understanding $\int_{F} E[X|\mathcal{F}]dP=\int_F E[X]dP$ in the definition of conditional expectation

Question

In the definition of conditional expectation, we require $\int_{F} E[X|\mathcal{F}]dP=\int_F X dP$ for all $F \in \mathcal{F}$, where $\mathcal{F}$ is a sub $\sigma$-algebra, and this condition implies $E[E[X|\mathcal{F}]]=E[X]$. What can go wrong if we alternatively define a conditional expectation to be a measurable random variable that satisfies $E[E[X|\mathcal{F}]]=E[X]$? Existence is definitely still true, so I think it is uniqueness is what we need to worry about, but I am having trouble thinking of a $\mathcal{F}$-measurable random variable satisfying $E[E[X|\mathcal{F}]]=E[X]$ that is not the conditional expectation using the correct definition.

Answer 1

Let$(\Omega, \mathcal A, P)$ be a probability space and $\mathcal F$ a sub-$\sigma$-algebra of $\mathcal A$. If $X:\Omega \to \mathbb R$ is a random variable, the conditional expectation $E[X|\mathcal F]$ is the "best $\mathcal F$-measurable function that approximates $X$" in the sense of the integral, that is, for all $F\in \mathcal F$ we have $$\int_FE[X|\mathcal F] dP = \int_F X dP$$ This idea is what my teacher uses to provide myself with an intuitive idea when I studied probability. It is supported by some examples: if you take $\mathcal F=\{\varnothing, \Omega\}$, then $E[X|\mathcal F]$ is a number (because every function that is $\{\varnothing, \Omega\}$-measurable must be a constant), and that constant is precisely $E[X]$. Thus we recover the idea that "the best number that can be used to summarize certain data is the mean".

• It is not difficult to think of a random variable $Y$ so that $E[Y]=E[X]$ but $Y \neq E[X|\mathcal F]$. For example, take $Y=E[X]$. To be more specific, let $$X:([0,1], \mathcal B, m) \to ([0,1], \mathcal B) \quad , \quad X(x)=x$$ where $\mathcal B$ is the Borel $\sigma$-algebra and $m$ is the Lebesgue measure. Pick $\mathcal F$ any sub-$\sigma$-algebra of $\mathcal B$. If $Y=\frac12$, then $E[X]=E[Y]$ but $\int_F Y dP \neq \int_F X dP$ for any interval $F=[0,a]$, $0<a<1$.