\mathcal{\Sigma}$-measurable mapping from $U$ to $\mathbb{R}$, defined as follows (from Wikipedia):

    Define measure Q on U to be the probability measure induced by $Y$ on $(U, \mathcal{\Sigma})$, as $Q(B) = P(Y^{−1}(B)), \forall B \in \mathcal{\Sigma}$.

    Define $E(X \vert Y)$ to be the integrable function $g:U \rightarrow \mathbb{R}$ such that

    $$ \int_{Y^{-1}(B)} X(\omega) \ d \operatorname{P} = \int_{B} g(u) \ d \operatorname{Q}, \forall B \in \mathcal{\Sigma}.$$

    If I am correct, this definition is related to the first one as: $$E(X \mid Y) \circ Y= E\left(X \mid Y^{-1} \left(\Sigma\right)\right). $$

I was wondering which of the above two is the definition of $E(X \mid Y)$?

Thanks and regards! References (links or books) will also be appreciated!

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  • $\begingroup$ How do you define $E(X|\sigma(Y))$? If it is the usual way, the random variable which satisfies $EX1_A=E[E(X|\sigma(Y)1_A]$ for $A\in\sigma(Y)$ and $1_A$ set indicator function, then two definitions are identical. $\endgroup$ Commented Feb 25, 2011 at 15:23
  • $\begingroup$ @mpiktas: For the definition, see en.wikipedia.org/wiki/Conditional_expectation#Formal_definition . Could you show me why they are are identical please? $\endgroup$ Commented Feb 25, 2011 at 15:27
  • $\begingroup$ $g$ is the measurable mapping from $(U,\Sigma)$ to $\mathbb{R}$, which means that it is a random variable, only its domain is a probability space $(U,\Sigma,P\circ Y^{-1})$. Since the measure on this probability space is induced by $Y$, distributional properties of $g$ and $E(X|\sigma(Y))$ will be the same. So strictly speaking these definitions define two different mathematical objects, but they have similar properties. $\endgroup$ Commented Feb 25, 2011 at 15:36

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I've only seen $E[X|Y]$ used to denote the first one. The second is a function $g$ such that $g(Y) = E[X|\sigma(Y)]$. (by the way, $g$ is only unique up to $Q$-null sets). Informally, you might write $g(y) = E[X|Y=y]$, which is actually correct when $Y$ is discrete (i.e. when $Q$ is atomic).

By definition, $E[X|\sigma(Y)]$ is a $\sigma(Y)$-measurable random variable that is supposed to answer the following: if I told you the value of $Y$, what would be your best estimate of the value of $X$? Your estimate would be different depending on the value of $Y$, so it should be some function $g$ of $Y$. Your second definition is referring to that function $g$.

If you want to prove that in the discrete case $g(y) = E[X|Y=y]$, call the right side $h(y)$. Then show that the random variable $h(Y)$ satisfies the conditions that uniquely define $E[X|\sigma(Y)]$: namely, $h(Y)$ is $\sigma(Y)$-measurable and for any $A \in \sigma(Y)$, $E[h(Y);A] = E[X;A]$. Note that in the discrete case, $A$ is necessarily a countable union of events of the form $\{Y=y_i\}$.

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  • $\begingroup$ Thanks! I was wondering what kinds of situation/applications one will use the first one, and what kinds of situation/applications one will use the second one? References (links or books) will also be appreciated! $\endgroup$ Commented Feb 25, 2011 at 6:57
  • $\begingroup$ I was wondering (1) why "The second is a function g such that g(Y)=E[X|Y]"? By the second definition, E[X|Y] is a mapping$: U \times \mathcal{F} \rightarrow \mathbb{R}$, and how can it be a function of r.v. Y? (2) How to show "g(y)=E[X|Y=y], which is actually correct when Y is discrete (i.e. when Q is atomic)"? Any reference? Thanks! $\endgroup$ Commented Feb 25, 2011 at 9:42
  • $\begingroup$ (1) I meant $E[X|Y]$ in sense 1. I've clarified. (2) I added a hint. I don't have any references for you offhand, sorry. $\endgroup$ Commented Feb 25, 2011 at 14:02

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