Natural Filtrations And Observation Histories

From a Process to Its Observation History

A filtration records how information accumulates over time. Given a process, one canonical choice is to let the information at time $t$ consist exactly of the observations of that process made by time $t$. This construction is useful as a baseline: any other filtration describing information available to an observer can be compared with it. It also makes precise the difference between seeing a process continuously and seeing only selected samples.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$. When Brownian motion is used below, $B_t$ denotes a standard Brownian motion on this space. Let $X=(X_t)_{t\ge0}$ be a process taking values in a measurable space $(E,\mathcal{E})$. The observations through time $t$ are the random variables $X_s$ for $0\le s\le t$; the sigma-field they generate is the raw natural filtration of $X$.

Definition: For each $t\ge0$, the raw natural filtration of $X$ is $$ \mathcal{F}^{X,0}_t=\sigma(X_s:0\le s\le t). $$ Here $\sigma(X_s:0\le s\le t)$ is the smallest sigma-field on $\Omega$ making each $X_s$, $s\le t$, measurable. Equivalently, it is generated by the events $X_s^{-1}(A)$, for $s\le t$ and $A\in\mathcal{E}$. “Raw” means that no completion by null sets or right-continuous modification is included in this definition.

The superscript $0$ emphasizes that this is the unaugmented construction. An ambient filtration can contain more information than the process history. For example, it may reveal an additional random variable at time zero. Conversely, a sampled observation history can contain less information than the full history of a continuously observed process.

The Minimality Property

Theorem: The family $(\mathcal{F}^{X,0}_t)_{t\ge0}$ is a filtration. Moreover, it is the smallest filtration $(\mathcal{G}_t)_{t\ge0}$ for which each observation $X_t$ is measurable with respect to $\mathcal{G}_t$. More precisely, if $X_s$ is $\mathcal{G}_s$-measurable for every $s$, then $\mathcal{F}^{X,0}_t\subseteq\mathcal{G}_t$ for every $t$.

Proof. If $0\le s\le t$, every observation used to generate $\mathcal{F}^{X,0}_s$ is also used to generate $\mathcal{F}^{X,0}_t$. Thus $\mathcal{F}^{X,0}_s\subseteq\mathcal{F}^{X,0}_t$, so the family is increasing and is a filtration. Now suppose $(\mathcal{G}_t)$ is a filtration and $X_r$ is $\mathcal{G}_r$-measurable for each $r$. For any $r\le t$, the filtration property gives $\mathcal{G}_r\subseteq\mathcal{G}_t$. Therefore every $X_r$ with $r\le t$ is $\mathcal{G}_t$-measurable. Since $\mathcal{F}^{X,0}_t$ is the smallest sigma-field making all those observations measurable, $\mathcal{F}^{X,0}_t\subseteq\mathcal{G}_t$. Conversely, $X_t$ is measurable with respect to $\mathcal{F}^{X,0}_t$ because it is one of the variables used to generate that sigma-field. These facts establish the claimed minimality. $\square$

This property explains why “natural” refers to the information generated by the process itself, not to all information on the probability space. If another filtration $(\mathcal{G}_t)$ already contains the process’s observations at their respective times, then it must contain the natural history as well. The inclusion can be strict.

Worked Example: A Two-Observation Coordinate Process

Let $\Omega=\{a,b,c,d\}$, and define a process by its two observations $X_0(a)=X_0(b)=0$, $X_0(c)=X_0(d)=1$, and $X_1(a)=X_1(c)=0$, $X_1(b)=X_1(d)=1$. Take $X_t=X_1$ for $t\ge1$. Find the natural sigma-fields at times $0$ and $1$.

At time zero, the only coordinate observation is $X_0$. Its two possible values have inverse images $\{a,b\}$ and $\{c,d\}$, so $\mathcal{F}^{X,0}_0=\{\varnothing,\{a,b\},\{c,d\},\Omega\}$.

At time one, both coordinates have been observed. Their ordered pairs are $(0,0)$ at $a$, $(0,1)$ at $b$, $(1,0)$ at $c$, and $(1,1)$ at $d$. Each outcome is therefore a singleton atom of the generated sigma-field, giving $\mathcal{F}^{X,0}_1=2^\Omega$, the set of all subsets of $\Omega$.

Check: $X_0$ alone cannot distinguish $a$ from $b$, so its sigma-field has four events, not sixteen. The second observation separates each pair: for example, $\{a\}=\{X_0=0\}\cap\{X_1=0\}$. The four singleton events and their unions give all $2^4=16$ subsets, as claimed.

Sampled Histories Versus Continuous Observation

Suppose a process is observed only at designated sampling times. At an intermediate time, the observer’s available information is generated by the samples already received, not by the unobserved values between samples. For Brownian motion, this distinction can be strict even when the sampling interval is short.

Worked Example: Brownian Motion Sampled Every Half Unit

Let $\Delta=\tfrac12$, and define the sampled history at time $t$ by $\mathcal{H}_t=\sigma(B_{k\Delta}: k\text{ is a nonnegative integer and }k\Delta\le t)$. Find this history at $t=\tfrac34$, and compare it with the full Brownian history $\mathcal{F}^{B,0}_{3/4}$.

The sample times not exceeding $\tfrac34$ are $0$ and $\tfrac12$, so $\mathcal{H}_{3/4}=\sigma(B_0,B_{1/2})=\sigma(B_{1/2})$, since $B_0=0$. In contrast, $\mathcal{F}^{B,0}_{3/4}$ contains $B_{3/4}$ as a measurable random variable.

