Confusion in understanding definition of conditional expectation

Question

Let me define it first:

Let $X$ an integrable random variable on a probability space $(\Omega, S, P)$, and $C$ is a sub-sigma algebra of $S$. Then there exists a unique $C$-measurable random variable $Y$ such that $\int_{E}YdP$=$\int_{E}XdP$ for all $E\in C$. This random variable $Y$ is called the expectation of $X$ given $C$ and denoted by $E(X/C)$.

What I know is if we have a random variable $X$ on a probability space then the expectation is given by $E[X]$=$\int_{\Omega}XdP$ now I want to understand what the above definition of conditional expectation is saying, like if someone can give some example related to the above definition then that will be helpful. Thanks

Answer 1

See chapter 4 of https://services.math.duke.edu/~rtd/PTE/PTE5_011119.pdf for more detail and examples. My favorite example is the one from undergrad probability: Let $X$ and $Y$ be discrete random variables. Then $$E(X \mid Y = y) = \sum_{x}xP(X = x \mid Y = y) = \frac{1}{P(Y = y)}\sum_{x}P(X = x, Y = y) = \frac{1}{P(Y = y)}E(X 1_{Y = y}).$$ This says that $E(X \mid Y = y)$ is the average of $X$ over the set where $Y = y$. In general, assuming $E(X^2) < \infty$, you can show that $E(X \mid Y)$ is the function $h(Y)$ of $Y$ that minimizes $E((h(Y) - X)^2)$. So in this sense, $E(X \mid Y)$ is the closest function of $Y$ to $X$.

Answer 2

A very simple example can be given with an standard dice. Let the probability space $(\Omega ,\mathcal{F},P)$ where $\Omega :=\{1,\ldots ,6\}$, $\mathcal{F}:=2^{\Omega }$ and $P(\{\omega \}):=\frac1{6}$ for every $\omega \in \Omega $.

Now suppose that the random variable $X:\Omega \to \mathbb{R}$ represent a dice, it means that $X(\omega )=\omega $ so $P( X=k)=\frac1{6}$ when $k\in\{1,\ldots ,6\}$, and is zero otherwise. Then a sub-$\sigma $-algebra of $\mathcal{F}$ is $\mathcal{G}:=\{\{1,3,5\},\{2,4,6\},\emptyset ,\Omega \}$, then we have that

$$ \mathrm{E}[X|\mathcal{G}](\omega )=\begin{cases} \frac1{P(\{1,2,3\})}\int_{\{1,3,5\}}X\,d P,&\text{ when }\omega \in\{1,3,5\}\\ \frac1{P(\{2,4,6\})}\int_{\{2,4,6\}}X\,d P,&\text{ when }\omega \in\{2,4,6\} \end{cases} $$

The previous relation follows from the fact that if $A\in \mathcal{G}$ is an atom of the probability space $(\Omega , \mathcal{G}, P)$ then $\mathrm{E}[X|\mathcal{G}]$ can be chosen to be constant in $A$, as there is no $B\subset A$ with $P(B)<P(A)$ and $P(B)\neq 0$. Then from the equality

$$ \int_{A}\mathrm{E}[X|\mathcal{G}]\,d P=\int_{A}X\,d P,\quad \text{ for every }A\in \mathcal{G} $$

and if $\mathrm{E}[X|\mathcal{G}]$ is constant in $A$, it follows that $\mathrm{E}[X|G](\omega )=\frac1{P(A)}\int_{A}X\,d P$ for every $\omega \in A$ (notice that we also can set the above instead of "for every $\omega \in A$" as "for almost every $\omega \in A$").