# Definition of conditional expectation with two random variables.

Most definitions of the conditional expectation use $$\sigma$$-algebra arguments:

Let $$X$$ be a random variable on a probability space $$(E, \mathcal{E}, \mathbb{P})$$ with $$\mathbb{E}| X|\le\infty$$ and $$\mathcal{E}^\prime$$ be sub-$$\sigma$$-algebra w.r.t. $$\mathcal{E}$$, the conditional expectation $$\mathbb{E}\big[X| \mathcal{E}^\prime\big]$$ is any random variable $$Y$$ on $$\mathcal{E}^\prime$$ such that for all $$A\in\mathcal{E}^\prime$$, $$\int_{A} Xd\mathbb{P}=\int_{A} Yd\mathbb{P}$$.

But I also encounter conditional expectations of two random variables $$X$$ and $$Y$$ which are simply like $$\mathbb{E}[X|Y]$$. Most of the time $$X$$ is a function of $$Y$$ (and another random variable). What is the formal definition of this version?

My personal definition is:

Given a random variables $$X$$ on $$(E, \mathcal{E}, \mathbb{P})$$ and another random variable $$Y$$ on $$(E, \mathcal{E}^\prime, \mathbb{P})$$ with $$\mathcal{E}^\prime\subseteq\mathcal{E}$$, the conditional expectation $$\mathbb{E}[X|Y]$$ is any random variable $$Z$$ on $$(E, \sigma(Y), \mathbb{P})$$ such that for all $$A\in\sigma(Y)$$, $$\int_{A} Xd\mathbb{P}=\int_{A} Zd\mathbb{P}$$.

$$\sigma(Y)$$ is the $$\sigma$$-algebra generated by $$Y$$. Is this correct?

• The first definition is from Rick Durrett's Probability Theory and Examples. Jan 8 at 9:00
• Not any random variable $Z$ on $(E, \mathcal{E}, \mathbb{P})$ such that for all $A\in\sigma(Y)$, $\int_{A} Xd\mathbb{P}=\int_{A} Zd\mathbb{P}$. $Z$ has to be measurable w.r.t. $\sigma (Y)$. Jan 8 at 9:03
• You should note that $Z=X$ always satisfies your requirement ! Jan 8 at 9:16
• Thanks. I've updated the definition and added another $\sigma$-algebra! Jan 8 at 9:28
• It is still wrong as the case $\mathcal E'=\mathcal E$ shows. Jan 8 at 9:34

Conditioning on $$\sigma$$-algebra or on random variable (or rather the $$\sigma$$-algebra generated by the random variable) amounts to the same thing. This is because any sub-$$\sigma$$-algebra is generated by a random variable (by random variable I mean measurable function).
Suppose $$(\Omega,\mathscr{F},\mu)$$ is a probability space, and let $$\mathcal{A}$$ a $$\sigma$$-algebra contained $$\mathscr{F}$$, and $$X:(\Omega,\mathscr{F})\rightarrow(\mathbb{R},\mathcal{B}(\mathbb{R}))$$ a numeric random variable. Define the function $$Y:\Omega\rightarrow\Omega$$ as $$Y(\omega)=\omega$$. Notice that $$Y$$ is a $$(\Omega,\mathscr{F})$$-$$(\Omega,\mathcal{A})$$ random variable for if $$A\in\mathcal{A}$$, $$Y^{-1}(A)=A\in\mathcal{A}\subset\mathscr{F}$$, furthermore, $$\sigma(Y)=\{Y^{-1}(A):A\in\mathcal{A}\}=\mathcal{A}$$. Hence $$\mathbb{E}[X|Y]:=\mathbb{E}[X|\sigma(Y)]=\mathbb{E}[X|\mathcal{A}]$$