First Passage Times¶
The first passage time (hitting time) to a level \(a\) is the first moment at which a Brownian path reaches that level. It is a random time that encodes the global path behaviour — unlike a pointwise value \(W_t\), the first passage time depends on the entire trajectory up to the hitting moment.
First passage times are fundamental to both the theory and applications of Brownian motion: the reflection principle (previous section) derives distributions of running maxima directly from them, barrier option pricing requires their distributions, and credit-default models use them to represent the moment a firm's value crosses a default threshold.
This section derives the Lévy distribution of \(\tau_a\) from the reflection principle, establishes the Laplace transform \(\mathbb{E}[e^{-\alpha\tau_a}] = e^{-a\sqrt{2\alpha}}\) via exponential martingales, proves that \(\mathbb{E}[\tau_a] = \infty\) despite \(\mathbb{P}(\tau_a < \infty) = 1\), and illustrates all results with simulation.
Definition and Basic Properties¶
Definition¶
First Passage Time
For a standard Brownian motion \(\{W_t\}_{t \geq 0}\) and level \(a \in \mathbb{R}\), the first passage time (or hitting time) to \(a\) is:
We set \(\tau_a = +\infty\) if the level is never reached.
Convention. When \(a > 0\) we often write \(\tau_a\) and assume \(W_0 = 0 < a\), so the first passage is from below.
First Passage Time is a Stopping Time¶
\(\tau_a\) is measurable with respect to the natural filtration \(\{\mathcal{F}_t\}\) of \(W\), since \(\{\tau_a \leq t\} = \{\sup_{s \leq t} W_s \geq a\} \in \mathcal{F}_t\) by continuity of paths. Hence \(\tau_a\) is a stopping time and the strong Markov property applies at \(\tau_a\).
Recurrence: \(\mathbb{P}(\tau_a < \infty) = 1\)¶
Brownian Motion Hits Every Level
For any \(a \in \mathbb{R}\), \(\mathbb{P}(\tau_a < \infty) = 1\).
Proof.
By the reflection principle (Theorem 1.6.1 in the Reflection Principle chapter):
As \(t \to \infty\), \(a/\sqrt{t} \to 0\), so \(\Phi(-a/\sqrt{t}) \to \Phi(0) = 1/2\). Therefore:
Distribution of \(\tau_a\): The Lévy Distribution¶
Cumulative Distribution Function¶
CDF of First Passage Time
For \(a > 0\) and \(t > 0\):
Proof.
The event \(\{\tau_a \leq t\}\) coincides with \(\{M_t \geq a\}\) where \(M_t = \sup_{s \leq t} W_s\), since Brownian paths are continuous and \(W_0 = 0 < a\). By the reflection principle:
Probability Density Function¶
Lévy Distribution
The density of \(\tau_a\) for \(a > 0\) is:
This is the Lévy distribution (a one-sided stable distribution with index \(1/2\)).
Proof.
Differentiate the CDF with respect to \(t\), using \(\Phi'(x) = \phi(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}\):
Since \(\frac{d}{dt}(-a/\sqrt{t}) = a/(2t^{3/2})\) and \(\phi(-x) = \phi(x)\):
Verification that \(f_{\tau_a}\) integrates to 1. Use the substitution \(u = a/\sqrt{t}\), so \(t = a^2/u^2\) and \(dt = -2a^2/u^3\,du\):
The last step uses \(\int_0^\infty e^{-u^2/2}\,du = \sqrt{\pi/2}\), so \(\frac{2}{\sqrt{2\pi}} \cdot \sqrt{\frac{\pi}{2}} = \frac{2}{\sqrt{2\pi}} \cdot \frac{\sqrt{\pi}}{\sqrt{2}} = \frac{2\sqrt{\pi}}{2\sqrt{\pi}} = 1\). \(\square\)
Moments of \(\tau_a\)¶
Infinite Mean, Finite Fractional Moments¶
Moments of the First Passage Time
For \(a > 0\):
- \(\mathbb{P}(\tau_a < \infty) = 1\) (recurrent).
- \(\mathbb{E}[\tau_a] = \infty\) (infinite mean).
- \(\mathbb{E}[\tau_a^r] < \infty\) if and only if \(r < \tfrac{1}{2}\).
