Proving Galmarino's Test

Question

Galmarino's Test gives a condition equivalent to being a stopping time. It says:

Let $X$ be a continuous stochastic process with index set $\mathbb{R}_+$ (i.e. each sample path is a continuous function of time). Let $\mathscr{F}$ be the filtration generated by $X$. Then a random time $T$ is a stopping time iff

for every pair of outcomes $\omega$ and $\omega'$, $T(\omega) = t$, $X_s(\omega) = X_s(\omega')$ for $s \leq t \implies T(\omega') = t$

The condition essentially says that the map $T$ restricted to $\{T \leq t\}$ factors through $(X_s)_{s \leq t}$, and I can prove that it is necessary by using a monotone class argument. I don't know how to prove the converse though.

Answer 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.

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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:

1. $A \in \mathcal{F}_t$
2. 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:

1. $T$ is a stopping time, i.e. $\{T \leq t\} \in \mathcal{F}_t$ for all $t \geq 0$.
2. 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.

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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.

Answer 2

I think saz's proof is wrong in the last step when he claims $T\cdot I_{\{T\le t\}}$ is $\mathcal{F}_t$ measurable by Lemma 2. Also, his proof only works for real valued stochastic process. In fact, Galmarino Test works for process taking value in any measurable space $(E,\mathcal{E})$.

Galmarino Test: Let $(X_t)_{t\ge0}$ be a $(E,\mathcal{E})$ valued canonical process defined on the canonical probability space $(\Omega,\mathcal{F},P)$, that is, $\Omega=E^{[0,+\infty)}$, $\mathcal{F}=\mathcal{E}^{[0,+\infty)}$ and $X_t$ is the coordinate map on $E^{[0,+\infty)}$: for any $t\ge0$: $X_t(\omega)=\omega(t)\;\forall \omega\in \Omega$. Let $\mathcal{F}_t=\sigma(X_s,0\le s\le t)$ for $t\ge0$ be the natural filtration. Suppose a map $T$ from $\Omega$ to $[0,+\infty]$ is a random time. Then the following two statements are equivalent:

(a) $T$ is a $\mathcal{F}_t$ stopping time

(b) For any $\omega,\omega'\in\Omega$ and any $t\ge0$, $T(\omega)\le t$ and $X_s(\omega)=X_s(\omega')\forall s\le t$ imply $T(\omega)=T(\omega')$.

Proof: For each $t\ge0$, define $a_t$ to be a map from $\Omega$ to $\Omega$ such that for any $\omega\in\Omega(=E^{[0,+\infty)}$), $a_t(\omega)(s)=\omega(s\wedge t)$ for $s\ge0$.

Loosely speaking, $a_t(\omega)$ keeps the part of $\omega$ before $t$ and reset part of $\omega$ after $t$ to be $\omega(t)$. By the definition of $a_t$, for any $\omega,\omega'\in\Omega$, $a_t(\omega)=a_t(\omega')$ is equivalent to $\omega(s)=\omega'(s)\forall s\le t$, which is also equivalent to $X_s(\omega)=X_s(\omega)\forall s\le t$ (because $X_s$ is the coordinate map). Thus statement (b) is equivalent to the follow statement:

(b') For any $\omega,\omega'\in\Omega$ and any $t\ge0$, $T(\omega)\le t$ and $a_t(\omega)=a_t(\omega')$ imply $T(\omega)=T(\omega')$.

Claim 1: (b') is equivalent to the following statement:

(b'') For any $\omega,\omega'\in\Omega$ and any $t\ge0$, $T(\omega)\le t$ and $a_t(\omega)=a_t(\omega')$ imply $T(\omega')\le t$.

Let's first show (b') implies (b''). Suppose (b') holds. Let $\omega,\omega'\in\Omega$ and $t\ge0$ such that $T(\omega)\le t$ and $a_t(\omega)=a_t(\omega')$. By (b'), $T(\omega')=T(\omega)$. Since we already have $T(\omega)\le t$ so $T(\omega')\le t$.

