In the theory of stochastic process, besides the $\sigma$-algebra $\mathcal {F}$, we have an increasing sequence of $\sigma$-algebras $\{{\mathcal {F}}_{{t}}\}_{{t\geq 0}} $ called filtration. According to Wikipedia,

a filtration is often used to represent the change in the set of events that can be measured, through gain or loss of information.

What is confusing me is that the probability measure $\mu$ is unchanged. So from the beginning, we have already known the probability of every events in every $\mathcal {F}_{t} $.

So why we use filtration and pretend we know nothing beyond $\mathcal {F}_{t} $ at time t?

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    $\begingroup$ What happens to the definition of martingale if you take $\mathcal F_t=\mathcal F$ for all $t$ ? $\endgroup$ Jun 3, 2020 at 12:23
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    $\begingroup$ If $\mathcal {F}_{t}$ = $\mathcal {F}$, then every $X_{s}$ is $\mathcal {F}_{t}$-measurable. {$X_{t}$} cannot be a martingale or all $X_{t}$ are the same. This is why do we need filtration. Thank you! $\endgroup$
    – Alfred
    Jun 6, 2020 at 12:56

1 Answer 1


You know the probability of every event in every $\mathcal{F}_t$, but the idea is that at time $t$ you know specifically which event you are in. For an easy example, you can think of flipping two fair coins. The sample space is $\Omega = \{ HH, HT, TH, TT\}$, and the terminal $\sigma$-algebra is $\mathcal{F} = 2^\Omega$. Then $\mathcal F_n$ will be what you know from observing the first $n$ coin tosses. $\mathcal F_0 = \{\emptyset, \Omega\}$ is just the trivial $\sigma$-algebra because you don't have any information (except whether or not the coins were actually tossed), so you can just say the probability that you end up in any particular event in $\mathcal F$. Now $\mathcal F_1 = \{\emptyset, \Omega, \{HH, HT\}, \{TH, TT \} \}$ because you will know exactly which of those events you are in after observing the first toss; e.g. if it was $H$, you're in $\{HH, HT\}$ (and $\Omega$). And $\mathcal F_2 = \mathcal F = 2^\Omega$ because after observing both coin tosses you know exactly what the outcome is, so you can say for sure whether or not you're in any particular subset of $\Omega$.


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