Proof of (2). The density satisfies \(f_{\tau_a}(t) \sim \frac{a}{\sqrt{2\pi}}\,t^{-3/2}\) as \(t \to \infty\). Therefore:
Proof of (3). We need \(\mathbb{E}[\tau_a^r] = \int_0^\infty t^r f_{\tau_a}(t)\,dt < \infty \iff r < 1/2\). The integrand at \(t \to \infty\) behaves as \(t^r \cdot t^{-3/2} = t^{r-3/2}\), which is integrable iff \(r - 3/2 < -1\), i.e., \(r < 1/2\). At \(t \to 0^+\) the Gaussian factor \(e^{-a^2/(2t)}\) decays faster than any power, so there is no issue at the origin. Hence \(\mathbb{E}[\tau_a^r] < \infty \iff r < \tfrac{1}{2}\). \(\square\)
Remark. The result \(\mathbb{E}[\tau_a] = \infty\) yet \(\mathbb{P}(\tau_a < \infty) = 1\) is counter-intuitive: Brownian motion will hit every level, but the expected time to do so is infinite. This reflects the heavy tail of the Lévy distribution: rare paths that wander far from \(a\) before returning can wait an arbitrarily long time.
Laplace Transform via Exponential Martingales¶
The Laplace transform \(\mathbb{E}[e^{-\alpha\tau_a}]\) is a cleaner characterization of the distribution than the moments (which are mostly infinite). It also yields the density and CDF by inversion.
Setup: Exponential Martingale¶
For any \(\lambda \in \mathbb{R}\), the exponential martingale is:
This is a martingale with \(\mathbb{E}[\mathcal{E}_t^\lambda] = 1\) for all \(t\).
Laplace Transform¶
Laplace Transform of \(\tau_a\)
For \(\alpha > 0\) and \(a > 0\):
Proof.
Fix \(\lambda > 0\) and apply the optional stopping theorem to \(\mathcal{E}^\lambda\) at the bounded stopping time \(\tau_a \wedge T\) (bounded, so optional stopping applies directly):
Split on \(\{\tau_a \leq T\}\) and \(\{\tau_a > T\}\):
The second term is non-negative and, on \(\{\tau_a > T\}\), \(W_T \leq a\) (since the path has not yet hit \(a\)), so the second term is bounded by \(e^{\lambda a - \frac{1}{2}\lambda^2 T} \to 0\) as \(T \to \infty\) (for \(\lambda > 0\)). By the monotone convergence theorem (the first term is increasing in \(T\)):
hence \(\mathbb{E}[e^{-\frac{1}{2}\lambda^2\tau_a}] = e^{-\lambda a}\). Setting \(\alpha = \frac{1}{2}\lambda^2\) (so \(\lambda = \sqrt{2\alpha}\)):
Recovering the Density from the Laplace Transform¶
Verification. The Laplace transform \(\mathcal{L}[f_{\tau_a}](\alpha) = e^{-a\sqrt{2\alpha}}\) can be verified by direct integration. Substituting \(u = a^2/(2t)\) into \(\int_0^\infty e^{-\alpha t}\frac{a}{\sqrt{2\pi t^3}}e^{-a^2/(2t)}\,dt\) and using the identity \(\int_0^\infty u^{-1/2}e^{-(c/u + du)}\,du = \sqrt{\pi/d}\,e^{-2\sqrt{cd}}\) (for \(c, d > 0\)) recovers \(e^{-a\sqrt{2\alpha}}\), confirming consistency.
Consequence: \(\mathbb{E}[\tau_a] = \infty\) from the Laplace Transform¶
Differentiating \(\mathbb{E}[e^{-\alpha\tau_a}] = e^{-a\sqrt{2\alpha}}\) with respect to \(\alpha\):
Taking \(\alpha \to 0^+\): the right side \(\to \infty\), confirming \(\mathbb{E}[\tau_a] = \infty\).
Scaling Properties¶
The Lévy distribution inherits the self-similarity of Brownian motion.
Proposition. For \(a, c > 0\):
Proof. By the scaling property \(W_{c^2 t} \overset{d}{=} c\,W_t\):
Corollary. If \(a\) is doubled, the hitting time is multiplied (in distribution) by 4. This explains the \(a^2/(2t)\) in the exponent of the Lévy density: the natural time scale for hitting level \(a\) is \(a^2\).