Next, we show (b'') implies (b'). Suppose (b'') is true. Let $\omega,\omega'\in\Omega$ and $t\ge0$ such that $T(\omega)\le t$ and $a_t(\omega)=a_t(\omega')$. We need to show $T(\omega)=T(\omega')$.

Define $r=T(\omega)$ so $r\le t$ (becauase $T(\omega)\le t$). Now we have $$T(\omega)\le r\quad (1)\quad \mbox{ and }a_r(\omega)=a_r(\omega')\quad (2)$$ (1) is simply because of the definition of $r$. (2) is because $a_t(\omega)=a_t(\omega')$, that is, $\omega(s)=\omega'(s)\forall s\le t$. Since $r\le t$, so of course we also have $\omega(s)=\omega'(s)\forall s\le r$, which is (2). By (1)(2) and using (b''), we have $T(\omega')\le r$ which is $$T(\omega')\le T(\omega)\quad (*)$$ Define $r'=T(\omega')$. So $r'=T(\omega')\le T(\omega)\le t$. Similarly, we have $$T(\omega')\le r'\quad (3)\quad \mbox{ and }a_{r'}(\omega)=a_{r'}(\omega')\quad (4)$$ By (3)(4) and using (b'') (we need to exchange the role of $\omega$ and $\omega'$ when using (b'')), we have $T(\omega)\le r'$, which is $T(\omega)\le T(\omega')$. This combined with (*) shows $T(\omega)=T(\omega')$.

Claim 2: $a_t$ is $\mathcal{F}_t/\mathcal{F}$ measurable for each $t\ge0$.

Recall $X_t$ is the coordinate map so $\mathcal{F}=\mathcal{E}^{[0,+\infty)}=\sigma(X_t,t\ge0)$. Now let's fix $t\ge0$. $a_t$ takes value in $\Omega=E^{[0,+\infty)}$ so to show $a_t$ is $\mathcal{F}_t/\mathcal{E}^{[0,+\infty)}$ measurable, we only need to show each coordinate of its coordinate is $\mathcal{F}_t/\mathcal{E}$ measurable, that is, to show $X_s\circ a_t$ is $\mathcal{F}_t/\mathcal{E}$ measurable for each $s\ge0$.

For each $s\ge0$, $X_s\circ a_t(\omega)=a_t(\omega)(s)=\omega(s\wedge t)=X_{s\wedge t}(\omega)$ is $\mathcal{F}_t/\mathcal{E}$ measurable because $X_{s\wedge t}$ is $\mathcal{F}_{s\wedge t}/\mathcal{E}$ measurable and $\mathcal{F}_{s\wedge t}\subseteq\mathcal{F}_t$. Proof of claim 2 is done.

Claim 3: $\mathcal{F}_t=\{a_t^{-1}(A)|A\in\mathcal{F}\}$ for each $t\ge0$.

For any $A\in\mathcal{F}$, $a_t^{-1}(A)\in\mathcal{F}_t$ (claim 2), so $\{a_t^{-1}(A)|A\in\mathcal{F}\}\subseteq\mathcal{F}_t$. Now we only need to show $\mathcal{F}_t\subseteq\{a_t^{-1}(A)|A\in\mathcal{F}\}$. For any $B\in\mathcal{F}_t=\sigma(X_s,s\le t)$, there exist $t'\in [0,t]$ and $S\in\mathcal{E}$ such that $B=X_{t'}^{-1}(S)$. Now for any $\omega\in\Omega$, $X_{t'}(\omega)=\omega(t')=\omega(t'\wedge t)=a_t(\omega)(t')=X_{t'}(a_t(\omega))$. This shows $X_{t'}=X_{t'}\circ a_t$. Thus $B=X_{t'}^{-1}(S)=(X_{t'}\circ a_t)^{-1}(S)=a_t^{-1}(X_{t'}^{-1}(S))\in \{a_t^{-1}(A)|A\in\mathcal{F}\}$ where the last step is because $X_{t'}$ is $\mathcal{F}_{t'}/\mathcal{E}$ measurable so $X_{t'}^{-1}(S)\in\mathcal{F}_{t'}\subseteq\mathcal{F}$. Therefore $\mathcal{F}_t\subseteq\{a_t^{-1}(A)|A\in\mathcal{F}\}$. Claim 3 is proved.

