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If $\tau$, $\rho$ are stopping times, then it is easily seen that $\tau+\rho$ is also a stopping time. However $\tau-\rho$ and $\tau\rho$ are not necessarily stopping times as it requires a "peak into the future", but I am not sure why this is so and would appreciate any feedback to facilitate my understanding.

For the difference case, I think it not as if $(X,\mathscr{F})$ is a measurable space and $f, \space g \space :(X,\mathscr{F})\rightarrow([0,\infty),\mathscr{B}([0,\infty))$ are measurable maps then if for some $x\in X$, $f(x)-g(x)<0$ then I say it does not make sense that $f-g$ is measurable as it maps some elements outside of $[0,\infty)$, and $(f-g)^{-1}(\mathscr{B}([0,\infty)))$ is not a sub $\sigma$-algebra of $\mathscr{F}$. Is my reasoning correct?

So in the case of stooping time, $\tau$ we have the added requirement that $1_{\{\tau\leq t\}}$ be adapted to the given filtration. So my question is why does the difference of stopping times require a "peak into the future"? I have the same qualms about the product case.

However in the case of the reciprocal, I think I see why, as if we take $t=1/2$, then $\{1/\tau\leq2\}=\{\tau\geq2\}\in\mathscr{F}_2$ and we have no way on knowing if the event is in $\mathscr{F}_{1/2}$

So any comments and answers would be really appreciated.

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Let $(\mathcal{F}_t)_{t \geq 0}$ be a filtration and $\tau$ an $\mathcal{F}_t$-stopping time, i.e. $\{\tau \leq t\} \in \mathcal{F}_t$ for all $t \geq 0.$

For a fixed constant $c >0$ the mapping $\varrho(\omega) := c$ defines an $\mathcal{F}_t$-stopping time. Then $\tau-\varrho = \tau-c$ is, in general, not a $\mathcal{F}_t$-stopping time since

$$\{\tau-\varrho \leq t\} = \{\tau \leq t+c\}$$

is in $\mathcal{F}_{t+c}$ but not necessarily in $\mathcal{F}_t$. A similar thing happens if we consider $\varrho \cdot \tau$ and $\varrho:=c >1$.

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    Thank you, this clears things up for me, but just a quick question about the measurability of $f-g$ am right to say that it makes no sense to ask about the measurability of the difference if it maps elements outside the co-domain in question? Also for stopping times $\tau$ and $\rho$; if for for some $T$ have $\mathcal{F}_{\tau},\space \mathcal{F}_{\rho}\subset\mathcal{F}_T$ can we infer that $\{\tau-\rho<0\}\in\mathcal{F}_T$ as $\tau$ and $\rho$ are $\mathcal{F}_T$-measurable? – 2017-02-25
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    @AnselB What exactly do you mean by "co-domain"? Regarding your second question: Yes, that's correct. If $\mathcal{F}_{\tau} \cup \mathcal{F}_{\varrho} \subseteq \mathcal{F}_T$, then $\tau$ and $\varrho$ are $\mathcal{F}_T$-measurable and so is its difference $\tau-\varrho$. – 2017-02-25
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    Entschuldigung. By codomain I just mean the target set of the maps, so here $f$ and $g$ map into $[0,\infty)$and are $\mathscr{F}/\mathscr{B}([0,\infty))$. However in the case where$f-g$ maps some elements to negative numbers, and if we have to only consider the mesurable space $([0,\infty),\mathscr{B}([0,\infty))$ and not the whole of $\mathbb{R}$ (as in stopping times), then my point is that we shouldn't even consider $f-g$ as it makes no sense to ask if it is measurable w.r.t. $([0,\infty),\mathscr{B}([0,\infty))$. Is this a correct statement? Thanks again. – 2017-02-25
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    @AnselB Yeah, sure, unless you assume $g \leq f$. – 2017-02-26
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Intuitively, one should think about a stopping time in conjunction with a process. It's nice to think about a process as a stock price, and a stopping time as a strategy for deciding when to sell.

For instance, imagine a stock whose price at time $t=0$ is 100. let $\tau$ be the first time the stock price hits 200 (i.e. $\tau = \inf\{t : X_t = 200\}$). Similarly $\rho$ be the first time the price reaches 50. So $\tau$ itself is a reasonable strategy: "sell when the price reaches or exceeds 200". You could actually do that, so $\tau$ is a stopping time. $\tau+5$ is also a reasonable strategy: "Wait until the price hits 200, wait 5 more days, then sell." So $\tau+5$ is also a stopping time. $\tau-5$ is not a stopping time: "Sell five days before the stock hits 200." If the stock hits 200 on day 8, you won't know that until day 8, at which point you'll see you should have sold on day 3, but by then it's too late.

So $\tau+\rho$ is a stopping time: "Wait until the stock has hit both prices, add the times, and sell at that time." If it hits 200 on day 8 and 50 on day 9, you are supposed to wait until day 17 to sell. You could actually do that. It's a stopping time, though a stupid one because there is no particular meaning to adding absolute times to get an absolute time.

$\tau - \rho$ is not a stopping time for reasons similar to $\tau-5$ (though it is still stupid because subtracting absolute times isn't meaningful as an absolute time). For $\tau\rho$, consider what happens if $\tau = 1/2$ and $\rho = 3/4$. (And it's even stupider because multiplying two times doesn't give a time at all; the units are wrong.)