# Does time remaining matter in NO Touch-ONE Touch probabilities?

I asked a question some days back and got an answer which I understand and make sense: Probability of touching short call strike and not touching touching short put strike of a short strangle?

However, the answer to that question brought up another related question to my mind.

With the probability of that event happening given as P(A ∩ Bcomplement)=P(A)−P(A∩B):

Is that the probability for the event with or without time still left to expiration?

Or the probability of the event only without time left in the trade?

NOTE: Because if there is still time in the trade, it can still touch B before the trade expires even though it touched A first and satisfied the condition

Referring to one touch down no touch up (OTD-NTU) option with expiry $$T$$ as interpreted in your previous question, it might help to formalize the payoff stated in the respective answer.

The option pays rebate \$1 at expiry $$T$$ if $$\boxed{ \tau_L \leq T \; {\rm and} \; \tau_L < \tau^H }$$ where: $$\tau_L = \min \; \{t \geq 0 : S_t \leq L \}$$ and $$\tau^H = \min \; \{t \geq 0 : S_t \geq H \},$$ with $$L< S_0 < H$$. Its price amounts to calculating (under $$Q$$ probability measure) $$E^Q\left[e^{-rT}1_{\{\tau_L \leq T\} \cap \{\tau_L < \tau^H\}}\right]=e^{-rT} Q(\{\tau_L \leq T\} \cap \{\tau_L < \tau^H\}),$$ where $$1_A$$ is $$1$$ if event $$A$$ takes place, and $$0$$ otherwise, and $$r$$ is a flat risk-free discount rate. The \$1 rebate pay can be made at touching (hitting) time $$\tau_L$$ too, in which case the price of the option is:

$$E^Q\left[e^{-r\tau_L} 1_{\{\tau_L \leq T\} \cap \{\tau_L < \tau^H\}}\right].$$

Its calculation is more complex (pay timing is random).

Back to your $$A$$ and $$B$$ events in the question, they are:

$$A = \{\tau_L \leq T \}, \; B = \{\tau^H \leq T \}.$$

So, the payoff for which

$$Q(A\cap B^c) = Q(\{\tau_L \leq T\} \cap \{\tau^H > T\})$$

would be its price is the one that would pay \$1 at expiry $$T$$ if $$\boxed{ \tau_L \leq T \; {\rm and} \; \tau^H > T.}$$ • Thanks for your reply, but it is a bit math-heavy for me but in summary you mean it would payoff only at the expiration? – user56826 Jun 11, 2021 at 20:03 • For the last boxed payoff yes. – ir7 Jun 11, 2021 at 20:37 • In general, no. For a NTD (no touch down) option there is only one pay date possible, expiry (or after expiry). For one touch down (OTD) option the pay date can be the hitting time ($\tau$) itself, that is, pay date is random. Same for OTD-NTU (in which case the option terminates immediately with 0 pay when the$H$gets hit before$L$does). This forces us to think in terms of the hitting time itself (its expectation, distribution, joint distribution with other hitting times etc.), not just POT probabilities of events like$\tau_L \leq T$where$T$is fixed,$Q(\tau \leq T)$. – ir7 Jun 11, 2021 at 20:37 • (Note that last probability can be seen as$Q(\tau_L \leq T)=Q(\min_{t\leq T} S_t < L)$, the probability of the minimum of stock price over the life of the option being less than$L\$)
– ir7
Jun 11, 2021 at 20:43