Conditional Expectation of X with respect to a $\sigma$-algebra $\mathcal{G}$

conditional probabilityconditional-expectationfiltrationsmeasure-theory

I read this definition from my lecture handout:

If X is a random variable on a countable probability space $(\Omega,\mathcal{F},\mathbb{P})$, then the conditional expectation $\mathbb{E}_\mathbb{P}(X|\mathcal{G})$ of X with respect to $\mathcal{G}$ is defined as a random variable which satisfies, for every $\omega\in A_i$,
$$\mathbb{E}_\mathbb{P}(X|\mathcal{G})(\omega)=\sum_{i\in I}\frac{\mathbb{E}_\mathbb{P}(X\mathbb{1}_{A_i})\mathbb{1}_{A_i}}{\mathbb{P}(A_i)}\quad\quad \forall A_i\in \mathcal{P}$$
$\mathcal{P}$ is a countable partition that generates $\mathcal{G}$.

And I have the following questions:

  1. Some of the other materials give this definition as a conditional expectation of X with respect to the whole $\sigma$-field $\mathcal{G}$, which without indicating the small omega, i.e.:$$\mathbb{E}_\mathbb{P}(X|\mathcal{G})=\sum_{i\in I}\frac{\mathbb{E}_\mathbb{P}(X\mathbb{1}_{A_i})\mathbb{1}_{A_i}}{\mathbb{P}(A_i)}.$$
    I personally support this definition, because $\mathbb{E}_\mathbb{P}(X|\mathcal{G})(\omega)$ seems to be a partial condition and it should be equivalent to $\mathbb{E}_\mathbb{P}(X|\mathcal{G})\mathbb{1}_{A_i}=\mathbb{E}_\mathbb{P}(X|A_{i})$. So which one is the correct definition?
  2. Why we have two indicator functions over there? As per my logic, $\mathbb{E}_\mathbb{P}(X|\mathcal{G})=\sum_{i\in I}\mathbb{E}_\mathbb{P}(X|A_i)=\sum_{i\in I}\frac{\mathbb{E}_\mathbb{P}(X\mathbb{1}_{A_i})}{\mathbb{P}(A_i)}.$ I don't know where is the other $\mathbb{1}_{A_i}$ comes from.

Best Answer

In what follows

  • $(\Omega ,\mathcal F, \mathbb P)$ is a probability space,

  • $X$ is a random variable defined on $\Omega $,

  • $\mathcal G\subseteq \mathcal F$ is a sub-$\sigma $-algebra generated by a countable partition $\{A_i\}_{i\in I}$ of $\Omega $, formed by $\mathcal F$-measurable sets.

The first thing to bear in mind is that the conditional expectation $\mathbb E_{\mathbb P}(X|\mathcal G)$ is another random variable (rather than just a number). Since random variables are nothing but functions defined on $\Omega $, the notation $$ \mathbb E_{\mathbb P}(X|\mathcal G)(\omega ) $$ makes perfect sense as it indicates the value taken by the function $\mathbb E_{\mathbb P}(X|\mathcal G)$ on a given point $\omega $ of $\Omega $.

The second important point is that $\mathbb E_{\mathbb P}(X|\mathcal G)$ is constant on every set $A_i$ (although its constant value may change among the various $A_i$).

By the definition of conditional expectation, the constant value taken on $A_i$ by $\mathbb E_{\mathbb P}(X|\mathcal G)$ is the expected value of $X$ on $A_i$, namely $$ \mathbb E_{\mathbb P}(X|A_i ) := \frac {\mathbb E_{\mathbb P}(X1_{A_i} )}{\mathbb P(A_i)}. $$

Denoting the above expected value simply by $e_i$, the question is how should we express the function which is supposed to take on the value $e_i$ on $A_i$. I guess the best way to do this is simply to write $$ \sum_{i\in I}e_i1_{A_i}, \tag 1 $$ observing that if $\omega $ lies in some $A_j$, then $1_{A_i}(\omega )$ vanishes for every $i$, except for $i=j$, in which case $1_{A_i}(\omega )=1$, so the above sum comes out as $e_j$, which is precisely what we expect.

Notice that there is no $\omega $ in expression (1) for the same reason many people consider it incorrect to say

"consider the function $\sin(x)$"

In fact, the function is called simply "$\sin$", whereas "$\sin(x)$" is meant to denote the value of the function $\sin$ at the given real number $x$.

According to this we therefore have that $$ \mathbb E_{\mathbb P}(X|\mathcal G)= \sum_{i\in I}e_i1_{A_i}, \tag 2 $$ which, upon substituting the appropriate value for $e_i$, is exactly what the OP writes in the first question.

If we want to explicitly indicate the dependency of these functions on a variable $\omega $, I'd write $$ \mathbb E_{\mathbb P}(X|\mathcal G) (\omega ) = \sum_{i\in I}e_i1_{A_i}(\omega ), \quad\forall \omega \in \Omega , \tag 3 $$ noticing that now $\omega $ shows up on both sides.

On the down side, I believe mixing the LHS of (3) with the RHS of (2) is incorrect so I agree 100% with the OP that their second formula is preferable over the first one.

Regarding the OP's second question, I see a problem in the sense that $$ \mathbb{E}_\mathbb{P}(X|\mathcal{G}) $$ is supposed to be a random variable, as already discussed, while $$ \sum_{i\in I}\mathbb{E}_\mathbb{P}(X|A_i) $$ strikes me as a number.

Inserting the missing $1_{A_i}$ on the last expression would make it a function, which is more in line with the functional nature of the conditional expectation.

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