Reflection Principle¶

Introduction¶

The reflection principle is a pathwise symmetry argument used to compute probabilities involving the running maximum of Brownian motion. From it we extract:

the distribution of the running maximum \(M_t\),
the joint distribution of \((M_t, W_t)\),
and (via \(\{\tau_a \le t\} = \{M_t \ge a\}\)) the gateway to first-passage results developed in § First Passage Times.

Recall (see § Brownian Motion): \(W_t \sim \mathcal{N}(0,t)\) with continuous paths.

Maximum of Brownian Motion¶

Statement and Geometric Idea¶

Let \(W_t\) be standard Brownian motion and \(a > 0\). Define the maximum up to time \(t\):

\[M_t := \sup_{0 \le s \le t} W_s\]

Theorem 1.6.1 (Reflection Principle for Maximum)

For any \(t > 0\) and \(a > 0\):

\[\boxed{ \mathbb{P}(M_t \ge a) = 2\mathbb{P}(W_t \ge a) }\]

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 \(a\).

Detailed Proof¶

Proof:

Define the first hitting time of level \(a\):

\[\tau_a := \inf\{s \ge 0 : W_s = a\}\]

We partition the event \(\{M_t \ge a\}\) based on where the path ends:

\[\{M_t \ge a\} = \{M_t \ge a, W_t \ge a\} \cup \{M_t \ge a, W_t < a\}\]

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 \(\{W_t \geq a\} \subseteq \{M_t \geq a\}\):

\[\mathbb{P}(M_t \ge a, W_t \ge a) = \mathbb{P}(W_t \ge a)\]

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:

\[\tilde{W}_s = \begin{cases} W_s & \text{if } s \le \tau_a \\ 2a - W_s & \text{if } s > \tau_a \end{cases}\]

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:

\[\tilde{W}_t = 2a - W_t\]

Bijection: The map \(W \mapsto \tilde{W}\) establishes a bijection:

\[\{M_t \ge a, W_t < a\} \leftrightarrow \{M_t \ge a, W_t > a\}\]

Each path ending at \(W_t = x < a\) corresponds to a reflected path ending at \(\tilde{W}_t = 2a - x > a\).

Step 3: Combine.

\[\mathbb{P}(M_t \ge a, W_t < a) = \mathbb{P}(M_t \ge a, W_t > a) = \mathbb{P}(W_t > a)\]

(The second equality uses the fact that if the reflected path ends above \(a\), the original path must have hit \(a\).)

Therefore:

\[\begin{array}{lll} \mathbb{P}(M_t \ge a) &=& \mathbb{P}(M_t \ge a, W_t \ge a) + \mathbb{P}(M_t \ge a, W_t < a)\\ &=& \mathbb{P}(W_t \ge a) + \mathbb{P}(W_t > a)\\ &=& 2\mathbb{P}(W_t \ge a) \quad \square\end{array}\]

Explicit Formula for the Maximum¶

Since \(W_t \sim \mathcal{N}(0, t)\):

\[\mathbb{P}(M_t \ge a) = 2\mathbb{P}(W_t \ge a) = 2\left[1 - \Phi\left(\frac{a}{\sqrt{t}}\right)\right] = 2\Phi\left(-\frac{a}{\sqrt{t}}\right)\]

where \(\Phi\) is the standard normal CDF.

Joint Distribution of Maximum and Endpoint¶

Statement¶

Theorem 1.6.2 (Reflection Principle for Joint Events)

For \(a > 0\) and \(b < a\):

\[\boxed{ \mathbb{P}(M_t \ge a, W_t \le b) = \mathbb{P}(W_t \ge 2a - b) }\]

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 \(\tau_a\):

\[\tilde{W}_t = 2a - W_t \ge 2a - b > a\]

The bijection maps:

\[\{M_t \ge a, W_t \le b\} \leftrightarrow \{W_t \ge 2a - b\}\]

