Prove the tower property for conditional expectation using filters

Question

While studying about martingales I came about the following proposition :
$$\mathbb{E}(\mathbb{E}(X|\mathcal{F}_n)|\mathcal{F}_m) = \mathbb{E}(X|\mathcal{F}_m)$$ if $m \le n$ where $\mathcal{F}_n$ is the filter defined on $X_0,...,X_n$ Is it possible to prove the same without using the field, measurability and indicator function arguments as given here : Steve Cheng's Notes ?
Note : I think $\mathbb{E}(X|\mathcal{F}_t)$ can be taken to be equal to $\mathbb{E}(X | X_0,X_1,X_2,.....,X_t)$ in a field-less world

Answer 1

The first equation is an application of the tower property. If $\mathcal H$ is a sub-$\sigma$-algebra of $\mathcal{G}$, then $E[X | \mathcal{H}] = E[E[X|\mathcal {G}] \mathcal{H}]$. There is no need for a filtration to use this property, but there is a need for $\sigma$-algebras.

As for your second question, the answer is a bit subtle. If you are talking about martingales, there must be a filtration. This is not inconsistent with taking $E[X|X_0,\ldots,X_n]$ though, because by definition $E[Z|X_0,\ldots,X_n] = E[Z| \sigma(X_0,\ldots,X_n)] = E[Z|\mathcal{F}^X_n]$. That is, if $X$ is a process, one can always consider the filtration generated by $X$, defined by $\mathcal{F}^X_n := \sigma(X_0,\ldots,X_n)$. In many situations this is the most natural choice, and some authors will talk about $X$ "being a martingale" without referring to a filtration. In these cases they always mean $X$ is a martingale relative to the filtration $\mathcal{F}^X$.