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Let $\Omega$ be the set of applications in $\mathbb{R}_{+}$ which are right continuous and have left limit (except in 0). For all $t\in\mathbb{R}_{+}$, we define the coordinated application $X_t$ as $X_t(\omega)=\omega(t)\quad (\omega \in \Omega)$. Given $a \in \mathbb{R}$, prove that the set $\{X_t=a\}$ is in $\mathcal{F}_t$ but not in $\mathcal{F}_{t-}$.

Here $\mathcal{F}_t$ is the natural filtration $\mathcal{F}_t=\{X_s: s\leq t\}$ and $\mathcal{F}_{t-}=\sigma\{\mathcal{F}_s :s<t\}$.

Intuitively is clear, but I don't know how to formalize it.

saz
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MaTeAr
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1 Answers1

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For these kind of assertions on filtrations which are intuitively clear but, somehow, hard to pin down, I have found Galmarino's test to be a useful tool. Typically, the intuition comes from the understanding that "the filtration $\mathcal{F}_t$ encodes the information up to time $t$" and Galmarino's test makes a rigorous statement out of this.

(Easy version of Galmarino's test) Let $(Y_t)_{t \geq 0}$ be a stochastic process with canonical filtration $\mathcal{F}_t^Y = \sigma(Y_s; s \leq t)$. For every $A \in \mathcal{F}_t^Y$ the following implication holds: If $\omega \in A$ and $\omega' \in \Omega$ are such that $Y_s(\omega)=Y_s(\omega')$ for all $s \leq t$, then $\omega' \in A$.

See this question for a proof.

Let's use this result to prove your assertion. Since $(X_t)_{t \geq 0}$ is càdlàg, the process $Y_t := X_{t-}$ is well-defined. Using the càdlàg property of the sample paths of $(X_t)_{t \geq 0}$ and the fact that $X_t = \lim_{s \downarrow t} X_{s-} = \lim_{s \downarrow t}Y_s$, we find that $$\mathcal{F}_t^Y = \sigma(X_{s-}; s \leq t) = \sigma(X_s; s < t) = \mathcal{F}_{t-}^X$$ and $$\forall s < t\::\: X_s(\omega)=X_{s}(\omega') \iff \forall s \leq t\::\: Y_s(\omega)=Y_{s}(\omega').$$ Applying the above theorem, we obtain that any $A \in \mathcal{F}_{t-}^X$ satisfies the following implication: If $\omega \in A$ and $\omega' \in \Omega$ are such that $X_s(\omega)=X_s(\omega')$ for all $s<t$, then $\omega' \in A$. This implication fails to hold for $A=\{X_t=a\}$. Take for instance,

$$\omega := a \cdot 1_{[t,\infty)} \quad \text{and} \quad \omega':=(a+1) 1_{[t,\infty)},$$

then $\omega \in A$ and $X_s(\omega)=0=X_s(\omega')$ for all $s<t$, but $\omega' \notin \{X_t=a\}$. Consequently, $\{X_t=a\} \notin \mathcal{F}_{t-}^X$.

saz
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