I am asked to prove that the minimum and sum of two stopping times are both stopping times themselves. Before engaging in the proofs, I am curious about why I am asked to prove these things. For the minimum case, it seems natural that people are interested in $n\wedge\tau$, where $\tau$ is some stopping time. For example, what is the expected wait time of the event that stock X reaches a face value of $Y before 2050? The meaning of the sum of stopping times, however, is not obvious to me. I looked at https://math.stackexchange.com/questions/232774/sum-of-two-stopping-times-is-a-stopping-time and https://math.stackexchange.com/questions/219366/minimum-of-two-stopping-times-is-a-stopping-time, but did not find a desirable explanation. My questions are then

  1. Is my interpretation of the minimum of stopping times correct?
  2. What does it mean by summing two stopping times? Is there a concrete example?
  3. It seems to me that a lot of things you can come up with using two stopping times $\nu$ and $\tau$ are also stopping times (despite the fact that I don't know what they mean). For example, $|\nu-\tau|_+$, $\max\{\nu,\tau\}$, etc. Again, I don't believe that there is a point in extending this exercise. Am I correct?
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    $\begingroup$ A fun little paper "Which Functions of Stopping Times are Stopping Times?" projecteuclid.org/download/pdf_1/euclid.aop/1176996983 ."Examples" of sum of stopping times: Proof of Lemma 2.1 in hec.unil.ch/halbrech_files/Impulse.pdf and proof of Lemma 5.2 in finmath.stanford.edu/seminars/documents/goldberg.pdf . I do not guarantee that you will find these "examples" to be satisfying if you have a practical bent. $\endgroup$ – Mark L. Stone Jul 20 '16 at 19:57
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    $\begingroup$ One practical reason that you might want to have practice at proving things are stopping times is in order to use results that hold for stopping times. It's often the case that a problem (a real, practical problem) might be tricky until you can show you can apply some neat little result that makes it straightforward. $\endgroup$ – Glen_b Jul 20 '16 at 23:19

1) Your interpretation is correct.

2) For summing stopping times, imagine you have some multidimensional random process where your stopping times are on the first, and second coordinate respectively passing some value. Then their sum could be interpreted as the resources expended on both coordinates. For example, you're waiting for two computers to finish a parallelized computation, and you pay a fixed cost for every second a given computer operates. Then the total operating cost is proportional to $\tau_1+\tau_2$, the stopping times of the first, second machine respectively.

3) Mark's answer is complete on this. For all intents and purposes, most things you do to transform stopping times will also result in a stopping time. The exceptions start coming up when you do an infinite number of such things, especially for a process whose filtration is not right-continuous.

Here's a seemingly bizarre example. Let $X_t=tB$, where $B=\pm 1$ with probability $1/2$ each. In other words, you flip a coin and then start moving in the direction it lands on. Note that $X_t=0$. Now let $\tau_n$ be defined be the hitting time of the set $[1/n,\infty)$ for each $n>0$. Convince (prove!) that this is a valid stopping time.

Unfortunately, $\tau:=\min_n \tau_n=\inf\{t: X_t\in (0,\infty)\}$ is NOT a valid stopping time. To see this, first notice that the filtration for $X_t$ is as usual: $F_t=\sigma(X_s: s\leq t)$. As well $F_0=\sigma(X_0)$ is trivial since $X$ is constant.

Notice that $\{\tau\leq 0\}=\cap_n \{\tau\leq 1/n\}$ is equivalent to $\{B=1\}$. On the other hand, the filtration of $X_t$ at $t=0$ is trivial, which means $\tau$ is not a stopping time. The key issue is here is lack of right-continuous filtrations along with the fact that $\tau$ is the hitting time of $(0,\infty)$ which is an open set. Contrast this with $[0,\infty)$, for which $\tau=0$ trivially.


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