# Trying to understand $\mathbb E[X\mid \mathcal F]$ and Martingale concept

Q1) Let $X$ a r.v. on a probability space $(\Omega ,\mathcal F,P)$ and $\mathcal G\subset \mathcal F$ a sub $\sigma -$algebra. What does $$\mathbb E[X\mid \mathcal G]\ \ ?$$ I understand what is $\mathbb E[X\mid Y]$ when $Y$ is an r.v., but not when it's a $\sigma -$algebra.

Q2) Let $(M_t)$ a stochastic process adapted to the filtration $\{\mathcal F_t\}_t$. I know that $(M_t)$ is a martingale if $\mathbb E[|M_t|]<\infty$ and $$\mathbb E[M_t\mid \mathcal F_s]=M_s,\quad s\leq t.$$ Could you explain me what does mean the last condition ? I would interpret is that when $t>s$, it's enough to know $M_s$ to now $M_t$ given $\mathcal F_s$, but I have no intuition neither any idea of what it can mean.

(Q1) I think it's helpful to think of it as an integral. For outcome $$\omega\in\Omega$$ we have that $$\mathbb E [X \mid \mathcal G](\omega)=\frac{1}{P(A)}\int_A X(x) P(dx)$$, where $$A$$ is the "smallest" set in $$\mathcal G$$ that contains $$\omega$$ (as long as $$P(A)\neq0$$). If $$X$$ is constant on $$A$$, then $$\mathbb E [X \mid \mathcal G](\omega)=X(\omega).$$

Another way to think of $$\mathcal G$$ is representing how accurately we can "resolve" the probability space. If we can distinguish between $$X(\omega_1)$$ and $$X(\omega_2)$$ for any two outcomes, then we have perfect resolution as far as $$X$$ is concerned. However, if $$X(\omega_1)\neq X(\omega_2)$$ but $$\mathcal G$$ doesn't contain any sets that have $$\omega_1$$ but not $$\omega_2$$ and vice versa (i.e. $$\omega_1\in A$$ implies $$\omega_2\in A$$ and vice versa), then if $$\mathcal G$$ defines our "resolution" we cannot distinguish between $$\omega_1$$ and $$\omega_2$$. Hence even though $$X(\omega_1)\neq X(\omega_2)$$ we have that $$\mathbb E [X \mid \mathcal G](\omega_1)=\mathbb E [X \mid \mathcal G](\omega_2)$$.

I find the picture on the Wikipedia article very helpful. Where you can see that $$\mathbb E [X \mid \mathcal G]$$ is just $$X$$ but averaged according to how well $$\mathcal G$$ resolves the probability space.

There are lots of other good answers to similar questions on MathSE as well if you search around.

(Q2) Regarding the Martingale property, consider a simple example: a gambler repeatedly flips a fair coin and gains $$1$$ dollar on heads and loses $$1$$ dollar on tails. Let $$X_n$$ be the gambler's balance after the $$n^{th}$$ coin flip. We can see that the expected winnings on the next flip is just the current balance of the gambler, but lets look at it regarding filtrations.

Let $$X_0=0$$ be the gamblers initial balance, then $$X_1$$ can be either $$-1$$ or $$1$$ hence $$\sigma(X_1)=\left\{\emptyset, \{X_1=-1\}, \{X_1=1\}, \Omega\right\}.$$ Note that $$\{X=i\}$$ means $$\{\omega\in\Omega \mid X(\omega)=i\}.$$

In general, we can show that $$\mathbb E[X_{n+m}\mid\sigma(X_1,\ldots,X_n)]=X_n.$$ This means that the distribution of the random variable $$\mathbb E[X_{n+m}\mid\sigma(X_1,\ldots,X_n)]$$ is the same as the distribution of $$X_n.$$

(Edit: actually it means a bit more than that: $$\mathbb E[X_{n+m}\mid\sigma(X_1,\ldots,X_n)]$$ and $$X_n$$ are the same random variable. This is a subtle, but important point. If I write $$X=Y$$ and perform the random experiment that gives $$X=5$$, then you know precisely that $$Y=5$$ as well, since they are the same random variable. But merely having the same distribution only gives $$P(X=5)=P(Y=5)$$, but they could be independent or have some arbitrary dependence structure, i.e. an arbitrary joint distribution---as long as it has the appropriate marginals.)

As a simple example, consider $$\mathbb E[X_2\mid\sigma(X_1)].$$ If $$X_1(\omega)=-1$$ then $$E[X_2\mid\sigma(X_1)](\omega)=-1$$ since $$X_2$$ has equal chances of being $$0$$ and $$-2$$, and if $$X_1(\omega)=1$$ then $$E[X_2\mid\sigma(X_1)](\omega)=1$$ since $$X_2$$ has equal chances of being $$0$$ and $$2$$. Thus $$E[X_2\mid\sigma(X_1)]$$ behaves exactly like $$X_1.$$ Of course, this is a very simple example. One could go on to show that $$E[X_3\mid\sigma(X_1)]=X_1,$$ $$E[X_3\mid\sigma(X_1,X_2)]=X_2,$$ etc. They aren't very difficult calculations and are quite instructive.

The intuition for Q1 I often found useful is that a $\Sigma$-algebra in some sense is a model for known information. The sets in the $\Sigma$-algebra are the ones you have detailed information about, and the sets outside of it are indistinguishable to you, i.e. you don't know what happened or not.

When you condition on a random variable, you are really conditioning on the $\Sigma$-algebra, induced by that random variable. Hence, the notion of conditional expectation you describe is more general than the one taught in the standard undergrad probability course...

As for Q2, the intuition behind the definition of a martingale is to describe a stochastic process which is expected on average to stay in the same place. For example, an unbiased random walk is an example of a martingale, as would be any Ito process with zero drift.

This is the idea behind the condition. The LHS is the expectation of where the process should be on average in some future time (think of $s$ as "now" in that context, and $t>s$ is some future time). The RHS is the value now. Thus, the condition forces the expected future location of the process to be its current location.

• thank you for your answer. I will read it in detail and give you a feed back if necessary. Commented Jul 21, 2016 at 16:47