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In some discussion regarding martingales, suppose we have a random process $\{X_k\}_{k \geq 0}$, authors use

$$\mathbb{E}[X_{k+1}|\mathcal{F}_k]$$

to denote the expectation with respect to a filtration, i.e., a increasing $\sigma$-algebra of the past random variables, $\mathcal{F}_k = \sigma(X_0, X_1, \ldots, X_k\}$.

A rough interpretation is that we are conditioning on increasing amount of information.

But why bother conditioning on a $\sigma$-algebra? (which is not unique, and can be very large, i.e., $\mathcal{F}_k = \text{power set}$, and contains things like $\varnothing$ which makes no sense in terms of information necessary)

Why not simply condition on the past random variables themselves? That is,

$$\mathbb{E}[X_{k+1}|X_k, X_{k-1}, \ldots, X_0]$$

Isn't the latter expression more interpretable and direct?

In what situation is conditioning on the filtration better than conditioning on the random variables themselves? Ultimately, I don't see why we need to condition on the filtration.

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  • $\begingroup$ You are probably thinking of discrete random variables. For general random variable the two conditional expectations are one and the same. What is your definition of $E(X_{k+1}|X_k,x_{k-1},...,X_0)$? $\endgroup$ – Kabo Murphy Jul 11 at 7:34
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Here is a case where it is more convenient to take the conditional expectation with respect to a filtration which may be an other one than the natural one: truncation arguments.

Suppose that we are studying a martingale differences sequence $\left(X_k\right)_{k\geqslant 1}$ with respect to a filtration $\left(\mathcal F_k\right)_{k\geqslant 0}$. For a fixed $R>0$, let $X_{k,R}:= X_k\mathbf 1\{\lvert X_k\rvert\leqslant R\}$. This is still measurable with respect to $\sigma(X_k)$ hence to $\sigma\left(X_1,\dots,X_k\right)$, but we may lose the martingale difference property (unless we are in the conditionally symmetric case).

Therefore, we define $$ D_{k,R}:= X_k\mathbf 1\{\lvert X_k\rvert\leqslant R\}-\mathbb E\left[X_k\mathbf 1\{\lvert X_k\rvert\leqslant R\}\mid \mathcal F_{k-1}\right]; $$ $$ D'_{k,R}:= X_k\mathbf 1\{\lvert X_k\rvert\gt R\}-\mathbb E\left[X_k\mathbf 1\{\lvert X_k\rvert\gt R\}\mid \mathcal F_{k-1}\right] $$ and $\left(D_{k,R},\mathcal F_k\right)$, $\left(D'_{k,R},\mathcal F_k\right)$ are both martingales differences sequences statisfying $X_k=D_{k,R}+D'_{k,R}$.

If we had taken instead the natural filtrations of $(X_k\mathbf 1\{\lvert X_k\rvert\leqslant R\})_{k\geqslant 1}$ (respectively $(X_k\mathbf 1\{\lvert X_k\rvert\gt R\})_{k\geqslant 1}$), that is, $$ D_{k,R}:=X_k\mathbf 1\{\lvert X_k\rvert\leqslant R\}-\mathbb E\left[X_k\mathbf 1\{\lvert X_k\rvert\leqslant R\}\mid \mathcal \sigma(X_{1,R},\dots ,X_{k-1,R})\right] $$ $$ D'_{k,R}:=X_k\mathbf 1\{\lvert X_k\rvert\gt R\}-\mathbb E\left[X_k\mathbf 1\{\lvert X_k\rvert\gt R\}\mid \mathcal \sigma(X_1-X_{1,R},\dots ,X_{k-1}-X_{k-1,R})\right] $$ then we would not be sure that the equality $X_k=D_{k,R}+D'_{k,R}$ holds.

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