= Solution
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> gives
$$
\mathbb P(T_a\leq t)=2\{1-\Phi(a/\sqrt t)\},\qquad
h_0(t)=\frac{a}{\sqrt{2\pi t^3}}e^{-a^2/(2t)}.
$$
Here $\Phi$ is the <standard normal distribution function>. To change the <drift> to $-b$, use the <Girsanov theorem> on each finite horizon, with density $L_t=\exp(-bB_t-b^2t/2)$. On $\{T_a\leq t\}$, <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 consequently
$$
h_b(s)=e^{-ab-b^2s/2}h_0(s)
=\frac{a}{\sqrt{2\pi s^3}}\exp\left(-\frac{(a+bs)^2}{2s}\right),\qquad s>0.
$$
There 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>:
$$
\begin{aligned}
\mathbb E[e^{-\theta T_{a,b}}]
&=e^{-ab}\int_0^\infty e^{-(\theta+b^2/2)s}h_0(s)\,ds\\
&=\boxed{\exp[-a(b+\sqrt{b^2+2\theta})]},\qquad\theta>0.
\end{aligned}
$$
The zero-discount limit is $1$ for $b\leq0$ 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>, define
$$
F_b(t)=e^{-2ab}\Phi\left(\frac{bt-a}{\sqrt t}\right)+\Phi\left(-\frac{a+bt}{\sqrt t}\right).
$$
Both terms tend to zero as $t\downarrow0$. Let $\phi$ be the <standard normal density>. The identity
$$
e^{-2ab}\phi\left(b\sqrt t-\frac a{\sqrt t}\right)
=\phi\left(b\sqrt t+\frac a{\sqrt t}\right)
$$
follows by expanding the two squares. Differentiating the two normal probabilities then gives
$$
F_b'(t)=\phi\left(b\sqrt t+\frac a{\sqrt t}\right)
\left[\frac b{2\sqrt t}+\frac a{2t^{3/2}}-\frac b{2\sqrt t}+\frac a{2t^{3/2}}\right]
=h_b(t).
$$
Since this is the previously derived hitting density and both distribution functions start at zero,
$$
\boxed{\mathbb P(T_{a,b}\leq t)=e^{-2ab}\Phi\left(\frac{bt-a}{\sqrt t}\right)+1-\Phi\left(\frac{a+bt}{\sqrt t}\right).}
$$
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>,
$$
S_t=S_0\exp\left((\rho-\sigma^2/2)t+\sigma W_t^Q\right).
$$
Set $T=t_0$ and
$$
a=\frac{\log(c/S_0)}\sigma>0,\qquad
b=\frac{\sigma^2/2-\rho}\sigma.
$$
Reaching 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 gives
$$
\boxed{V_0=e^{-\rho T}\left[\Phi\left(\frac{a+bT}{\sqrt T}\right)-e^{-2ab}\Phi\left(\frac{bT-a}{\sqrt T}\right)\right].}
$$
This 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.
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