Reflection Principle¶
Introduction¶
In Brownian Motion Foundations, we briefly introduced the reflection principle (Theorem 1.3.19) and established that:
This elegant result exploits the symmetric structure of Brownian motion to evaluate probabilities of path-dependent events—such as hitting a barrier or attaining a maximum—by constructing cleverly reflected sample paths.
The reflection principle is one of the most powerful tools in Brownian motion theory, with applications spanning:
- Barrier options in mathematical finance (knock-in, knock-out options)
- First passage time distributions for risk management
- Survival probabilities in credit risk modeling
- Drawdown analysis for portfolio risk assessment
This section provides a comprehensive treatment:
- Geometric reflection arguments for maximum and joint events
- First passage time distribution (Lévy distribution)
- Alternative proof via exponential martingales and Laplace transforms
- Joint distribution of maximum and endpoint \((M_t, W_t)\)
- Applications to mathematical finance
Throughout, we include Python visualizations that illustrate the geometric intuition behind the reflection principle.
Reflection Principle¶
1. Statement and Geometric Idea¶
Let \(W_t\) be standard Brownian motion and \(a > 0\). Define the maximum up to time \(t\):
Theorem 1.6.1 (Reflection Principle for Maximum)
For any \(t > 0\) and \(a > 0\):
Geometric Idea:
The proof constructs a pathwise bijection between two sets of paths:
- Set A: Paths that hit level \(a\) before time \(t\) and end below \(a\)
- Set B: Paths that hit level \(a\) before time \(t\) and end above \(a\)
The bijection works by reflecting the portion of the path after the first hitting time \(\tau_a\) across the level \(a\).
2. Detailed Proof¶
Proof:
Define the first hitting time of level \(a\):
We partition the event \(\{M_t \ge a\}\) based on where the path ends:
Step 1: Event where path ends above \(a\).
Since \(W_0 = 0 < a\) and paths are continuous, any path with \(W_t \geq a\) must have crossed \(a\) at some earlier time, so \(\{W_t \geq a\} \subseteq \{M_t \geq a\}\):
Step 2: Event where path ends below \(a\).
For paths that hit \(a\) at time \(\tau_a < t\) but end at \(W_t < a\), we construct the reflected path:
Key observation: By the strong Markov property, after hitting \(a\), the process \(W_{\tau_a + s} - a\) is a Brownian motion independent of \(\mathcal{F}_{\tau_a}\). The reflection \(2a - W_s\) has the same distribution as \(W_s\) for \(s > \tau_a\).
Therefore, the reflected endpoint is:
Bijection: The map \(W \mapsto \tilde{W}\) establishes a bijection:
Each path ending at \(W_t = x < a\) corresponds to a reflected path ending at \(\tilde{W}_t = 2a - x > a\).
Step 3: Combine.
(The second equality uses the fact that if the reflected path ends above \(a\), the original path must have hit \(a\).)
Therefore:
3. Explicit Formula¶
Since \(W_t \sim \mathcal{N}(0, t)\):
where \(\Phi\) is the standard normal CDF.
4. Python Visualization¶
The following code visualizes the reflection principle by showing an original path that hits level \(a\) and its reflected counterpart.
