I'm quite new to time series analysis. I'm reading a text book, where the conditional expected value of a process $(X_t)$ is written as $E[X_t|\mathcal{F}_{t-1}]$, where $\mathcal{F}_{t-1}$ is called the "natural filtration", which apparently (from what I understood) is a sigma-algebra that contains all the past observations in the series.

Can someone explain in simple terms, why does $\mathcal{F}_{t-1}$ need to be a sigma-algebra? Can't we just say that it's the collection of all $X_s$ for $s<t$?

  • $\begingroup$ There are many possible path $X_s$. The filtration is the "collection" of the paths $\endgroup$
    – dodo
    Nov 4, 2023 at 18:46

1 Answer 1


I'll admit that from a practical point of view, your intuition is correct. $E[X_t|\mathcal{F}_{t-1}]$ is the expected value of $X_t$ given the past history. The use of a sigma algebra is a more rigorous/formalized way to describe it. There's explanation of the interpretation on Wikipedia.


$$E[X_t|(X_s):s<t]$$ is a random variable, not an expectation, since it depends on the values of $(X_s):s<t$. We can therefore say that $E[X_t|(X_s):s<t]$ is measurable with respect to $\sigma((X_s):s<t)$, which is just the filtration $\mathcal{F}_{t-1}$ (i.e., sigma algebra formed from subsets of infinite sample sequences $X_n$ that share a particular subsequence $(X_s=x_s):s<t$ of particular values). So, these are cylinder sets based on the first $t-1$ values.

So, that's the interpretation of the notation. It's a little confusing because we also see conditional expectations with respect to random variables or a particular value. This is just a very formal way to state it.

$\mathcal{F}$ Example

As you mentioned in your comments, we can construct a simple filtration using the Bernoulli process. This is actually how I came to understand filtrations as well, and its the simple mental model I keep handy when reading abstract theorems.

Let $X_t \in \{0,1\}$ What is the filtration at say, $t=1$?

Well, its the sigma algebra formed from the $t$-tuples of all possible assignments to $(X_t):t<2$:


What about $t=2$? We'll for each 1-tuple that formed $\mathcal{F}_1$ we can append one of the possible values of $X_2$ to it to get a set of 2-tuples. The sigma algebra over these 2-tuples results in 13 sets (events):


And so on for higher values of $t$. As you can see, filtrations can get big very fast. However, you can see why the terminology makes sense...if you think of the set of infinite sequences of 0's and 1's, each such assignment (to infinite variables!) constitutes a single sample point from the sample space of the Bernoulli process. In this context, we see that what a filtration does is define progressively finer sets of events, where previous events are split into several, more refined, events. So we are going from coarse to fine in terms of how we form our sigma algebras. This is a lot like a filter, which goes from a coarse mesh to a fine mesh to pull out particles.

I should clarify one thing here: the above is not strictly correct, because I really need to form the sigma algebra not on the tuples, but on the inverse image of the tuple back onto the underlying sample space, since random variables are measurable wrt a measure on some underlying sample space. However, that would add a lot of notation and not much more understanding. We almost never see the actual sample space (we just assume it exists), we just see the random variables, so what most of us do is turn the range of the random variable into a sample space and then treat observations as direct views of this new sample space. So, for the above example, we simply take the sample space to be $\Omega =\{0,1\}$ with the associated probability measure $P(\omega)=0.5$. Then, $X_t$ becomes a trivial random variable (i.e., its just the identity function).

Sorry for the theoretical aside, but I figured someone was going to bring it up and its good to at least have exposure to the full probabilistic interpreation, even though most of us don't concern ourselves with this mystical $\Omega$ that forms the domain for $X(\omega)$.

Now, this is a simple process, but we can still meaningfully describe $E[X_t|\mathcal{F}_{t-1}]$. From the definition of conditional expectation with respect to a sigma algebra, the conditional expectation means:

$$E[X_t|\mathcal{F}_{t-1}] \;\mathrm{exists}\; \iff E[X_t|S]=\int_S X_t dP_{X_t}\; \forall S \in \mathcal{F}_{t-1}$$

So, part of the issue with understanding conditional expectations on sigma algebras is that its a definition not a result of the usual conditional expectation with respect to an event. What you would need to do is verify that the definition holds for this process. This is easy for the Bernoulli process since $E[X_t|\mathcal{F}_{t-1}]=0.5$ since the $X_t$ are iid. The conditional expectation for more complex stochastic processes is not so simple to calculate.

But, bottom line is that the conditional expectation wrt a sigma algebra defines a new random variable, its not just a simple value like a normal expectation. Since its a random variable, its needs to be measureable wrt some sigma algebra. In this case, we are providing the filtration.

  • $\begingroup$ Thanks. Let me check if I understood correctly...is $\mathcal{F}_{t-1}$ is a collection of fixed (not random) values? $\endgroup$
    – gbow28
    Oct 23, 2015 at 19:35
  • $\begingroup$ $\mathcal{F}_{t-1}$ is a sigma algebra, which by definition is a collection of measurable sets, not values. It does no harm to replace it by the sigma algebra generated by $X_{t-1}$, though, which might help clarify its role in the theory. $\endgroup$
    – whuber
    Oct 23, 2015 at 19:45
  • $\begingroup$ Could you please give me a simplistic example of what this sigma-algebra $\mathcal{F}_{t-1}$ would look like? For example, assume that $(X_t)$ is a process that takes only values 0 or 1 at any given time $t$. $\endgroup$
    – gbow28
    Oct 23, 2015 at 19:49
  • $\begingroup$ Another clarifying question: does $\mathcal{F}_{t-1}$ contain only the time period $t-1$, and not everything prior to $t$? I thought it was the latter one. $\endgroup$
    – gbow28
    Oct 23, 2015 at 20:10
  • $\begingroup$ I posted such an example at stats.stackexchange.com/a/123754. $\endgroup$
    – whuber
    Oct 23, 2015 at 21:50

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