Exponential-Martingale Transforms for One-Sided Hitting Times

The Hitting-Time Transform

A one-sided first-passage time records whether a diffusion reaches a specified level and, if it does, how long that takes. Discounting by $e^{-\lambda\tau}$ makes the quantity $\mathbb{E}[e^{-\lambda\tau}\mathbf{1}_{\{\tau<\infty\}}]$ sensitive to both the chance of reaching the level and the time required. An exponential martingale gives this transform directly, without needing the first-passage density.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$, and assume $B$ is a standard Brownian motion with respect to this filtration: its paths are continuous, it is adapted, and for $s<t$, $B_t-B_s$ is independent of $\mathcal{F}_s$ and has distribution $N(0,t-s)$. This compatibility with the filtration is important for the conditional-expectation and martingale arguments below.

Definition: Fix $x<a$, a drift $\mu\in\mathbb{R}$, and a volatility $\sigma>0$. Set $X_t=x+\mu t+\sigma B_t$ and define the upper-level hitting time $\tau=\inf\{t\ge0:X_t\ge a\}$, with $\inf\varnothing=\infty$. For $\lambda\ge0$, the discounted hitting transform is $u_\lambda(x)=\mathbb{E}[e^{-\lambda\tau}\mathbf{1}_{\{\tau<\infty\}}]$, where the product is defined to be zero on $\{\tau=\infty\}$.

Continuity of $X$ means that on $\{\tau<\infty\}$, $X_\tau=a$. We use the Itô integral relative to $\mathbb{P}$; its integrands are progressively measurable processes $H$ satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$ for each finite $T$. In particular, the exponential process used below has an integrand in this class on every finite time interval.

Constructing the Exponential Martingale

For $\lambda>0$, choose $k$ to solve $\mu k+\tfrac12\sigma^2 k^2=\lambda$. The positive root is $k_\lambda=(\sqrt{\mu^2+2\lambda\sigma^2}-\mu)/\sigma^2>0$. Consider $Y_t=e^{-\lambda t}e^{k_\lambda(X_t-x)}$. The choice of $k_\lambda$ cancels its drift.

Proposition: For $\lambda>0$, $Y$ is a true martingale on every finite time interval, with $\mathbb{E}[Y_t\mid\mathcal{F}_s]=Y_s$ for $0\le s\le t$.

Proof. Apply Itô’s formula to $f(t,z)=e^{-\lambda t+k_\lambda z}$, evaluated at $z=X_t-x$. Its derivatives are $f_t=-\lambda f$, $f_z=k_\lambda f$, and $f_{zz}=k_\lambda^2 f$. Since $d(X_t-x)=\mu\,dt+\sigma\,dB_t$, and $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$, Itô’s formula gives

dY_t =\left(-\lambda+\mu k_\lambda+\tfrac12\sigma^2 k_\lambda^2\right)Y_t\,dt +\sigma k_\lambda Y_t\,dB_t =\sigma k_\lambda Y_t\,dB_t.

Thus $Y_t=1+\int_0^t \sigma k_\lambda Y_s\,dB_s$. For finite $T$, the integrand is progressively measurable. Also $X_s-x=\mu s+\sigma B_s$, so the normal exponential moment gives $\mathbb{E}[Y_s^2]=e^{(-2\lambda+2\mu k_\lambda+2\sigma^2k_\lambda^2)s}$, which is finite and bounded for $0\le s\le T$. Therefore $\mathbb{E}\int_0^T(\sigma k_\lambda Y_s)^2\,ds<\infty$. The Itô integral is a square-integrable martingale, so its conditional expectation of future increments given $\mathcal{F}_s$ is zero. Consequently $\mathbb{E}[Y_t\mid\mathcal{F}_s]=Y_s$. This verifies a true martingale, not merely a local martingale. $\square$

Optional Stopping at a Truncated Hitting Time

The hitting time $\tau$ is a stopping time because $X$ is continuous and adapted. For fixed $t$, $\tau\wedge t$ is bounded. The bounded optional-stopping theorem for a martingale therefore gives $\mathbb{E}[Y_{\tau\wedge t}]=\mathbb{E}[Y_0]=1$. On $\{\tau\le t\}$, continuity gives $Y_{\tau\wedge t}=e^{-\lambda\tau}e^{k_\lambda(a-x)}$. On $\{\tau>t\}$, the process has not reached $a$, so $X_t<a$ and $0\le Y_{\tau\wedge t}=e^{-\lambda t}e^{k_\lambda(X_t-x)}\le e^{-\lambda t}e^{k_\lambda(a-x)}$. This bound tends to zero for $\lambda>0$.

Theorem: For $x<a$, $\sigma>0$, $\mu\in\mathbb{R}$, and $\lambda>0$, the one-sided hitting-time transform is $$ \mathbb{E}\!\left[e^{-\lambda\tau}\mathbf{1}_{\{\tau<\infty\}}\right] =\exp\!\left(-\frac{a-x}{\sigma^2}\big(\sqrt{\mu^2+2\lambda\sigma^2}-\mu\big)\right). $$

Proof. Split the expectation of $Y_{\tau\wedge t}$ over $\{\tau\le t\}$ and $\{\tau>t\}$. The first part is $e^{k_\lambda(a-x)}\mathbb{E}[e^{-\lambda\tau}\mathbf{1}_{\{\tau\le t\}}]$. The second part is bounded above by $e^{-\lambda t}e^{k_\lambda(a-x)}$. As $t\to\infty$, the second part tends to zero. The first part converges, by monotone convergence, to $e^{k_\lambda(a-x)}\mathbb{E}[e^{-\lambda\tau}\mathbf{1}_{\{\tau<\infty\}}]$. Since the total expectation is $1$, this limit equals $1$. Solving for the transform gives $e^{-k_\lambda(a-x)}$; substituting the displayed positive root for $k_\lambda$ yields the formula. $\square$

To include $\lambda=0$, let $\lambda\downarrow0$. The variables $e^{-\lambda\tau}\mathbf{1}_{\{\tau<\infty\}}$ increase to $\mathbf{1}_{\{\tau<\infty\}}$, so monotone convergence identifies the limiting transform with the probability of ever hitting the level.