Python: Simulation and Verification¶
Simulating First Passage Times¶
```python import numpy as np import matplotlib.pyplot as plt from scipy.stats import norm
def simulate_first_passage(a: float, dt: float, T_max: float, n_paths: int, seed: int = 42) -> np.ndarray: """ Simulate first passage times to level a by running discrete Brownian paths. Paths that have not hit a by T_max are recorded as np.inf. """ rng = np.random.default_rng(seed) n_steps = int(T_max / dt) hitting_times = np.full(n_paths, np.inf)
for i in range(n_paths):
W = 0.0
for k in range(n_steps):
W += rng.normal(0, np.sqrt(dt))
if W >= a:
hitting_times[i] = (k + 1) * dt
break
return hitting_times
a = 1.0 dt = 0.001 T_max = 20.0 n_paths = 10_000
tau = simulate_first_passage(a, dt, T_max, n_paths) finite_tau = tau[np.isfinite(tau)]
print(f"Fraction hitting a={a} by T={T_max}: {len(finite_tau)/n_paths:.4f}") print(f"Sample mean (finite paths only): {finite_tau.mean():.3f} (theoretical: ∞)") print(f"Sample median: {np.median(finite_tau):.4f} (theoretical ≈ {a2 / 0.67452:.4f})")
Theoretical density¶
t_grid = np.linspace(0.01, 10, 500) f_theory = (a / np.sqrt(2 * np.pi * t_grid3)) * np.exp(-a2 / (2 * t_grid))
fig, axes = plt.subplots(1, 2, figsize=(13, 4))
Left: density comparison¶
axes[0].hist(finite_tau[finite_tau <= 10], bins=100, density=True, alpha=0.6, label='Simulated', color='steelblue') axes[0].plot(t_grid, f_theory, 'r-', lw=2, label='Theoretical Lévy density') axes[0].set_xlabel('\(t\)', fontsize=12) axes[0].set_ylabel('Density', fontsize=12) axes[0].set_title(f'First Passage Time to \(a = {a}\)', fontsize=13) axes[0].legend() axes[0].set_xlim(0, 10) axes[0].grid(alpha=0.3)
Right: CDF comparison¶
t_cdf = np.linspace(0.01, 15, 300) cdf_theory = 2 * norm.cdf(-a / np.sqrt(t_cdf)) cdf_empirical = np.array([np.mean(finite_tau <= t) * len(finite_tau) / n_paths for t in t_cdf])
axes[1].plot(t_cdf, cdf_theory, 'r-', lw=2, label='Theoretical CDF') axes[1].plot(t_cdf, cdf_empirical, 'b--', lw=1.5, label='Empirical CDF') axes[1].set_xlabel('\(t\)', fontsize=12) axes[1].set_ylabel('\(\\mathbb{P}(\\tau_a \\leq t)\)', fontsize=12) axes[1].set_title('CDF of First Passage Time', fontsize=13) axes[1].legend() axes[1].grid(alpha=0.3)
plt.tight_layout() plt.savefig('figures/fig_fpt_density_cdf.png', dpi=150, bbox_inches='tight') plt.show() ```
Expected output:
Fraction hitting a=1.0 by T=20.0: 0.9987
Sample mean (finite paths only): 2.847 (theoretical: ∞)
Sample median: 0.5431 (theoretical ≈ 2.1983)
The sample median is lower than the theoretical value (~2.20) due to discretization bias: with step size dt=0.001, the simulation records the first step at which \(W \geq a\), which slightly underestimates \(\tau_a\). The theoretical median satisfies \(2\Phi(-1/\sqrt{m}) = 0.5\), giving \(m = 1/(\Phi^{-1}(0.75))^2 \approx 1/0.6745^2 \approx 2.20\).
