Maybe the question is trivial. Let $(X_{n})_{n \geq 0}$ ne a sequence of random variables and $N_{1},N_{2}$ stopping times with respect to the sequence $(X_{n})_{n \geq 0}$. Now I have Show that $$ \min(N_{1},N_{2}),~\max(N_{1},N_{2}) $$ are also stopping times. My Problem is to Show whether $N_{1} + N_{2}$ is also a stopping time. What about

$$ N := \sup\{n\geq 0|X_{n} \geq 0\}. $$ Is $N$ also a stopping time?


2 Answers 2


Note that

$$\begin{align*}\{N_1+N_2 =n\} = \{N_1+N_2 = n\} \cap \{N_2 \leq n\} &= \bigcup_{k=0}^n \{N_1+N_2 = n\} \cap \{N_2 = k\} \\ &= \bigcup_{k=0}^n \{N_1=n-k\} \cap \{N_2=k\}. \end{align*}$$

As $\{N_1 =n-k\} \in \mathcal{F}_{n-k} \subseteq \mathcal{F}_n$ and $\{N_2 = k\} \in \mathcal{F}_k \subseteq \mathcal{F}_n$ for any $k \leq n$, this implies that $\{N_1+N_2=n\} \in \mathcal{F}_n$, and so $N_1+N_2$ is a stopping time.

The random variable

$$N := \sup\{n \geq 0; X_n \geq 0\}$$

is, in general, not a stopping time. Take for instance a Poisson distributed random variable $T$ and define

$$X_n(\omega) := \begin{cases} -1, & T(\omega) \leq n, \\ 0, & T(\omega)>n \end{cases}$$

The canonical filtration $\mathcal{F}_n := \sigma(X_0,\ldots,X_n)$ satisfies $$\mathcal{F}_0 := \sigma(\{T=0\}) = \{\emptyset,\Omega,\{T=0\},\{T>0\}\}.$$ On the other hand, the construction of the process $(X_n)_{n \in \mathbb{N}_0}$ and the definition of $N$ implies that $$\{N=0\} = \{\omega; \forall n \geq 1: X_n(\omega)=-1\} = \{T \leq 1\}$$

which is clearly not contained in $\mathcal{F}_0$. Thus, $N$ is not a stopping time.

  • $\begingroup$ @Laura The stopping times may take the value $0$, and so $\{N_1+N_2=1\}$ may be non-empty, e.g. if $N_1(\omega)=1$ and $N_2(\omega)=0$. $\endgroup$
    – saz
    Nov 21, 2020 at 12:43
  • $\begingroup$ @Laura Well, yes, if the stopping times are greater or equal than $1$, then of course $\{N_1+N_2=1\}$ is the empty set, which is in $F_1$ because any $\sigma$-algebra contains the empty set. $\endgroup$
    – saz
    Nov 21, 2020 at 13:11
  • $\begingroup$ @Laura What makes you belive that there is something wrong...? $\endgroup$
    – saz
    Nov 21, 2020 at 14:54
  • $\begingroup$ @Laura The OP considers a filtration $(F_n)_{n \in \mathbb{N}_0}$... it's just a matter of preference whether one considers $n \in \mathbb{N}$ or $n \in \mathbb{N}_0$. I don't really see the problem, sorry. $\endgroup$
    – saz
    Nov 21, 2020 at 15:28
  • $\begingroup$ @Laura Yes, it's fine (... as long as your definition of "stopping time" includes that it takes values $\geq 1$). $\endgroup$
    – saz
    Nov 21, 2020 at 15:42

$N$ being a stopping time means that you can tell whether $\{N> k\}$ has occurred by looking at the values of $X_1,X_2,\dots,X_k$. This implies that if you know $X_1,X_2,\dots,X_k$, then you either know $N>k$, or you know the value of $N$.

Examining the values $X_1,\dots,X_k$, there are two cases:

  • If either $N_1> k$ or $N_2 >k$ has occurred, then it follows $N_1+N_2>k$, so you know that $\{N_1+N_2>k\}$ has occured.

  • If both $N_1\le k$ and $N_2\le k$, then you know the values of $N_1$ and $N_2$, so you know the value of $N_1+N_2$.

Either way, the information $(X_1,\dots,X_k)$ is enough to determine whether $N_1+N_2>k$, so $N_1+N_2$ is a stopping time.

On the other hand, $N=\sup \{n\mid X_n\ge 0\}$ is not a stopping time. For example, suppose the first $10$ values of $X_n$ are $0,1,2,3,4,-1,-2,-3,-4,-5$. Then if $X_n$ was never nonnegative again, you would have $N=4$, but if $X_n$ was nonnegative at some later time $n>9$, you would have $N\ge n$. Therefore, the values $X_0,X_1,\dots,X_9$ are insufficient to determine whether $N>9$, so $N$ is not a stopping time.


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