```python import matplotlib.pyplot as plt import numpy as np
Parameters¶
a = 2 T = 5 num_steps = 1000 dt = T / num_steps
Search for a valid path (one that hits level a)¶
np.random.seed(47) # Fixed seed for reproducibility found = False seed = 47
while not found: np.random.seed(seed) dW = np.random.normal(0, np.sqrt(dt), size=num_steps) path = np.cumsum(np.insert(dW, 0, 0)) t = np.linspace(0, T, num_steps + 1)
hits_a = np.where(path >= a)[0]
if len(hits_a) > 0:
hit_index = hits_a[0]
found = True
else:
seed += 1
Reflect path after hitting a¶
reflected_path = path.copy() reflected_path[hit_index + 1:] = 2 * a - path[hit_index + 1:]
Plotting¶
fig, ax = plt.subplots(figsize=(12, 6))
ax.plot(t, path, label='Original Path', lw=3, alpha=0.4, color="blue") ax.plot(t, reflected_path, linestyle='-', lw=1.5, label=f'Reflected Path (after hitting {a})', color="red")
Markers¶
ax.scatter(t[hit_index], path[hit_index], s=100, color='green', zorder=5, label=f'First hit at level {a}') ax.scatter(t[-1], path[-1], s=100, color='blue', zorder=5, label=f'Original endpoint: {path[-1]:.2f}') ax.scatter(t[-1], reflected_path[-1], s=100, color='red', zorder=5, label=f'Reflected endpoint: {reflected_path[-1]:.2f}')
Reference line¶
ax.axhline(a, color='black', linestyle='--', linewidth=2, label=f'Barrier level {a}')
Formatting¶
ax.set_title(f'Reflection Principle: Path Hitting Level {a}', fontsize=14, fontweight='bold') ax.set_xlabel('Time', fontsize=12) ax.set_ylabel('Value', fontsize=12) ax.legend(loc='upper left', fontsize=10) ax.grid(alpha=0.3)
plt.tight_layout() plt.savefig('figures/fig09_reflection_basic.png', dpi=150, bbox_inches='tight') plt.show()
print(f"Original path ends at: {path[-1]:.4f}") print(f"Reflected path ends at: {reflected_path[-1]:.4f}") print(f"Sum: {path[-1] + reflected_path[-1]:.4f} (should equal {2*a})") ```
Output:
Original path ends at: 2.2453
Reflected path ends at: 1.7547
Sum: 4.0000 (should equal 4)
Interpretation:
- Blue path: Original Brownian motion that hits \(a\) at the green dot
- Red path: Reflected version after hitting \(a\)
- Note: \(W_t + \tilde{W}_t = 2a\) (the endpoints are symmetric about level \(a\))
- Exactly one of the original or reflected path ends above level \(a\)
Joint Distribution¶
1. Statement¶
Theorem 1.6.2 (Reflection Principle for Joint Events)
For \(a > 0\) and \(b < a\):
Proof:
The same reflection argument applies. For paths that hit level \(a\) at time \(\tau_a\) and end at \(W_t \le b < a\), reflect the portion after \(\tau_a\):
The bijection maps:
Therefore:
2. Explicit Formula¶
Since \(W_t \sim \mathcal{N}(0, t)\):
3. Python Visualization¶
This code shows a path that hits \(a\) and ends below \(b < a\), along with its reflection.
```python import matplotlib.pyplot as plt import numpy as np
Parameters¶
a = 2.0 b = -1.5 T = 5 num_steps = 1000 dt = T / num_steps
Search for a valid path (hits a, ends below b)¶
np.random.seed(91) # Fixed seed for reproducibility found = False seed = 91
while not found: np.random.seed(seed) dW = np.random.normal(0, np.sqrt(dt), size=num_steps) path = np.cumsum(np.insert(dW, 0, 0)) t = np.linspace(0, T, num_steps + 1)
hits_a = np.where(path >= a)[0]
if len(hits_a) > 0 and path[-1] < b:
hit_index = hits_a[0]
found = True
else:
seed += 1
Reflect path after hitting a¶
reflected_path = path.copy() reflected_path[hit_index + 1:] = 2 * a - path[hit_index + 1:]
Plotting¶
fig, ax = plt.subplots(figsize=(12, 6))
ax.plot(t, path, label='Original Path', lw=3, alpha=0.4, color="blue") ax.plot(t, reflected_path, linestyle='-', lw=1.5, label=f'Reflected Path (after hitting {a})', color="red")
Markers¶
ax.scatter(t[hit_index], path[hit_index], s=100, color='green', zorder=5, label=f'First hit at level {a}') ax.scatter(t[-1], path[-1], s=100, color='blue', zorder=5, label=f'Original endpoint < {b}') ax.scatter(t[-1], reflected_path[-1], s=100, color='red', zorder=5, label=f'Reflected endpoint > {2*a - b:.1f}')