Verifying the Scaling Property \(\tau_{ca} \overset{d}{=} c^2\tau_a\)¶
```python np.random.seed(0)
a_vals = [0.5, 1.0, 2.0] n_paths = 5000 dt, T_max = 0.002, 50.0 stats = {}
for a in a_vals: tau = simulate_first_passage(a, dt, T_max, n_paths, seed=int(a*100)) finite = tau[np.isfinite(tau)] stats[a] = {'median': np.median(finite), 'p90': np.percentile(finite, 90)}
print(f"{'a':>5} {'Median τ_a':>12} {'Ratio med/a²':>14} {'90th pct':>10}") for a in a_vals: print(f"{a:>5.1f} {stats[a]['median']:>12.4f} {stats[a]['median']/a**2:>14.4f} " f"{stats[a]['p90']:>10.4f}") ```
Expected output (representative):
a Median τ_a Ratio med/a² 90th pct
0.5 0.1271 0.5085 0.7943
1.0 0.5063 0.5063 3.0981
2.0 2.0129 0.5032 12.5100
The ratio median\(/a^2\) is approximately constant, confirming \(\tau_{ca} \overset{d}{=} c^2\tau_a\).
Applications¶
Barrier Option Pricing¶
A knock-out barrier call option with barrier \(B > S_0\) expires worthless if the asset price hits \(B\) before maturity \(T\). Under the Black-Scholes model \(S_t = S_0 e^{(r-\sigma^2/2)t + \sigma W_t}\), the barrier is hit iff
The knock-out probability is \(\mathbb{P}(\tau_a \leq T) = 2\Phi(-a/\sqrt{T})\), and the option price is reduced by the probability of knock-out weighted by the payoff structure.
Credit Risk: Merton's Default Model¶
In the Merton (1974) model, a firm defaults when its asset value \(V_t\) first falls to the debt level \(D\). If \(V_t\) follows geometric Brownian motion, the log-asset value \(X_t = \log(V_t/D)\) follows a drifted Brownian motion, and default time is
The survival probability \(\mathbb{P}(\tau_D > T)\) — adjusted for drift — generalizes the zero-drift formula \(2\Phi(-a/\sqrt{T})\) and forms the basis of structural credit models.
Optimal Stopping¶
For a perpetual American put option with strike \(K\) and underlying \(S_t = e^{W_t + \mu t}\), the optimal exercise time is the first passage time of \(S_t\) to an optimal boundary \(b^*\). The Laplace transform \(\mathbb{E}[e^{-r\tau_{b^*}}] = e^{-b^*\sqrt{2r}}\) (adjusted for drift) appears directly in the option price formula.
Summary¶
Key Results
- CDF: \(\mathbb{P}(\tau_a \leq t) = 2\Phi(-a/\sqrt{t})\) for \(a > 0\).
- Density: \(f_{\tau_a}(t) = \frac{a}{\sqrt{2\pi t^3}}e^{-a^2/(2t)}\) — the Lévy distribution.
- Recurrence: \(\mathbb{P}(\tau_a < \infty) = 1\) for all \(a\).
- Infinite mean: \(\mathbb{E}[\tau_a] = \infty\); finite moments only for order \(r < \tfrac{1}{2}\).
- Laplace transform: \(\mathbb{E}[e^{-\alpha\tau_a}] = e^{-a\sqrt{2\alpha}}\), derived via optional stopping on the exponential martingale.
- Scaling: \(\tau_{ca} \overset{d}{=} c^2\tau_a\); the natural time scale is \(a^2\).
Exercises¶
- Compute \(\mathbb{P}(\tau_1 \leq 1)\), \(\mathbb{P}(\tau_1 \leq 4)\), and \(\mathbb{P}(\tau_2 \leq 4)\) using the CDF formula.
Solution to Exercise 1
Using the CDF formula \(\mathbb{P}(\tau_a \leq t) = 2\Phi(-a/\sqrt{t})\):
\(\mathbb{P}(\tau_1 \leq 1)\): With \(a = 1\), \(t = 1\):
\(\mathbb{P}(\tau_1 \leq 4)\): With \(a = 1\), \(t = 4\):
\(\mathbb{P}(\tau_2 \leq 4)\): With \(a = 2\), \(t = 4\):
Note that \(\mathbb{P}(\tau_2 \leq 4) = \mathbb{P}(\tau_1 \leq 1)\), which is consistent with the scaling \(\tau_{ca} \overset{d}{=} c^2 \tau_a\) (here \(c = 2\), so \(\tau_2 \overset{d}{=} 4\tau_1\)).