Now we are ready to prove the theorem. Since (b) is equivalent to (b') which is also equivalent to (b''), now we only need to show (a) and (b'') are equivalent.

Step 1: Show (a) implies (b''). Suppose (a) is true. Let $\omega,\omega'\in\Omega$ and $t\ge0$ such that $T(\omega)\le t$ and $a_t(\omega)=a_t(\omega')$. We need to show $T(\omega')\le t$.

Since $T(\omega)\le t$ and by (a) and claim 3, $\omega\in\{T\le t\}\in\mathcal{F}_t=\{a_t^{-1}(A)|A\in\mathcal{F}\}$. Thus there exists $S\in\mathcal{F}$ such that $\{T\le t\}=a_t^{-1}(S)$. So $\omega\in a_t^{-1}(S)$. Thus $a_t(\omega)\in A$. Because $a_t(\omega)=a_t(\omega')$, $a_t(\omega')\in A$ which implies that $\omega'\in a_t^{-1}(A)=\{T\le t\}$ so $T(\omega')\le t$.

Step 2: Show (b'') implies (a). Suppose (b'') is true. We need to show $\{T\le t\}\in\mathcal{F}_t$ for any $t\ge0$. Now fix $t\ge0$. We claim $\{T\le t\}=a_t^{-1}(\{T\le t\})$.

To show this claim, we first show $\{T\le t\}\subseteq a_t^{-1}(\{T\le t\})$. For any $\omega\in\{T\le t\}$. So $T(\omega)\le t$. Define $\omega'=a_t(\omega)$ so $a_t(\omega')=a_t(a_t(\omega))=a_t(\omega)$ where in the last step we used the fact $a_t\circ a_t$ by the definition of $a_t$. Now we have $T(\omega)\le t$ and $a_t(\omega)=a_t(\omega')$ and thus by (b'') $T(\omega')\le t$ so $T(a_t(\omega))\le t$, thus $a_t(\omega)\in \{T\le t\}$ hence $\omega\in a_t^{-1}(\{T\le t\})$. Therefore $\{T\le t\}\subseteq a_t^{-1}(\{T\le t\})$ is proved. Next, we show $a_t^{-1}(\{T\le t\})\subseteq\{T\le t\}$. For any $\omega'\in a_t^{-1}(\{T\le t\})$, $a_t(\omega')\in\{T\le t\}$ thus $T(a_t(\omega'))\le t$. Define $\omega=a_t(\omega')$ so $T(\omega)\le t$ and $a_t(\omega)=a_t(a_t(\omega'))=a_t(\omega')$. By (b''), $T(\omega')\le t$. Hence $\omega'\in\{T\le t\}$. $a_t^{-1}(\{T\le t\})\subseteq\{T\le t\}$ is proved.

Therefore, we have proved $\{T\le t\}=a_t^{-1}(\{T\le t\})$. Now it's clear $\{T\le t\}=a_t^{-1}(\{T\le t\})\in\mathcal{F}_t$ because $\{T\le t\}\in\mathcal{F}$ ($T$ is a random time) and $a_t$ is $\mathcal{F}_t/\mathcal{F}$ measurable (claim 2).

Answer 3

Got a hint from http://wt.iam.uni-bonn.de/fileadmin/WT/Inhalt/people/Karl-Theodor_Sturm/Lectures/vorlesungWS0809/sheet1.pdf

Consider the problem in the canonical space. Define

$\alpha_t(\omega(\cdot)) = \omega(\cdot \wedge t)$

Using the monotone class theorem, we can show that the mapping:

$\alpha_t: (\Omega,{\mathcal F}_t^X) \to (\Omega,\mathcal{F})$

is measurable. And using the condition, we can show that

$ (T\le t) = \alpha_t^{-1}(T\le t)$

Therefore, $(T\le t)\in \mathcal{F}_t^X$.