Reference lines¶
ax.axhline(a, color='green', linestyle='--', linewidth=2, label=f'Upper barrier: {a}') ax.axhline(b, color='blue', linestyle='--', linewidth=2, label=f'Lower threshold: {b}') ax.axhline(2a - b, color='red', linestyle='--', linewidth=2, label=f'Reflection level: {2a - b:.1f}')
Formatting¶
ax.set_title(f'Reflection Principle: Hit {a}, End Below {b}', fontsize=14, fontweight='bold') ax.set_xlabel('Time', fontsize=12) ax.set_ylabel('Value', fontsize=12) ax.legend(loc='upper left', fontsize=9) ax.grid(alpha=0.3)
plt.tight_layout() plt.savefig('figures/fig10_reflection_joint.png', dpi=150, bbox_inches='tight') plt.show()
print(f"Original path ends at: {path[-1]:.4f} (below {b})") print(f"Reflected path ends at: {reflected_path[-1]:.4f} (above {2*a - b:.1f})") ```
Output:
Original path ends at: -1.9391 (below -1.5)
Reflected path ends at: 5.9391 (above 5.5)
Interpretation:
- Original path (blue) hits upper barrier \(a\) and ends below lower level \(b\)
- Reflected path (red) ends above the mirror level \(2a - b\)
- This bijection proves \(\mathbb{P}(M_t \ge a, W_t \le b) = \mathbb{P}(W_t \ge 2a - b)\)
First Passage Time¶
1. Distribution via Reflection Principle¶
Define the first passage time (first hitting time) to level \(a > 0\):
Theorem 1.6.3 (CDF of First Passage Time)
The cumulative distribution function of \(\tau_a\) is:
where \(\Phi\) is the standard normal CDF.
Proof:
The event \(\{\tau_a \le t\}\) is equivalent to \(\{M_t \ge a\}\) (the maximum reaches \(a\) by time \(t\)).
By the reflection principle (Theorem 1.6.1):
2. Probability Density Function¶
Theorem 1.6.4 (PDF of First Passage Time - Lévy Distribution)
The probability density function of \(\tau_a\) is:
This is called the Lévy distribution (or inverse Gaussian with zero drift).
Proof:
Differentiate the CDF with respect to \(t\):
Using the chain rule and \(\Phi'(x) = \phi(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}\):
3. Properties of First Passage Time¶
Proposition 1.6.5
The first passage time \(\tau_a\) satisfies:
- \(\mathbb{P}(\tau_a < \infty) = 1\) (recurrence of Brownian motion)
- \(\mathbb{E}[\tau_a] = \infty\) (infinite expected hitting time)
- \(\text{Var}(\tau_a) = \infty\) (infinite variance)
Proof of (2):
The tail behavior of the density is:
The integral:
The exponent \(-1/2\) is marginally non-integrable at infinity. \(\square\)
Remark: This paradoxical result—certain to hit, but taking infinite time on average—reflects the heavy-tailed nature of the Lévy distribution.
Alternative Derivation via Martingales¶
1. Laplace Transform Method¶
We now derive the distribution of \(\tau_a\) using exponential martingales and optional stopping.
Theorem 1.6.6 (Laplace Transform of \(\tau_a\))
For \(\alpha > 0\):
Proof:
Step 1: Exponential martingale.
For any \(\lambda \in \mathbb{R}\), the process:
is a martingale with respect to the natural filtration \(\mathcal{F}_t = \sigma(W_s : s \le t)\).
Step 2: Optional stopping.
Fix \(\lambda > 0\). Since \(\tau_a \wedge T\) is a bounded stopping time, the optional stopping theorem applies directly (no further integrability conditions needed):
At time \(\tau_a \wedge T\):
Step 3: Split the expectation.
The second term is non-negative, so:
Step 4: Let \(T \to \infty\).
By monotone convergence (since \(\mathbb{P}(\tau_a < \infty) = 1\)):
Therefore:
Step 5: Change variables.
Let \(\alpha = \frac{1}{2}\lambda^2\), so \(\lambda = \sqrt{2\alpha}\):
2. Moments via Laplace Transform¶
Corollary 1.6.7
Differentiating the Laplace transform:
Taking \(\alpha \to 0\):
This confirms \(\mathbb{E}[\tau_a] = \infty\).
Proof:
Therefore:
Joint Density of Maximum and Endpoint¶
We now derive the complete joint density \(f_{M_t, W_t}(m, w)\).
