6

We have a notion of natural filtrations, which intuitively represents the history of the process as the process evolves over time.

We also have a notion of filtrations in general, which are increasing sequence of sub-sigma algebras.

Naturally, the latter concept is more abstract than the former, and I am having trouble getting a concrete grip on the latter.

In particular, if we have a stochastic process X, and a filter F, I tend to look at F as a natural filtration (although we only know it's a filtration in general, and not necessarily a natural one) of some other process Y. Can we do that?

As to why I am doing what I am doing, in many practical scenarios (like quantitative finance, which I am studying), we would be directly observing the process Y (say Y is the share price process) and hence our information would be the natural filtration of Y, but we might be interested in a slightly different process X (which might be the log of the share price or some other functional transformation say). In this scenario, the natural filtration of Y is simply a filtration from the perspective of X, and not a natural one.

Thanks a lot in advance!

Dhruv Gupta
  • 369

2 Answers2

4

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space equipped with a filtration $(\mathcal F_t)_{t\in\mathbb R_+}$.

For all $t\in\mathbb R_+$, let $Y_t:(\Omega,\mathcal F)\to(\Omega,\mathcal F_t)$ be defined for all $\omega\in\Omega$ by $Y_t(\omega)=\omega$. Then we clearly have $\sigma(Y_s,s\le t)=\mathcal F_t$, hence $(\mathcal F_t)_{t\in\mathbb R_+}$ is the natural filtration of the stochastic process $Y:(\Omega,\mathcal F)\to(\Omega^{\mathbb R_+},\bigotimes_{t\in\mathbb R_+}\mathcal F_t),\omega\mapsto(\omega)_{t\in\mathbb R_+}$.

So for any filtration $(\mathcal F_t)_{\mathbb R_+}$, there exists a measurable space $(E,\mathcal E)$ and a stochastic process $Y:\Omega\to E$ such that for all $t\in\mathbb R_+$, $\mathcal F_t=\sigma(Y_s,s\le t)$.

A much more interesting question would be to know whether this is true when $(E,\mathcal E)$ is fixed. For instance, for any filtration $(\mathcal F_t)_{t\in\mathbb R_+}$, does there exists a real-valued process $(Y_t:\Omega\to(\mathbb R,\mathcal B(\mathbb R))_{t\in\mathbb R_+}$ such that $(\mathcal F_t)_{t\in\mathbb R_+}$ is the natural filtration of $Y$? The answer is no, as shown is this post Is every sigma-algebra generated by some random variable?.

In many pratical scenarios like quantitative finance, as you said, you are very likely to deal with the natural filtration of an observable process $Y$ and you are interested in a different process $X$. But when $X$ is a one-to-one and onto function of $Y$, then the natural filtrations of $X$ and $Y$ are the same.

Will
  • 8,358
  • Very clear example! Thanks a ton! – Dhruv Gupta Mar 18 '19 at 05:26
  • @Will. Could you please elaborate on this expression: $Y_t(w) = w$. – Chris Jan 30 '21 at 18:10
  • It's just the definition of $Y_t$, whose purpose was to illustrate that any filtration can be seen as the natural filtration of some process $Y$. But here $Y$ takes values in $\Omega$, which might seem a bit strange. If we look for a real-valued random process $Y$, then the answer is negative in general. – Will Jan 30 '21 at 22:29
  • @Will I find this really interesting, but have difficulty understanding $Y$ as a process. I'm used to processes being of the form $\Omega\times\mathbb{R}_+\to(E,\mathcal{E}), (\omega,t)\mapsto Y_t(\omega)$. But with $Y$ in this form, what would $\mathcal{E}$ be? Thanks! – Bart Feb 22 '25 at 22:31
  • 1
    Your definition makes sense if all $Y_t$ share the same codomain $(E,\mathcal E)$. In my example, I used a specific codomain $(E,\mathcal E_t)$ for each $Y_t$. So the process I defined does not satisfy your definition of a process. – Will Feb 23 '25 at 06:50
0

Inspired by Will's answer I tried to think of one where the codomain of the process is constant. For convenience I consider the discrete time case.

Let $\mathbb{N}^*$ denote the set of integers $t>0$. Let $E = \bigcup_{t\in\mathbb{N}^*}\Omega^t$, where $\Omega^t$ is the $t$-ary Cartesian power of $\Omega$, and let $\mathcal{E}$ be the $\sigma$-field generated by the collection \begin{equation*} \mathcal{C} = \{C_1\times\cdots\times C_t: C_1\in\mathcal{F}_1,\ldots,C_t\in\mathcal{F}_t, t\in\mathbb{N}^*\}. \end{equation*} Let $(Y_t)_{t\in\mathbb{N}^*}$ be the stochastic process defined for each $t\in\mathbb{N}^*$ by \begin{equation*} Y_t:\Omega\ni \omega\mapsto\underbrace{(\omega,\ldots,\omega)}_{\text{$t$-fold}} \in\Omega^t. \end{equation*} Fix $t\in\mathbb{N}^*$. For any $u\in\mathbb{N}^*$ and $C_1\in\mathcal{F}_1,\ldots,C_u\in\mathcal{F}_u$, if $t = u$ then \begin{equation*} Y_t^{-1}(C_1\times\cdots\times C_u) = \bigcap^t_{s=1}C_s, \end{equation*} and otherwise $Y_t^{-1}(C_1\times\cdots\times C_u) = \varnothing$, so $Y_t^{-1}(\mathcal{C}) = \mathcal{F}_t$. Since $\sigma(Y_t^{-1}(\mathcal{C})) = Y_t^{-1}(\sigma(\mathcal{C}))$ we obtain\begin{equation*} \mathcal{F}_t = \sigma(Y_t^{-1}(\mathcal{C})) = Y_t^{-1}(\sigma(\mathcal{C})) = Y_t^{-1}(\mathcal{E}) . \end{equation*} As this holds for each $t\in\mathbb{N}^*$, we find for all $t\in\mathbb{N}^*$ that \begin{equation*} \mathcal{F}_t = \sigma\bigg(\bigcup^t_{s=1}\mathcal{F}_s\bigg) = \sigma\bigg(\bigcup^t_{s=1}Y_s^{-1}(\mathcal{E})\bigg) = \sigma(Y_s^{-1}(B):1\leq s\leq t, B\in\mathcal{E}). \end{equation*} So $(\mathcal{F}_t)_{t\in\mathbb{N}^*}$ is the natural filtration of $(Y_t)_{t\in\mathbb{N}^*}$.

Bart
  • 1,272
  • I, uh, perhaps I misread, because I was skimming too quickly. I'll retract my comment by deleting it... – Linas May 17 '25 at 03:59