Comparing two Stopping Times is in Sigma Field up to either Stopping Time

Last updated Nov 1, 2022

Let $\mathcal{F} = {\mathcal{B}_{n} : n \in \mathbb{N}}$ be a Discrete-Time Filtration over $\Omega$ and let $\nu, \nu’ : \Omega \to \bar{\mathbb{N}}$ be Stopping Times on $\mathcal{F}$. Then the following statements hold

1. $[\nu < \nu’] \in \mathcal{B}{\nu}\cap \mathcal{B}{\nu’}$
2. $[\nu = \nu’] \in \mathcal{B}{\nu}\cap \mathcal{B}{\nu’}$
3. $[\nu \leq \nu’] \in \mathcal{B}{\nu}\cap \mathcal{B}{\nu’}$

We show the following sequence of results

1. $[\nu < \nu’] \in \mathcal{B}_{\nu}$
2. $[\nu = \nu’] \in \mathcal{B}_\nu$

Then we have the following implications

1. $[\nu \leq \nu’] \in \mathcal{B}{\nu}$ since $[\nu < \nu’] \cup [\nu = \nu’] = [\nu \leq \nu’]$ and $\mathcal{B}\nu$ is a Sigma Algebra.
2. $[\nu < \nu’] = [\nu \geq \nu’]^{C} = [\nu’ \leq \nu]^{C} \in \mathcal{B}{\nu’}$ since $[\nu’ \leq \nu] \in \mathcal{B}{\nu’}$ by implication (1) and $\mathcal{B}_{\nu’}$ is a Sigma Algebra. This establishes statement (1). $\checkmark$
3. $[\nu = \nu’] = \mathcal{B}_{\nu’}$ by symmetry and result (2). This establishes statement (2). $\checkmark$
4. $[\nu \leq \nu’] \in \mathcal{B}{\nu’}$ since $[\nu \leq \nu’] = [\nu’ < \nu]^{C} \in \mathcal{B}{\nu’}$ by statement (1) and because $\mathcal{B}_{\nu’}$ is a Sigma Algebra. This establishes statement (3). $\checkmark$

Now to establish results (1) and (2). Observe that for $n \in \mathbb{N}$ $$[\nu < \nu’] \cap [\nu = n] = [n < \nu’] \in \mathcal{B}{n}$$ by condition (3) in Equivalent Conditions for being a Stopping Time. Thus $[\nu < \nu’] \in \mathcal{B}{\nu}$. $\checkmark$

Next, observe that for $n \in \mathbb{N}$ $$[\nu = \nu’] \cap [\nu = n] = [n = \nu’] \in \mathcal{B}{n}$$ by definition of a Stopping Time. Thus $[\nu = \nu’] \in \mathcal{B}{\nu}$. $\checkmark$.

This completes the proof. $\blacksquare$

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