Corollary: For $x<a$, the probability of hitting the upper level is $$ \mathbb{P}(\tau<\infty)= \begin{cases} 1, & \mu\ge0,\\[2pt] \exp\!\left(\dfrac{2\mu(a-x)}{\sigma^2}\right), & \mu<0. \end{cases} $$ If the process starts at or above the level, so $x\ge a$, then $\tau=0$ and the transform is $1$ for every $\lambda\ge0$.

Proof. In the transform formula, $\sqrt{\mu^2+2\lambda\sigma^2}$ tends to $|\mu|$ as $\lambda\downarrow0$. If $\mu\ge0$, then $|\mu|-\mu=0$, and the limiting transform is $1$. If $\mu<0$, then $|\mu|-\mu=-2\mu$, and the limiting transform is $\exp(2\mu(a-x)/\sigma^2)$. For $x\ge a$, the defining infimum includes time zero, so the stated transform is immediate. $\square$

Worked Transform Calculations

Worked Example: Negative Drift and a Positive Chance of Non-Hitting

Take $x=0$, $a=2$, $\mu=-1$, $\sigma=1$, and $\lambda=3$. Here $a-x=2$. The root equation is $-k+\tfrac12k^2=3$, or $k^2-2k-6=0$; its positive root is $k=1+\sqrt7$. Thus

\mathbb{E}[e^{-3\tau}\mathbf{1}_{\{\tau<\infty\}}] =e^{-2(1+\sqrt7)}.

The probability of ever hitting the level is the zero-discount limit, $\mathbb{P}(\tau<\infty)=e^{2(-1)(2)}=e^{-4}$; therefore the probability of never hitting it is $1-e^{-4}$. Check: $k=1+\sqrt7$ satisfies $-k+\tfrac12k^2=-(1+\sqrt7)+\tfrac12(8+2\sqrt7)=3$, so it solves the required drift-cancellation equation. Also $e^{-2(1+\sqrt7)}<e^{-4}$, as expected since discounting can only reduce the hitting probability.

Worked Example: Zero Drift with Nonunit Volatility

Take $x=1$, $a=4$, $\mu=0$, $\sigma=2$, and $\lambda=2$. The distance is $a-x=3$, and the root is $k=\sqrt{2\lambda\sigma^2}/\sigma^2=\sqrt{16}/4=1$. Hence

\mathbb{E}[e^{-2\tau}\mathbf{1}_{\{\tau<\infty\}}]=e^{-3}.

At zero discount, the hitting probability is $1$. Check: with these parameters, $\mu k+\tfrac12\sigma^2k^2=0+\tfrac12(4)(1)=2=\lambda$. The transform $e^{-3}$ lies strictly between $0$ and the hitting probability $1$, as a positive discount rate requires.

Worked Example: Positive Drift

Take $x=0$, $a=1$, $\mu=1$, $\sigma=1$, and $\lambda=1$. The positive root is $k=\sqrt{1+2}-1=\sqrt3-1$, giving

\mathbb{E}[e^{-\tau}\mathbf{1}_{\{\tau<\infty\}}]=e^{-(\sqrt3-1)}.

The hitting probability is $1$, since $\mu>0$. Check: $\mu k+\tfrac12\sigma^2k^2=(\sqrt3-1)+\tfrac12(\sqrt3-1)^2=(\sqrt3-1)+(2-\sqrt3)=1$, which is the chosen $\lambda$. Thus the exponent uses a root that satisfies the defining equation, and the discounted transform is less than the eventual hitting probability.

Why the Argument Works—and a Common Pitfall

The key stopping step is bounded: it uses $\tau\wedge t$, not optional stopping directly at an unbounded hitting time. The residual term on $\{\tau>t\}$ vanishes because $X_t<a$ there and the factor $e^{-\lambda t}$ decays. This is why the proof is immediate for $\lambda>0$. At $\lambda=0$, that deterministic decay is absent; taking the decreasing-discount limit instead handles the non-hitting event without assuming that $\tau$ is finite.

Common pitfall: using adaptedness as if it guaranteed Brownian increments independent of the past. If one writes $\mathbb{E}[B_t-B_s\mid\mathcal{F}_s]=0$ from adaptedness alone, the argument is invalid. For example, a filtration that already contains the entire Brownian path can make future increments known at time $s$. In this tutorial the filtration is assumed to make $B$ a Brownian motion relative to $(\mathcal{F}_t)$; that assumption supports the Itô integral’s martingale property and hence the optional-stopping argument.

Another distinction matters: the exponential process is a true martingale here because its stochastic-integral integrand is square-integrable on each finite time interval. A drift-cancelled Itô differential by itself would establish only a local-martingale calculation; it would not, without an integrability argument, justify setting the expectation equal to its initial value.

Check Your Understanding

Use the root equation and the stopping argument to justify each answer; in probability statements, distinguish hitting from non-hitting.

For $x<a$ and $\mu<0$, express the probability of never hitting $a$ in terms of $a-x$, $\mu$, and $\sigma$.
For $\mu=0$, explain why the zero-discount transform gives probability one of hitting the upper level.
Write the equation defining $k_\lambda$, and identify which root is used when $\lambda>0$.
Why is $\tau\wedge t$ suitable for optional stopping even when $\tau$ might be infinite?
What filtration assumption is needed to use the Brownian martingale and conditional-expectation properties in this derivation?