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A cash-at-maturity upper-barrier digital option pays a fixed amount if the stock touches an upper barrier by maturity. In the Black-Scholes model, write $a=\log (c/S_0)/\sigma$ and $b=-(r-{\sigma }^2/2)/\sigma$. Its unit-payoff price is $e^{-rT}$ times the linear-boundary Brownian first-passage distribution at $T$. This differs from paying at the hitting time, when the discount factor is random, and from an ordinary terminal-price digital option.

A contingent claim paying one at maturity if a specified price path has remained below an upper barrier, and zero otherwise. This survival payoff differs from a terminal digital put option, which tests only the terminal price. In the dividend-free Black-Scholes model, its value is the discount factor times the risk-neutral probability of avoiding the barrier, obtainable from the linear-boundary Brownian first-passage distribution.

For the Brownian first-passage time to a line, set $\lambda =\theta >0$ and choose $h=b+\sqrt{b^2+2\lambda }>0$, so $h^2/2-bh=\lambda$. The Exponential martingale for Brownian motion $M_s=e^{h{W}_s-h^{2}s/2}$, stopped at $T_{a,b}$, obeysThe optional stopping theorem at the bounded time $s\wedge T_{a,b}$ gives expectation one. The contribution from $\{T_{a,b}>s\}$ is at most $e^{ha-\lambda s}$, and tends to zero. Taking $e^{-\lambda \infty }=0$ therefore yieldsLetting $\theta \downarrow 0$ shows that the hitting probability is one for $b\le 0$ and $e^{-2ab}$ for $b>0$; in the latter case the hitting-time law has mass at infinity.

To derive the finite-time linear-boundary Brownian first-passage distribution, put $Z_s=W_s-bs$. This is Brownian motion with drift $\mu =-b$, and the event is $\{{\sup }_{s\le t}{Z}_s\ge a\}$. For driftless Brownian motion, the Brownian reflection principle says that the joint endpoint density on this event equals the normal density $p_t(x)$ when $x\ge a$ and the reflected density $p_t(2a-x)$ when $x<a$, where $p_t(x)=(2\pi t{)}^{-1/2}{e}^{-x^2/(2t)}$. The constant-drift change of measure weights these densities by $e^{\mu x-{\mu }^{2}t/2}$. Thus the crossing probability isCompleting the square gives respectively $\Phi ((\mu t-a)/\sqrt{t})$ and $e^{2\mu a}\Phi ((-\mu t-a)/\sqrt{t})$. Substituting $\mu =-b$ provesThe density weighting follows directly from the Exponential martingale for Brownian motion, or the Girsanov theorem, on the fixed finite horizon; it does not require eventual hitting.

For the barrier digital call, under the risk-neutral measure in the Black-Scholes model,Assume $\sigma >0$, put $a=\log (c/S_0)/\sigma >0$ and $b=-(\rho -{\sigma }^2/2)/\sigma$. The upper stock barrier is reached exactly when $W_s^Q$ reaches $a+bs$. The payment is at the fixed maturity $t_0$, not at the random hitting time. Hence its price isIf $\sigma =0$, use the deterministic path $S_s=S_0{e}^{\rho s}$ instead: the value is $e^{-\rho t_0}$ if that path reaches $c$ by $t_0$, and zero otherwise. The nondegenerate formula concerns the usual positive-volatility model.

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.