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A put with a prescribed exercise rule: exercise at the first touch of $B<K$, or at expiry if no touch occurs. Its value is bounded above by the American put option, which optimizes over exercise times. With zero interest rate, its value equals the down-and-out put value plus $(K-B)$ times a cash-at-expiry one-touch option value. At positive interest rate the rebate is paid at touch, so replacing it by an expiry payment would generally be incorrect.

A contract paying a fixed amount at expiry if specified barriers have not been touched. Its complement, with the same expiry payment convention, is a one-touch option. Paying upon first touch rather than at expiry changes the discounting and gives a different contract.

A down-and-out digital call with strike at the barrier pays one exactly when the barrier is avoided until expiry. Reflection therefore gives its alive priceThe complementary contract pays one at expiry when the barrier has been hit. Its price isThis is a cash-at-expiry one-touch option, not cash paid at the first touch. After the barrier has already been hit, its price is $e^{-r\tau }$ regardless of the current spot. In particular this branch applies immediately for $S\le B$. Current spot alone does not encode the full path history.

Here is the Brownian change-of-drift form of the Girsanov theorem. Let $W$ be a standard Brownian motion on $[0,T]$ and $\theta$ predictable with ${\int }_0^T{\theta }_s^{2}ds<\infty$ almost surely. Suppose the stochastic exponentialis a true martingale with $\mathbb{E}{Z}_T=1$. Define $dQ=Z_{T}dP$. Then $Q$ is equivalent to $P$ andThe same statement holds in $d$ dimensions with scalar products, squared norms and a vector drift. Novikov's condition $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T{\theta }_s^{2}ds)<\infty$ is a standard sufficient condition for the true-martingale hypothesis. Bounded $\theta$ is enough for all applications below.

For the proof, Itô formula gives $d{Z}_t=-Z_t{\theta }_{t}d{W}_t$. The Itô product rule gives the cancellationThe two drift terms cancel because $d[Z,W{]}_t=-Z_t{\theta }_{t}dt$. Stop $W^Q$ when its absolute value reaches $n$, at a time ${\tau }_n$ capped by $T$. The same product calculation makes $Z_t{W}_{t\wedge {\tau }_n}^Q$ a local martingale under $P$. Its absolute value is at most $n{Z}_t$; the density martingale is uniformly integrable over this finite horizon, so the stopped product is a true martingale. Bayes' conditional identity then givesAs $n$ increases, these stopping times exhaust the horizon by continuity. Thus $W^Q$ is a continuous local martingale under $Q$. Adding a finite-variation drift does not change quadratic variation, so $[W^Q{]}_t=t$.

To prove the Brownian conclusion rather than just assert it, for every real $u$ apply Itô formula to $F_t=\exp (iu{W}_t^Q+u^{2}t/2)$. Its differential is $iu{F}_{t}d{W}_t^Q$, and its modulus is bounded on the fixed horizon, so it is a true martingale. ConsequentlyThe conditional characteristic function identifies a centered normal increment with variance $t-s$, independent of the past. Together with continuity and $W_0^Q=0$, this proves the Brownian assertion. For vector $u$, the same argument uses $\Vert u{\Vert }^2$ and gives the multivariate version. If $|\theta |\le K$, stopping $Z$ and applying Itô formula to $Z^2$ gives a uniform second-moment bound $\mathbb{E}{Z}_{t\wedge \tau }^2\le e^{K^{2}t}$ by Gronwall inequality; this proves uniform integrability of the stopped densities and the needed true-martingale property in the bounded case.

In the Black-Scholes model with its usual augmented Brownian filtration, write $d{S}_t=S_t(\alpha dt+\sigma d{W}_t)$, $B_t=e^{\rho t}$, with $\sigma >0$. Take $\theta =(\alpha -\rho )/\sigma$, the market price of risk. Under $Q$, the preceding change of drift givesso $S_t/B_t$ is a martingale. The Martingale representation theorem then makes integrable Brownian-market claims attainable and gives their risk-neutral pricing as discounted conditional expectations. The physical drift $\alpha$ has disappeared from the price dynamics.

For the joint terminal-value and maximum law, first use the Brownian reflection principle at level $a>0$. Reflecting a path after its first hit of $a$ preserves its law as a Brownian motion and sends a terminal value $y<a$ to $2a-y>a$; the symmetry and strong Markov property of the post-hit increments justify this map. Thus, with ${\phi }_t(y)=(2\pi t{)}^{-1/2}{e}^{-y^2/(2t)}$,Take the constant-drift exponential density $e^{\mu W_t-{\mu }^{2}t/2}$. By Girsanov theorem, the coordinate process under this tilted measure has the law of $W_s+\mu s$ under $P$. Consequently, for $t>0$ and $x\le a$,Completing the squares givesIntegrating provesHere $\Phi$ is the standard normal distribution function. The first variable is the terminal Brownian value, distinct from its running maximum.

Set $T=t_0$, $a=\log (b/S_0)/\sigma >0$ and $\mu =(\rho -{\sigma }^2/2)/\sigma$. Under the risk-neutral measure, $\log (S_s/S_0)/\sigma =W_s^Q+\mu s$. Taking $x=a$ in the joint law gives the probability of not hitting the upper level. Complementing it and discounting the unit payment at maturity gives the one-touch option priceThe exponential weight is also $(b/S_0{)}^{2\rho /{\sigma }^2-1}$. The payment occurs at $T$, even if the barrier was hit earlier, so its discount factor is $e^{-\rho T}$, not a hitting-time discount factor.

Under the Black-Scholes model pricing measure, write $\log S_t/\sigma =B_t+ct$, where $c=(r-{\sigma }^2/2)/\sigma$. Brownian continuity and immediate crossing on reaching a level make the strict up-crossing time equal almost surely to the hitting time $H_a$. Its finite-time law has no atom at $T$, so the convention before rather than at expiry does not affect the price.

For truncated discounted Brownian first passage, since payment occurs at the hitting time, discount there, not at expiry. The one-touch option price isSet $d=\sqrt{c^2+2r}=|r+{\sigma }^2/2|/\sigma$. The expression under the square root is nonnegative for every real $r$, by this Black-Scholes identity. The discounted density satisfies $e^{-rt}{h}_c(t)=e^{a(c-d)}{h}_d(t)$. Applying part i yields the closed priceThis formula also covers $d=0$ and negative rates. When $r\ge 0$, one may use $d=(r+{\sigma }^2/2)/\sigma$, so the two prefactors simplify to $e^{-\sigma a}$ and $e^{2ra/\sigma }$. At $r=0$ the expression is the undiscounted hitting probability, as a consistency check.