1. Main Result¶
Theorem 1.6.8 (Joint PDF of Maximum and Endpoint)
For \(m > 0\) and \(w \le m\):
2. Derivation¶
Step 1: CDF via reflection.
From Theorem 1.6.2, for \(w < m\):
Therefore:
Step 2: Differentiate to get the joint PDF.
First, differentiate with respect to \(w\):
where \(\phi(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}\).
Now differentiate with respect to \(m\):
3. Conditional Distribution¶
Corollary 1.6.9 (Conditional PDF of Maximum Given Endpoint)
Given \(W_t = w\), the conditional density of \(M_t\) is:
for \(m \ge w\).
Proof:
Simplify the exponentials:
Therefore:
Applications¶
1. Application: Barrier Options¶
A knock-out barrier option pays off only if the underlying asset never crosses a barrier \(B > S_0\) before maturity.
Under the Black-Scholes model:
The asset hits the barrier if:
By the reflection principle:
and the survival probability (option remains alive) is:
2. Application: Survival Probabilities¶
In credit risk, the default time can be modeled as the first passage of a log-asset value to a default boundary.
If default occurs when the firm value hits level \(D\):
The survival probability is:
3. Application: Drawdown Analysis¶
The maximum drawdown from peak to trough is:
The joint distribution \(f_{M_t, W_t}(m, w)\) allows computation of drawdown probabilities for portfolio risk management.
4. Application: Perpetual American Options¶
For a perpetual American put with strike \(K\), the optimal exercise boundary \(b^*\) satisfies:
Using the Laplace transform \(\mathbb{E}[e^{-\alpha \tau_a}] = e^{-a\sqrt{2\alpha}}\), one can derive the optimal stopping rule.
Summary¶
The reflection principle is a fundamental tool for analyzing path-dependent properties of Brownian motion:
- Maximum distribution: \(\mathbb{P}(M_t \ge a) = 2\mathbb{P}(W_t \ge a)\)
- First passage time: \(f_{\tau_a}(t) = \frac{a}{\sqrt{2\pi t^3}} e^{-a^2/(2t)}\) (Lévy distribution)
- Laplace transform: \(\mathbb{E}[e^{-\alpha \tau_a}] = e^{-a\sqrt{2\alpha}}\)
- Joint distribution: \(f_{M_t, W_t}(m, w) = \frac{2(2m-w)}{t\sqrt{2\pi t}} e^{-(2m-w)^2/(2t)}\)
Key insights:
- Geometric reflection exploits symmetry to create path bijections
- Exponential martingales provide alternative derivations via optional stopping
- Applications span barrier options, credit risk, and portfolio management
Looking ahead:
- Girsanov theorem (Chapter 1.8): Change of measure for barrier options
- Optimal stopping (advanced): American option pricing
- Local time (advanced): Fine structure of hitting times
Exercises¶
Reflection Principle¶
Let \(M_t := \sup_{0 \le s \le t} W_s\).
- Use the reflection principle to compute \(\mathbb{P}(M_t \ge a)\) for \(a > 0\).
Solution to Exercise 1
By the reflection principle (Theorem 1.6.1), for \(a > 0\):
Proof: Partition \(\{M_t \ge a\}\) into paths ending above and below \(a\):
The first term equals \(\mathbb{P}(W_t \ge a)\) since continuous paths starting at \(0\) must cross \(a\) to end above \(a\). For the second term, the reflection argument at \(\tau_a\) maps \(\{M_t \ge a, W_t < a\}\) bijectively onto \(\{M_t \ge a, W_t > a\} = \{W_t > a\}\), preserving probability by the strong Markov property. Therefore:
Since \(\mathbb{P}(W_t = a) = 0\) for continuous distributions.
- Deduce the distribution of \(M_t\) (find the CDF and PDF).
Solution to Exercise 2
CDF: From Exercise 1, for \(a > 0\):
For \(a \le 0\): \(\mathbb{P}(M_t \le a) = 0\) since \(M_t \ge W_0 = 0\).
PDF: Differentiate the CDF with respect to \(a\) (for \(a > 0\)):
for \(a > 0\), and \(f_{M_t}(a) = 0\) for \(a < 0\). This is twice the density of \(|W_t|\), which makes sense since \(M_t \overset{d}{=} |W_t|\) (a consequence of the reflection principle).
