Is $\{\tau+\sigma<k\}\in \mathcal{F}_k$?

Question

Let me assume that $\tau, \sigma$ are optional times w.r.t $\Bbb{F}:=(\mathcal{F}_t)_{t\geq 0}$ which means that for all $k\geq 0$ we have $\{\tau<k\}\in \mathcal{F}_k$ and $\{\sigma<k\}\in \mathcal{F}_k$. I want to show that $\{\tau+\sigma<k\}\in \mathcal{F}_k$.

My idea was as following. Let me first remark that $\{\tau\wedge\sigma<k\}=\{\tau<k\}\cup \{\sigma<k\}\in \mathcal{F}_k$. Now let me define $\tau':=(\tau \wedge k)+1_{\{\tau<k\}}$ and $\sigma':=(\sigma \wedge k)+1_{\{\sigma<k\}}$. Then clearly $\tau', \sigma'$ are both $\mathcal{F}_k$-measurable by definition and hence $\tau'+\sigma'$ is also $\mathcal{F}_k$-measurable. It would be nice if one could say that $\{\tau+\sigma<k\}=\{\tau'+\sigma'<k\}$. But I see no way to prove this. Can someone give me a hint?

Edit I know in that $\{\tau+\sigma\leq k\}\in \mathcal{F}_k$ maybe this can also be helpful.

Answer 1

You can write $$\{\sigma+\tau<k\}=\bigcup_{q_{1},q_{2},\in\Bbb{Q}\cap[0,\infty)\,:\,q_{1}+q_{2}<k}\bigg(\{\sigma<q_{1}\}\cap\{\tau<q_{2}\}\bigg)$$

Now $\{\sigma<q_{1}\}$ and $\{\tau<q_{2}\}$ are both in $\mathcal{F_{k}}$ as $q_{1},q_{2}<k$ for all $q_{1},q_{2}$ . And hence so is the countable union.

(Try to recall the trick used to show that sum of two Borel measurable functions is Borel measurable. That's precisely what I am using here)

Answer 2

One way to demonstrate the condition is as follows:

Let's assume that $\tau$ and $\sigma$ are optional times with respect to the filtration $\mathbb{F} := (\mathcal{F}_{t})_{t \geq 0}$, which means that for all $k \geq 0$, we have ${\tau < k} \in \mathcal{F}_k$ and ${\sigma < k} \in \mathcal{F}_k$. We want to show that $\{\tau + \sigma < k\} \in \mathcal{F}_k$.

Since $\tau$ and $\sigma$ are optional times, we have:

$$\{\tau + \sigma < k\} = \bigcup_{n=1}^\infty \left(\left\{\tau < \frac{k}{2n}\right\} \cap \left\{\sigma < \frac{k}{2n}\right\}\right)$$

Note that $\{\tau < \frac{k}{2n}\} \in \mathcal{F}_{\frac{k}{2n}}$ and $\{\sigma < \frac{k}{2n}\} \in \mathcal{F}_{\frac{k}{2n}}$. Since $\mathcal{F}$ is a filtration, we have $\{\tau < \frac{k}{2n}\} \in \mathcal{F}_k$ and $\{\sigma < \frac{k}{2n}\} \in \mathcal{F}_k$ for all $n$. Therefore, $\{\tau + \sigma < k\} \in \mathcal{F}_k$.

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Another way to show this is:

Let $k \geq 0$ be fixed. Since $\tau$ and $\sigma$ are optional times, we have that $\{\tau < k\} \in \mathcal{F}_k$ and $\{\sigma < k\} \in \mathcal{F}_k$.

Now, consider the set $\{\tau + \sigma < k\}$. For any $\omega \in \{\tau + \sigma < k\}$, there exists $n \in \mathbb{N}$ such that $\tau(\omega) \leq n$ and $\sigma(\omega) \leq k - n$. Thus, we have

$$\{\tau + \sigma < k\} \subseteq \bigcup_{n=0}^{k} \{\tau = n\} \cap \{\sigma < k - n\}$$

Since ${\tau = n} \in \mathcal{F}n$ and $\{\sigma < k-n\} \in \mathcal{F}_{k-n}$, it follows that $\{\tau = n\} \cap \{\sigma < k - n\} \in \mathcal{F}n \otimes \mathcal{F}_{k-n} \subseteq \mathcal{F}_k$, where $\mathcal{F}_n \otimes \mathcal{F}_{k-n}$ denotes the product $\sigma$-algebra.

Therefore, we have

$$\{\tau + \sigma < k\} \subseteq \bigcup_{n=0}^{k} \{\tau = n\} \cap \{\sigma < k - n\} \in \mathcal{F}_k$$, which implies that $\{\tau + \sigma < k\} \in \mathcal{F}_k$.

Since $k \geq 0$ was arbitrary, we have shown that $\{\tau + \sigma < k\} \in \mathcal{F}_k$ for all $k \geq 0$, which implies that $\tau + \sigma$ is an optional time.

To conclude, we have shown that if $\tau$ and $\sigma$ are optional times with respect to the filtration $(\mathcal{F}_t)_{t\geq 0}$, then $\tau+\sigma$ is also an optional time. This result has important applications in stochastic calculus and stochastic processes, where it allows us to show that certain stopping times are indeed well-defined and have the desired properties.

Particularly I prefer the first way