The increment $B_{3/4}-B_{1/2}$ is independent of the Brownian history through $\tfrac12$, hence independent of $\mathcal{H}_{3/4}$, and has distribution $N(0,\tfrac14)$. Consequently, $\mathbb{E}[B_{3/4}\mid\mathcal{H}_{3/4}]=B_{1/2}$ and $\mathbb{E}[(B_{3/4}-B_{1/2})^2\mid\mathcal{H}_{3/4}]=\tfrac14$. If $B_{3/4}$ were $\mathcal{H}_{3/4}$-measurable, then its difference from $B_{1/2}$ would be measurable with respect to $\mathcal{H}_{3/4}$, and its conditional variance would be zero. The computed conditional variance is instead $\tfrac14>0$. Thus $B_{3/4}$ is not measurable with respect to the sampled history, and $\mathcal{H}_{3/4}$ is strictly smaller than $\mathcal{F}^{B,0}_{3/4}$.

Check: the elapsed time after the last sample is $\tfrac34-\tfrac12=\tfrac14$, exactly the variance of the independent Brownian increment. At $t=\tfrac12$, the sampled history is still not generally the full continuous history: the latter includes all $B_s$ for $s\le\tfrac12$, not only the endpoint sample.

The same minimality argument applies to sampled observations when they are organized by their arrival times. The sigma-field at time $t$ is generated by the samples that have arrived by then. It is not appropriate to insert an unobserved value merely because the underlying process has a value at every intermediate time.

Adding Information Beyond the Process

To compare filtrations, let $\mathcal{F}^{B,0}_t=\sigma(B_s:0\le s\le t)$ be Brownian motion’s raw natural filtration. If a random variable $Y$ is revealed at time zero, one can define $\mathcal{G}_t=\sigma(\mathcal{F}^{B,0}_t,Y)$. This remains increasing because the Brownian histories increase and the same information $\sigma(Y)$ is included at every time. When $Y$ concerns the future of the process, $\mathcal{G}_t$ can be strictly larger than its natural history.

Worked Example: Revealing a Future Brownian Value

For $t\ge0$, set $\mathcal{G}_t=\sigma(\mathcal{F}^{B,0}_t,B_1)$. Verify that this is a filtration and that it contains strictly more information than the Brownian natural filtration at $t=\tfrac12$.

If $s\le t$, then $\mathcal{F}^{B,0}_s\subseteq\mathcal{F}^{B,0}_t$. Adjoining $\sigma(B_1)$ to both gives $\mathcal{G}_s\subseteq\mathcal{G}_t$, so $(\mathcal{G}_t)$ is a filtration. By construction $B_1$ is $\mathcal{G}_{1/2}$-measurable.

The increment $B_1-B_{1/2}$ is independent of $\mathcal{F}^{B,0}_{1/2}$, has mean zero, and has variance $1-\tfrac12=\tfrac12$. Therefore $\mathbb{E}[B_1\mid\mathcal{F}^{B,0}_{1/2}]=B_{1/2}$ and $\mathbb{E}[(B_1-B_{1/2})^2\mid\mathcal{F}^{B,0}_{1/2}]=\tfrac12$. If $B_1$ were measurable with respect to $\mathcal{F}^{B,0}_{1/2}$, then $B_1-B_{1/2}$ would be measurable there and this conditional second moment would be zero. It is $\tfrac12$, so $B_1\notin\mathcal{F}^{B,0}_{1/2}$. Since $B_1\in\mathcal{G}_{1/2}$, the inclusion $\mathcal{F}^{B,0}_{1/2}\subseteq\mathcal{G}_{1/2}$ is strict.

Check: at time $0$, $B_0=0$ makes the raw Brownian natural sigma-field trivial, while $\mathcal{G}_0=\sigma(B_1)$ includes the nonconstant future value. At time $1$, $B_1$ is already in the natural history, so this particular extra-information construction adds no new variable $B_1$ then.

Two Filtrations and a Common Pitfall

For a process $X$ and a filtration $(\mathcal{G}_t)$, the inclusion $\mathcal{F}^{X,0}_t\subseteq\mathcal{G}_t$ says that the observations of $X$ through time $t$ are available in $\mathcal{G}_t$. The theorem gives a direct way to check this: verify that each observation $X_s$, $s\le t$, is measurable in $\mathcal{G}_t$. The natural filtration is the least such choice, but it need not be the filtration used in a particular model.

One common error is to define the information at time $t$ as $\sigma(X_t)$, using only the current value. The natural filtration uses the entire history $\sigma(X_s:s\le t)$. In the four-outcome example, $\sigma(X_1)$ has atoms $\{a,c\}$ and $\{b,d\}$, so it does not distinguish all outcomes. The actual history at time one also contains $X_0$, and the pair $(X_0,X_1)$ distinguishes $a,b,c,d$; consequently $\mathcal{F}^{X,0}_1=2^\Omega$. Replacing the history by the current value loses information.

Finally, the raw natural filtration should not be silently identified with its completed or right-continuous versions. Completion adds null events, and right-continuous augmentation imposes an additional continuity property on the filtration. Those are distinct operations; the definition here uses only the observations up to the stated time.

Check Your Understanding

For each question, distinguish the observations available by time $t$ from any information added by a larger filtration.

Write the raw natural filtration of a process $X$ at time $t$ using a sigma-field generated by its observations.
Why does $\mathcal{F}^{X,0}_s\subseteq\mathcal{F}^{X,0}_t$ hold whenever $s\le t$?
For Brownian motion sampled at times $k/2$, which observations generate the sampled history at $t=3/4$?
In the future-revelation example, why does the conditional variance show that $B_1$ is not measurable with respect to $\mathcal{F}^{B,0}_{1/2}$?
In the finite coordinate example, what information is lost if one uses $\sigma(X_1)$ instead of the full history at time one?