- Compute \(\mathbb{P}(|W_t| \ge a)\) using symmetry.
Solution to Exercise 3
By symmetry of Brownian motion, \(W_t \overset{d}{=} -W_t\), so \(\mathbb{P}(W_t \ge a) = \mathbb{P}(W_t \le -a) = \mathbb{P}(-W_t \ge a)\).
for \(a > 0\). Note this equals \(\mathbb{P}(M_t \ge a)\) from Exercise 1, confirming \(M_t \overset{d}{=} |W_t|\).
- Show that the joint density integrates to 1: \(\int_0^\infty \int_{-\infty}^m f_{M_t, W_t}(m, w) \, dw \, dm = 1\).
Solution to Exercise 4
We need to show \(\int_0^\infty \int_{-\infty}^m f_{M_t, W_t}(m, w)\,dw\,dm = 1\) where:
for \(m > 0\) and \(w \le m\).
Inner integral (over \(w\) for fixed \(m > 0\)): Substitute \(u = (2m - w)/\sqrt{t}\), so \(w = 2m - u\sqrt{t}\) and \(dw = -\sqrt{t}\,du\). When \(w = -\infty\), \(u = +\infty\); when \(w = m\), \(u = m/\sqrt{t}\):
Outer integral (over \(m\)):
More directly: \(\int_0^\infty e^{-m^2/(2t)}\,dm = \sqrt{t}\int_0^\infty e^{-v^2/2}\,dv = \sqrt{t}\sqrt{\pi/2}\). So \(\frac{2}{\sqrt{2\pi}}\sqrt{t}\sqrt{\pi/2} = \frac{2\sqrt{t}\sqrt{\pi}}{\sqrt{2}\sqrt{2\pi}} = 1\).
Hitting Times¶
Hitting Times¶
Define the stopping time \(\tau_a := \inf\{ t \ge 0 : W_t = a \}\) for \(a > 0\).
- Show that \(\mathbb{P}(\tau_a < \infty) = 1\) using the reflection principle.
Solution to Exercise 5
By the reflection principle, \(\mathbb{P}(\tau_a \le t) = \mathbb{P}(M_t \ge a) = 2\Phi(-a/\sqrt{t})\).
As \(t \to \infty\):
Therefore Brownian motion hits every positive level with probability 1. By symmetry (\(-W_t\) is also a Brownian motion), it also hits every negative level with probability 1.
- Compute \(\mathbb{E}[\tau_a]\), or explain why it is infinite. (Hint: Use the Laplace transform.)
Solution to Exercise 6
The density of \(\tau_a\) is \(f_{\tau_a}(t) = \frac{a}{\sqrt{2\pi t^3}} e^{-a^2/(2t)}\).
For large \(t\), \(e^{-a^2/(2t)} \to 1\), so the integrand behaves as \(t^{-1/2}\), which is not integrable at \(\infty\). Therefore \(\mathbb{E}[\tau_a] = \infty\).
Via Laplace transform: \(\mathbb{E}[\tau_a] = -\frac{d}{d\alpha}\mathbb{E}[e^{-\alpha\tau_a}]\big|_{\alpha=0} = \frac{a}{\sqrt{2\alpha}}e^{-a\sqrt{2\alpha}}\big|_{\alpha \to 0^+} = \infty\).
- Discuss how the recurrence of Brownian motion is reflected in these results.
Solution to Exercise 7
These results display a paradoxical property of recurrence for Brownian motion:
- \(\mathbb{P}(\tau_a < \infty) = 1\): Brownian motion certainly hits every level. This is recurrence — the process returns to (and exceeds) every value.
- \(\mathbb{E}[\tau_a] = \infty\): Despite certain hitting, the average time is infinite. This occurs because the Lévy distribution has a heavy tail \(f_{\tau_a}(t) \sim t^{-3/2}\), meaning there are rare paths that wander far from \(a\) for an extremely long time before finally hitting it.