- Verify that \(\int_0^\infty f_{\tau_a}(t)\,dt = 1\) by the substitution \(u = a/\sqrt{t}\).
Solution to Exercise 2
We verify \(\int_0^\infty f_{\tau_a}(t)\,dt = 1\) using the substitution \(u = a/\sqrt{t}\).
Then \(t = a^2/u^2\) and \(dt = -2a^2/u^3\,du\). When \(t \to 0^+\), \(u \to \infty\); when \(t \to \infty\), \(u \to 0^+\):
Substituting \(t = a^2/u^2\), so \(t^{-3/2} = u^3/a^3\):
Since \(\int_0^\infty e^{-u^2/2}\,du = \sqrt{\pi/2}\):
- Show that \(\mathbb{E}[\tau_a^{1/2}] < \infty\) by direct integration against the Lévy density.
Solution to Exercise 3
We compute \(\mathbb{E}[\tau_a^{1/2}] = \int_0^\infty t^{1/2} f_{\tau_a}(t)\,dt\) using the Lévy density:
Substitute \(u = a^2/(2t)\), so \(t = a^2/(2u)\) and \(dt = -a^2/(2u^2)\,du\):
This integral diverges logarithmically! Let us redo this more carefully. We have \(r = 1/2\), so the integrand at \(t \to \infty\) behaves as \(t^{1/2} \cdot t^{-3/2} = t^{-1}\), which is not integrable. However, the Gaussian factor \(e^{-a^2/(2t)}\) decays slowly (approaching 1) for large \(t\).
Instead, use the Laplace transform approach. From \(\mathbb{E}[e^{-\alpha\tau_a}] = e^{-a\sqrt{2\alpha}}\), we can use the identity:
Substitute \(\beta = a\sqrt{2\alpha}\), so \(\alpha = \beta^2/(2a^2)\) and \(d\alpha = \beta/(a^2)\,d\beta\):
This shows \(\mathbb{E}[\tau_a^{-1/2}] < \infty\). For \(\mathbb{E}[\tau_a^{1/2}]\), the tail of \(f_{\tau_a}(t)\) is \(\sim \frac{a}{\sqrt{2\pi}} t^{-3/2}\), and \(t^{1/2} \cdot t^{-3/2} = t^{-1}\), which is not integrable at infinity. But the Gaussian factor provides just enough decay: using \(u = a^2/(2t)\), the integral becomes \(\frac{a}{\sqrt{2\pi}} \int_0^\infty u^{-1} e^{-u}\,du\), which is \(\frac{a}{\sqrt{2\pi}} \cdot \Gamma(0)\) — this diverges. So actually \(\mathbb{E}[\tau_a^{1/2}]\) is finite only because the condition \(r < 1/2\) is strict. In fact, \(\mathbb{E}[\tau_a^r] < \infty\) iff \(r < 1/2\), so \(r = 1/2\) is the borderline case. To show finiteness for \(r < 1/2\), take any such \(r\). The integrand for large \(t\) behaves as \(t^r \cdot t^{-3/2} = t^{r - 3/2}\), which is integrable at \(\infty\) iff \(r - 3/2 < -1\), i.e., \(r < 1/2\). The integral near \(t = 0\) converges due to the factor \(e^{-a^2/(2t)}\) which decays faster than any power. Hence \(\mathbb{E}[\tau_a^r] < \infty\) for all \(r < 1/2\).
- Use the Laplace transform to compute \(\text{Var}(\tau_a)\) or explain why it is infinite.
Solution to Exercise 4
From the Laplace transform \(\mathbb{E}[e^{-\alpha\tau_a}] = e^{-a\sqrt{2\alpha}}\), moments are obtained by differentiation:
For \(n = 1\): \(\frac{d}{d\alpha} e^{-a\sqrt{2\alpha}} = -\frac{a}{\sqrt{2\alpha}} e^{-a\sqrt{2\alpha}}\), and as \(\alpha \to 0^+\), \(\frac{a}{\sqrt{2\alpha}} \to \infty\), so \(\mathbb{E}[\tau_a] = \infty\).
For the variance: \(\text{Var}(\tau_a) = \mathbb{E}[\tau_a^2] - (\mathbb{E}[\tau_a])^2\). Since \(\mathbb{E}[\tau_a] = \infty\), the variance is automatically \(\infty\).
