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Let $M_t={\sup }_{s\le t}{W}_s$ and write ${\varphi }_t(y)=(2\pi t{)}^{-1/2}{e}^{-y^2/(2t)}$ for the Brownian terminal density. By the Brownian reflection principle, for $y<a$ the reflected paths that hit $a$ and end at $y$ have terminal density ${\varphi }_t(2a-y)$. HenceThe Girsanov theorem changes zero drift into drift $\nu$ using the finite-horizon density $e^{\nu W_t-{\nu }^{2}t/2}$. The finite-horizon maximum of Brownian motion with drift therefore has distribution functionIn the first integral term, complete the square to get a normal density with mean $\nu t$. In the second, substitute $z=2a-y$ and complete the square with mean $-\nu t$; the prefactor is $e^{2a\nu }$. Here $\Phi$ is the standard normal distribution function, $a>0$ and $t>0$.

Let $T_{a,b}=\inf \{s\ge 0:W_s=a+bs\}$, with $b$ real. Survival means that $W_s-bs$ stays below $a$. Tilt by $e^{\theta W_t-{\theta }^{2}t/2}$; under the tilted probability, $W_s$ has drift $\theta$. Applying the maximum formula with drift $\theta -b$ gives the exponential terminal moment below a linear Brownian boundaryThis holds for every real $\theta$. The maximum distribution has no atom at the positive barrier, so strict or weak survival inequalities give the same probability here.

For the hitting-time moment, first use drift $-b$ in the maximum formula and take the complementary probability:If $z_1=(-a-bt)/\sqrt{t}$ and $z_2=(-a+bt)/\sqrt{t}$, their standard normal densities satisfy $\varphi (z_1)=e^{-2ab}\varphi (z_2)$. Differentiating the two terms, with $z_1^{\prime }=(a-bt)/(2t^{3/2})$ and $z_2^{\prime }=(a+bt)/(2t^{3/2})$, yields the drifted Brownian first-passage densityIts total mass need not be one, because the upward-sloping boundary may never be reached.

For $2\theta \le b^2$, put $\kappa =\sqrt{b^2-2\theta }\ge 0$. Multiplication by $e^{\theta s}$ gives $e^{\theta s}{f}_b(s)=e^{a(\kappa -b)}{f}_{\kappa }(s)$. Integrating to $t$ produces truncated discounted Brownian first passage, with negative discount rate allowed when the square root remains real:This includes the equality case $\kappa =0$, where it becomes $2e^{-ab}\Phi (-a/\sqrt{t})$. The expression integrates only finite hitting times up to $t$; no value is assigned to an exponential at an infinite hitting time.

For the cash-at-hit digital call, use the risk-neutral measure in the Black-Scholes model and denote expiry by $T=t_0$. ThenHitting $c>S_0$ corresponds to $W_s^Q=a+bs$, where $a=\log (c/S_0)/\sigma >0$ and $b=({\sigma }^2/2-\rho )/\sigma$. The unit-cash payment is made at the hitting time, so risk-neutral valuation discounts at that time, not at expiry. Set $\theta =-\rho$ andThe identity on the right verifies the square-root condition even for negative interest rates. The time-zero price isThis is in pounds for the unit-pound contract. It differs from a maturity-paid digital call option, whose discounting and exercise event are different.