The recurrence is weaker than for, say, a random walk on \(\mathbb{Z}\): Brownian motion in \(\mathbb{R}\) is recurrent (returns to every neighborhood of every point) but not positive recurrent (expected return time is infinite). In higher dimensions (\(d \ge 3\)), Brownian motion is transient and \(\mathbb{P}(\tau_a < \infty) < 1\) for suitable definitions of "hitting."
- Verify that \(\int_0^\infty f_{\tau_a}(t) \, dt = 1\) for the Lévy distribution \(f_{\tau_a}(t) = \frac{a}{\sqrt{2\pi t^3}} e^{-a^2/(2t)}\).
Solution to Exercise 8
We verify \(\int_0^\infty f_{\tau_a}(t)\,dt = 1\) where \(f_{\tau_a}(t) = \frac{a}{\sqrt{2\pi t^3}} e^{-a^2/(2t)}\).
Substitute \(u = a/\sqrt{t}\), so \(t = a^2/u^2\) and \(dt = -2a^2/u^3\,du\):
Since \(\int_0^\infty e^{-u^2/2}\,du = \sqrt{\pi/2}\):
Joint Distribution¶
Joint Distribution¶
- Show that \(\mathbb{P}(\tau_a = \tau_b) = 0\) for \(a \neq b\), \(a, b > 0\). (Hint: Suppose \(\tau_a = \tau_b = \tau\). Then \(W_\tau = a\) and \(W_\tau = b\) simultaneously. Why is this impossible when \(a \neq b\)?)
Solution to Exercise 9
Suppose \(\tau_a = \tau_b = \tau\) for some realization. At time \(\tau\), continuity of Brownian paths requires \(W_\tau = a\) (since \(\tau = \tau_a\)) and \(W_\tau = b\) (since \(\tau = \tau_b\)). But \(a \neq b\), so \(W_\tau\) cannot equal both simultaneously.
More rigorously, without loss of generality assume \(0 < a < b\). Then \(\tau_a \le \tau_b\) a.s. (the path must hit \(a\) before reaching \(b\) since it starts at \(0\) and is continuous). The event \(\{\tau_a = \tau_b\}\) requires \(W_{\tau_a} = b\), but \(W_{\tau_a} = a \neq b\). Therefore \(\{\tau_a = \tau_b\}\) is contained in a null set, and \(\mathbb{P}(\tau_a = \tau_b) = 0\).
- Compute \(\mathbb{E}[M_T | W_T = w]\) using the conditional density \(f_{M_t|W_t}(m|w)\).
Solution to Exercise 10
Using \(f_{M_t|W_t}(m|w) = \frac{2(2m-w)}{t} e^{-2m(m-w)/t}\) for \(m \ge \max(w, 0)\):
For \(w \ge 0\), substitute \(v = m - w/2\) (so \(2m - w = 2v\) and \(m = v + w/2\)), and let \(c = 2/T\):
This integral does not simplify to a closed form in elementary functions for general \(w\). However, for \(w = 0\):
Using \(\int_0^\infty x^2 e^{-\beta x^2}\,dx = \frac{\sqrt{\pi}}{4\beta^{3/2}}\) with \(\beta = 2/T\):
Laplace Transform¶
Laplace Transform¶
- Verify the Laplace transform formula \(\mathbb{E}[e^{-\alpha \tau_a}] = e^{-a\sqrt{2\alpha}}\) by direct integration:
Solution to Exercise 11
We compute \(\int_0^\infty e^{-\alpha t} f_{\tau_a}(t)\,dt = \frac{a}{\sqrt{2\pi}}\int_0^\infty t^{-3/2} e^{-\alpha t - a^2/(2t)}\,dt\).
Use the identity: for \(c, d > 0\),
Setting \(c = \alpha\) and \(d = a^2/2\):
- Use the Laplace transform to show that \(\mathbb{E}[\tau_a^{1/2}] < \infty\) but \(\mathbb{E}[\tau_a] = \infty\).
Solution to Exercise 12
From \(\mathbb{E}[e^{-\alpha\tau_a}] = e^{-a\sqrt{2\alpha}}\), the behavior near \(\alpha = 0\) encodes the moments.