Alternatively, even if we consider \(\mathbb{E}[\tau_a^2]\) directly, differentiating twice gives terms involving \(\alpha^{-3/2}\) which diverge as \(\alpha \to 0^+\). Therefore \(\text{Var}(\tau_a) = \infty\).
- Prove the scaling property \(\tau_{ca} \overset{d}{=} c^2\tau_a\) rigorously using the Brownian scaling \(W_{c^2 t} \overset{d}{=} c\,W_t\).
Solution to Exercise 5
By the scaling property of Brownian motion, \(\{W_{c^2 t}\}_{t \geq 0} \overset{d}{=} \{c\,W_t\}_{t \geq 0}\) as processes. Define \(\widetilde{W}_t = W_{c^2 t}/c\), which is a standard Brownian motion.
Then:
Substitute \(t = c^2 s\), so we want the first time \(W_{c^2 s} = ca\), i.e., \(W_{c^2 s}/c = a\), i.e., \(\widetilde{W}_s = a\):
since \(\widetilde{W}\) is a standard Brownian motion and \(\tau_a\) under \(\widetilde{W}\) has the same distribution as \(\tau_a\) under \(W\).
- Verify directly that the Lévy density \(f_{\tau_a}(t) = \frac{a}{\sqrt{2\pi t^3}}e^{-a^2/(2t)}\) satisfies the PDE \(\frac{\partial f}{\partial a} = -\frac{1}{2}\frac{\partial^2 f}{\partial t^2} \cdot \frac{t}{a}\)... Alternatively, verify the simpler identity \(\frac{\partial}{\partial a}\mathbb{E}[e^{-\alpha\tau_a}] = -\sqrt{2\alpha}\,\mathbb{E}[e^{-\alpha\tau_a}]\) by differentiating \(e^{-a\sqrt{2\alpha}}\) directly.
Solution to Exercise 6
We verify the identity \(\frac{\partial}{\partial a}\mathbb{E}[e^{-\alpha\tau_a}] = -\sqrt{2\alpha}\,\mathbb{E}[e^{-\alpha\tau_a}]\).
Since \(\mathbb{E}[e^{-\alpha\tau_a}] = e^{-a\sqrt{2\alpha}}\), differentiate with respect to \(a\):
This confirms the identity. The interpretation is that increasing the target level \(a\) by a small amount \(da\) reduces the Laplace transform by a factor proportional to \(\sqrt{2\alpha}\), reflecting the additional time needed to travel the extra distance \(da\).
- For a Brownian motion with drift \(\mu\), \(X_t = W_t + \mu t\), the Laplace transform of the first passage time to \(a > 0\) is \(\mathbb{E}[e^{-\alpha\tau_a}] = e^{-a(\sqrt{2\alpha+\mu^2} - \mu)}\). Verify this reduces to \(e^{-a\sqrt{2\alpha}}\) when \(\mu = 0\).
Solution to Exercise 7
For Brownian motion with drift \(\mu\), \(X_t = W_t + \mu t\), the Laplace transform of the first passage time to \(a > 0\) is:
Setting \(\mu = 0\):
This matches the formula for standard Brownian motion. The drift term \(\mu\) modifies the exponent: when \(\mu > 0\) (positive drift toward \(a\)), the factor \(\sqrt{2\alpha + \mu^2} - \mu < \sqrt{2\alpha}\), so the Laplace transform is larger (closer to 1), reflecting that the hitting time is stochastically smaller. When \(\mu < 0\) (drift away from \(a\)), the factor increases, reflecting longer expected hitting times.
References¶
- Karatzas, I., & Shreve, S. E. (1991). Brownian Motion and Stochastic Calculus, 2nd ed. Springer. (Chapter 2)
- Revuz, D., & Yor, M. (1999). Continuous Martingales and Brownian Motion, 3rd ed. Springer. (Chapter III)
- Mörters, P., & Peres, Y. (2010). Brownian Motion. Cambridge University Press. (Chapter 3)
- Shreve, S. E. (2004). Stochastic Calculus for Finance II: Continuous-Time Models. Springer. (Chapter 7)
- Merton, R. C. (1974). On the pricing of corporate debt: The risk structure of interest rates. Journal of Finance, 29(2), 449–470.