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If $0<B<K$, every surviving terminal price is already below the strike, so this is a barrier digital put paying one at expiry exactly when the upper barrier has been avoided. To use the Black-Scholes reflection symmetry, cap the free digital-put strike at $B$, rather than continuing to use $K$:Indeed, $D_B$ has terminal payoff one throughout the alive region, while its reflected argument exceeds $B$ and hence has zero terminal payoff. The barrier values again cancel. Using $D_K$ in both terms would generally leave an unwanted reflected terminal payoff near the barrier.

Writing $b=r-{\sigma }^2/2$ makes the discounted survival probability explicit:The price is independent of $K$ in this case. Both cases combine by replacing the free strike with $\min (K,B)$ in the up-and-out digital put formula.

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.

A digital put option extinguished if the stock reaches an upper barrier $B$. Under the dividend-free Black-Scholes model, conditional on not having hit the barrier before the current time, let $0<S<B$, $\tau =T-t>0$, $m=\min (K,B)$, $\gamma =1-2r/{\sigma }^2$, andThen its price isThe Black-Scholes reflection symmetry proves the pricing equation, and the two terms cancel at $S=B$. At expiry the reflected argument is above $B\ge m$, so its payoff is zero, while $S<m$ is equivalent to $S<K$ inside the alive region. Thus the correct free strike is capped at the barrier. For $K\ge B$, the value is the discounted upper-barrier survival probability, independent of $K$, as for a barrier digital put. For $K>B$ its terminal data have a corner discontinuity at the absorbing barrier. Set the price to zero after a hit; the reflected expression on the other side is not a valid contract value.