I come across an interesting question about barrier option as shown below.

Two barrier options are given with the same parameters including the barrier level. The first one is knocked out when it first hits the barrier. The second one is knocked out when it hits the same barrier the second time.

Question: Under the Black-Scholes framework, what is the relation of these two option prices?

(My thought: it probably can be said that the second option is more valuable because it is less likely to be knocked out. But what else?)

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    $\begingroup$ My intuition is that the two prices will be the same in a Black-Scholes framework. Conditional on a Brownian motion hitting some level $b$ at some time $t$, the expected number of crossings of $b$ in $[t, t + \Delta t]$ is infinite for any $\Delta t > 0$. I.e. the second option would knock-out immediately after the first. Search for "law of iterated logarithm". $\endgroup$ – LocalVolatility Mar 17 '19 at 23:40
  • $\begingroup$ @LocalVolatility Your explanation is tempting. Somehow I am not convinced. That the returns of underlying follow geometric Brownian motion should not mean the prices follow exactly the same path. When first one hits barrier, the second one would have either 1) passed the barrier which makes it valued zero or 2) not hit the barrier which makes it valued some other number. We don't know which is more likely in general. $\endgroup$ – Yi Bao Mar 18 '19 at 0:46
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    $\begingroup$ If they are options on the same stock, of course the underlying follows the same path. $\endgroup$ – Alex C Mar 18 '19 at 1:42
  • $\begingroup$ @AlexC Indeed. I was befuddled by the obvious. Thanks. $\endgroup$ – Yi Bao Mar 18 '19 at 13:34
  • $\begingroup$ @LocalVolatility I understand now. Thank you very much. $\endgroup$ – Yi Bao Mar 18 '19 at 14:05

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