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For $a>0$, real $b$ and standard Brownian motion, the first-passage time to a linear boundary hasHere $\Phi$ is the standard normal distribution function. Differentiation gives $a\exp (-(a+bt{)}^2/(2t))/\sqrt{2\pi t^3}$, the drifted Brownian first-passage density. Its Laplace transform is $\exp [-a(b+\sqrt{b^2+2\theta })]$ for $\theta >0$. There is mass $1-e^{-2ab}$ at infinity when $b>0$; for $b\le 0$ the hitting time is finite almost surely.

Subtract the boundary slope from the Brownian motion: $X_t=B_t-bt$. Then $T_{a,b}$ is the first-passage time of $X$ to the fixed level $a$. The zero-drift Brownian reflection principle givesHere $\Phi$ is the standard normal distribution function. To change the drift to $-b$, use the Girsanov theorem on each finite horizon, with density $L_t=\exp (-b{B}_t-b^{2}t/2)$. On $\{T_a\le t\}$, conditional expectation at the stopped time gives $\mathbb{E}[L_t\mid {\mathcal{F}}_{T_a}]=L_{T_a}$, because the density is a true martingale. Since $B_{T_a}=a$, the drifted Brownian first-passage density is consequentlyThere may be additional mass at infinite hitting time; the displayed density only describes finite passage. With $e^{-\theta \infty }=0$, apply the given zero-drift Brownian first-passage Laplace transform:The zero-discount limit is $1$ for $b\le 0$ and $e^{-2ab}$ for $b>0$, so the hitting probability is less than one precisely when the line moves away with positive slope.

For a direct verification of the linear-boundary Brownian first-passage distribution, defineBoth terms tend to zero as $t\downarrow 0$. Let $\varphi$ be the standard normal density. The identityfollows by expanding the two squares. Differentiating the two normal probabilities then givesSince this is the previously derived hitting density and both distribution functions start at zero,This proof applies to every real $b$, including the defective distribution when $b>0$.

For the barrier digital put, use the dividend-free Black-Scholes model with constant interest rate $\rho$ and spot volatility $\sigma >0$. Under the risk-neutral measure,Set $T=t_0$ andReaching the barrier $c$ is equivalent to $W_t^Q$ reaching $a+bt$. The contingent claim pays one only on the survival event $\{T_{a,b}>T\}$. Risk-neutral pricing therefore givesThis is a path-survival price, not the price of a terminal digital put option. Continuity of the stock paths and the absence of an atom in the finite-horizon maximum imply that alternative touching conventions change no price. As $c\to \infty$, the value tends to $e^{-\rho T}$; as $c\downarrow S_0$, it tends to zero. Both checks agree with the survival interpretation.

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.

For zero drift, the reflection principle gives the known Brownian first-passage time densityTo obtain a general drift, on a fixed finite horizon use the exponential martingale $M_t=e^{c{B}_t-c^{2}t/2}$ as change-of-measure density. The Novikov condition holds for constant $c$, and the Girsanov theorem says that the coordinate process $B$ has drift $c$ under the new measure. Thus the hitting law of $B$ under that measure is the law of $B+ct$ under the original measure. Conditional expectation of $M_T$ at a hitting time bounded by $T$ equals its stopped value, by the martingale optional sampling theorem. At $H_a=t$ this value is $e^{ca-c^{2}t/2}$. The drifted Brownian first-passage density is obtained by multiplying $h_0$, giving $\boxed{h_c(t)=a{e}^{-(a-ct{)}^2/(2t)}/\sqrt{2\pi t^3}}$.

To verify the integrated formula, put $u=(a-cT)/\sqrt{T}$ and $v=(-a-cT)/\sqrt{T}$. The identity $e^{2ac}\varphi (v)=\varphi (u)$ for the standard normal density givesBoth terms vanish as $T\downarrow 0$, so $\boxed{\mathbb{P}(H_a\le T)=\overline{\Phi }((a-cT)/\sqrt{T})+e^{2ac}\Phi ((-a-cT)/\sqrt{T})}$. The distribution has total mass one for $c\ge 0$ and mass $e^{2ac}$ for $c<0$; in the latter case the remaining probability is an atom at infinity.

When $d=\sqrt{c^2+2r}$ is real and nonnegative, the discounted drifted Brownian first-passage density is $e^{a(c-d)}{h}_d(t)$. Integrating yields $e^{a(c-d)}\Phi ((dT-a)/\sqrt{T})+e^{a(c+d)}\Phi ((-a-dT)/\sqrt{T})$. In a Black-Scholes model with $c=(r-{\sigma }^2/2)/\sigma$, the identity $c^2+2r=(r+{\sigma }^2/2{)}^2/{\sigma }^2$ ensures this formula covers every real interest rate.