Proving Galmarino's Test

Solution 1:

Throughout my answer, $(X_t)_{t \geq 0}$ denotes a stochastic process with continuous sample paths on the canonical space, i.e.

$$X_t(\omega) := \omega(t), \qquad \omega \in \Omega := C([0,\infty)),$$

As usual we denote by

$$\mathcal{F}_t := \sigma(X_s; s \leq t)$$

the canonical filtation of $(X_t)_{t \geq 0}$ and set $\mathcal{F} := \sigma(X_s; s \geq 0)$. Moreover, we define $a_t: C([0,\infty)) \to C([0,\infty))$ by

$$a_t(\omega)(s) := \omega(s \wedge t), \qquad s \geq 0, \omega \in C[0,\infty).$$

For the proof of Galmarino's test we need an auxiliary result.

Lemma: $$\mathcal{F}_t = a_t^{-1}(\mathcal{F}) \qquad \text{for all $t \geq 0$.}$$

Proof: First, we check that $a_t: (\Omega,\mathcal{F}_t) \to (\Omega,\mathcal{F})$ is measurable. Since $\mathcal{F} = \sigma(X_s; s \geq 0)$, this is equivalent to $$X_s \circ a_t: (\Omega,\mathcal{F}_t) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$$ being measurable for each $s \geq 0$. This, however, follows directly from the identity $$X_s \circ a_t(\omega) = (a_t(\omega))(s) = \omega(s \wedge t) = X_{s \wedge t}(\omega).$$ The measurability of $a_t$ yields $a_t^{-1}(\mathcal{F}) \subseteq \mathcal{F}_t$. To prove $\mathcal{F}_t \subseteq a_t^{-1}(\mathcal{F})$, it suffices to show that $\omega \mapsto X_s(\omega)$ is $a_t^{-1}(\mathcal{F})/\mathcal{B}(\mathbb{R})$-measurable for all $s \leq t$. Since

$$X_s(\omega) = \omega(s) = \omega(s \wedge t) = a_t(\omega)(s)= X_s(a_t(\omega))$$

for all $s \leq t$ we get

$$\{X_s \in B\} = \{X_s(a_t) \in B\} = \{a_t \in X_s^{-1}(B)\} \in a_t^{-1}(\mathcal{F})$$

for any $B \in \mathcal{B}(\mathbb{R})$. This shows that $X_s$ is indeed $a_t^{-1}(\mathcal{F})/\mathcal{B}(\mathbb{R})$-measurable for $s \leq t$.

Corollary 1: (Baby version of Galmarino's test) For any set $A \in \mathcal{F}$ the following statements are equivalent:

$A \in \mathcal{F}_t$

If $\omega \in A$, $\omega' \in \Omega$ are such that $X_s(\omega)=X_s(\omega')$ for all $s \leq t$, then $\omega' \in A$.

Proof: "1. $\implies$ 2.": Let $A \in \mathcal{F}_t$. By the above lemma, there exists $C \in \mathcal{F}$ such that $A=a_t^{-1}(C)$. Now if $\omega \in A$ and $\omega' \in \Omega$ are such that $X_s(\omega)=X_s(\omega')$ for all $s \leq t$, then $a_t(\omega)=a_t(\omega')$, and so $$1_A(\omega') = 1_{a_t^{-1}(C)}(\omega') = 1_C(a_t(\omega')) = 1_C(a_t(\omega))=1_A(\omega),$$ i.e. $\omega' \in A$.

"2. $\implies$ 1.": It follows our assumption that we have $$\omega \in A \iff a_t(\omega) \in A, $$ and so $$1_A(\omega) = 1_A(a_t(\omega)) = 1_{a_t^{-1}(A)}(\omega)$$ for all $\omega \in \Omega$, i.e. $A = a_t^{-1}(A)$. It follows from the above lemma that $A \in \mathcal{F}_t$.

Corollary 2: (Galmarino's test) For any random time $T$ the following statements are equivalent:

$T$ is a stopping time, i.e. $\{T \leq t\} \in \mathcal{F}_t$ for all $t \geq 0$.

If $\omega, \omega' \in \Omega$ are such that $T(\omega)=t$ and $X_s(\omega)= X_s(\omega')$ for all $s \leq t$, then $T(\omega')=t$.

Proof: "1. $\implies$ 2." Since $$\{T = t\} = \{T \leq t\} \backslash \bigcup_{k \in \mathbb{N}} \{T \leq t-1/k\} \in \mathcal{F}_t,$$ it follows from Corollary 1 that for any $\omega \in \{T = t\}$ the implication $$\{\forall s \leq t=T(\omega): \, \, X_s(\omega) = X_s(\omega')\} \implies \omega' \in \{T = t\}$$ holds which proves the assertion.

"2. $\implies$ 1." Set $A := \{T \leq t\}$. If $\omega \in A$ and $\omega' \in \Omega$ are such that $X_s(\omega) = X_s(\omega')$ for all $s \leq t$, then it follows from our assumption that $T(\omega')=t$, and so $\omega' \in A$. Applying Corollary 1 yields $\{T \leq t\} = A \in \mathcal{F}_t$.

Remarks:

Note that we haven't used the continuity of the sample paths; so, in fact, the claim does not hold only true for stochastic processes with continuous sample paths.
In 2. we may replace $T(\omega)=t$ and $T(\omega')=t$ by $T(\omega) \leq t$ and $T(\omega') \leq t$, respectively.

Thanks to @Shashi who helped a lot to improve this answer; (s)he come up with the idea to prove Galmarino's test using Corollary 1.