\(\mathbb{E}[\tau_a^{1/2}] < \infty\): By the relation \(\mathbb{E}[\tau_a^r] = \frac{1}{\Gamma(-r)}\int_0^\infty \alpha^{-r-1}(1 - \mathbb{E}[e^{-\alpha\tau_a}])\,d\alpha\) (for \(0 < r < 1\)), or more directly from the density: the tail \(f_{\tau_a}(t) \sim \frac{a}{\sqrt{2\pi}} t^{-3/2}\) gives \(\int t^{1/2} \cdot t^{-3/2}\,dt = \int t^{-1}\,dt\), which diverges logarithmically. However, the exact integral \(\int_0^\infty t^{1/2} f_{\tau_a}(t)\,dt\) includes the Gaussian factor \(e^{-a^2/(2t)}\) which provides convergence near \(t = \infty\)... In fact, \(\mathbb{E}[\tau_a^r] < \infty\) if and only if \(r < 1/2\). For \(r\) strictly less than \(1/2\): the integrand \(t^r \cdot t^{-3/2} = t^{r-3/2}\) at infinity, integrable since \(r - 3/2 < -1\).
\(\mathbb{E}[\tau_a] = \infty\): For \(r = 1\), the integrand behaves as \(t^{1-3/2} = t^{-1/2}\) at infinity, which is not integrable. Therefore \(\mathbb{E}[\tau_a] = \infty\).
The Laplace transform makes this transparent: expanding \(e^{-a\sqrt{2\alpha}} \approx 1 - a\sqrt{2\alpha} + \cdots\) near \(\alpha = 0\), the leading correction is \(\sqrt{\alpha}\), not \(\alpha\). If \(\mathbb{E}[\tau_a]\) were finite, we would need \(e^{-a\sqrt{2\alpha}} \approx 1 - \alpha\mathbb{E}[\tau_a] + \cdots\), but the \(\sqrt{\alpha}\) singularity prevents this.
Applications¶
Applications¶
- For a knock-in barrier option that activates when the asset first hits level \(B > S_0\), derive the activation probability by time \(T\) using the reflection principle.
Solution to Exercise 13
A knock-in barrier option activates when the asset first hits level \(B > S_0\). Under the Black-Scholes model \(S_t = S_0 e^{(r - \sigma^2/2)t + \sigma W_t}\), the barrier is hit by time \(T\) if:
However, this simplification is not exact because the drift term \((r - \sigma^2/2)s\) varies with \(s\). For the simplified case (ignoring drift, i.e., using the martingale measure where the log-price drift is \(-\sigma^2/2\)), define:
The activation probability is:
With drift \(\mu = (r - \sigma^2/2)/\sigma\), the Girsanov-adjusted formula gives a more complex expression involving both \(\Phi(-a/\sqrt{T})\) and correction terms from the drift.
- (Drawdown) The drawdown at time \(t\) is \(DD_t = M_t - W_t\). Using the joint density, compute \(\mathbb{P}(DD_T > d)\) for fixed \(d > 0\).
Solution to Exercise 14
The drawdown \(DD_T = M_T - W_T\) where \(M_T = \sup_{0 \le s \le T} W_s\). We want \(\mathbb{P}(DD_T > d) = \mathbb{P}(M_T - W_T > d)\).
Using the joint density \(f_{M_T, W_T}(m, w)\):
Substituting \(f_{M_T, W_T}(m, w) = \frac{2(2m-w)}{T\sqrt{2\pi T}} e^{-(2m-w)^2/(2T)}\) and setting \(v = 2m - w\) (so \(w = 2m - v\) and the condition \(w \le m - d\) becomes \(v \ge m + d\)):
The inner integral evaluates as:
Therefore:
Substitute \(u = (m + d)/\sqrt{T}\):
This shows that the drawdown \(DD_T\) has the same distribution as \(|W_T|\), i.e., \(DD_T \overset{d}{=} M_T \overset{d}{=} |W_T|\).
References¶
- Karatzas, I., & Shreve, S. E. (1991). Brownian Motion and Stochastic Calculus, 2nd ed. Springer. (Chapter 3, Section 6)
- Revuz, D., & Yor, M. (1999). Continuous Martingales and Brownian Motion, 3rd ed. Springer. (Chapter VI)
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