Let $(\mathcal F_n)_n$ a filtration. We say that a process $(X_n)_{n\in \mathbb N}$ is a predictable process for the filtration if $X_0$ is $\mathcal F_0$ measurable and $X_n$ is $\mathcal F_{n-1}$ measurable for all $n>0$.

I don't really understand what such a property means. Could someone give me some insight?


1 Answer 1


The classical interpretation of $\sigma$-algebras is information.

A nice example comes from quantitative finance. Suppose we have a certain amount of money $V_n$ at a certain (discrete) time $t_n$. Suppose we decide to invest a certain percentage $\alpha_n$ of this money in a risky title which has value $S_n$ at time $t_n$ and put the remaining part $\beta_n$ in our bank account.

$S_n$ can be modeled as a random variable, but so can $\alpha_n$, $\beta_n$: our fund's allocation varies based on how the title's value fluctuates.

You can conceptualize the sigma algebra $\mathcal F_n$ generated by $S_n$ as the information from the values of $S$ in the first $n$ periods.

What makes $S_n$ and $\alpha_n$ different is that we have control of the amount of money $\alpha_{n+1}$ that we want to invest at the time $t_n$ in the title $S$ because this decision must be made before $t_{n+1}$. In other words, the value of $\alpha_{n+1}$ must depend exclusively on a measurable function of the first $n$ values of $S$.

We translate this concept to math as

$$\alpha_{n+1}\in \mathcal F_n$$

i. e. we impose that $\alpha_n$ is a predictable process: based on the information at time $t_n$, we must already know what value $\alpha_{n+1}$ has: $$\mathbb E[\alpha_{n+1}\mid \mathcal F_n]=\alpha_{n+1}$$

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    $\begingroup$ for anyone wondering, the last line is equivalent to being predictable. $\endgroup$ Jul 10, 2020 at 14:02

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