Therefore:

\[\mathbb{P}(M_t \ge a, W_t \le b) = \mathbb{P}(W_t \ge 2a - b) \quad \square\]

Explicit Formula for the Joint Tail¶

Since \(W_t \sim \mathcal{N}(0, t)\):

\[\boxed{ \mathbb{P}(M_t \ge a, W_t \le b) = 1 - \Phi\left(\frac{2a - b}{\sqrt{t}}\right) = \Phi\left(\frac{b - 2a}{\sqrt{t}}\right) }\]

First passage times: handled separately

Recall (see § First Passage Times): the first hitting time \(\tau_a := \inf\{t \ge 0 : W_t = a\}\) satisfies \(\{\tau_a \le t\} = \{M_t \ge a\}\), so its distribution, density, moments, and Laplace transform follow from the maximum identity above. The full development is in that section; this page handles only the geometric/symmetry side.

Joint Density of Maximum and Endpoint¶

We now derive the complete joint density \(f_{M_t, W_t}(m, w)\).

Main Result¶

Theorem 1.6.8 (Joint PDF of Maximum and Endpoint)

For \(m > 0\) and \(w \le m\):

\[\boxed{ f_{M_t, W_t}(m, w) = \frac{2(2m - w)}{t\sqrt{2\pi t}} \exp\left(-\frac{(2m - w)^2}{2t}\right) }\]

Derivation¶

Step 1: CDF via reflection.

From Theorem 1.6.2, for \(w < m\):

\[\mathbb{P}(M_t \ge m, W_t \le w) = \mathbb{P}(W_t \ge 2m - w)\]

Therefore:

\[\begin{array}{lll} \mathbb{P}(M_t \le m, W_t \le w) &=&\displaystyle \mathbb{P}(W_t \le w) - \mathbb{P}(W_t \ge 2m - w)\\ &=&\displaystyle \Phi\left(\frac{w}{\sqrt{t}}\right) - \left[1 - \Phi\left(\frac{2m - w}{\sqrt{t}}\right)\right]\\ &=&\displaystyle \Phi\left(\frac{w}{\sqrt{t}}\right) + \Phi\left(\frac{2m - w}{\sqrt{t}}\right) - 1 \end{array}\]

Step 2: Differentiate to get the joint PDF.

\[f_{M_t, W_t}(m, w) = \frac{\partial^2}{\partial m \partial w} \mathbb{P}(M_t \le m, W_t \le w)\]

First, differentiate with respect to \(w\):

\[\frac{\partial}{\partial w}\left[\Phi\left(\frac{w}{\sqrt{t}}\right) + \Phi\left(\frac{2m - w}{\sqrt{t}}\right) - 1\right] = \phi\left(\frac{w}{\sqrt{t}}\right) \cdot \frac{1}{\sqrt{t}} - \phi\left(\frac{2m - w}{\sqrt{t}}\right) \cdot \frac{1}{\sqrt{t}}\]

where \(\phi(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}\).

Now differentiate with respect to \(m\):

\[\begin{array}{lll} \frac{\partial}{\partial m}\left[-\phi\left(\frac{2m - w}{\sqrt{t}}\right) \cdot \frac{1}{\sqrt{t}}\right] &=&\displaystyle -\frac{1}{\sqrt{t}} \cdot \phi'\left(\frac{2m - w}{\sqrt{t}}\right) \cdot \frac{2}{\sqrt{t}}\quad (\because\phi'(x) = -x\phi(x))\\ &=&\displaystyle -\frac{2}{t} \cdot \left(-\frac{2m - w}{\sqrt{t}}\right) \phi\left(\frac{2m - w}{\sqrt{t}}\right)\\ &=&\displaystyle \frac{2(2m - w)}{t\sqrt{t}} \cdot \frac{1}{\sqrt{2\pi}} e^{-(2m-w)^2/(2t)}\\ &=&\displaystyle \frac{2(2m - w)}{t\sqrt{2\pi t}} \exp\left(-\frac{(2m - w)^2}{2t}\right) \quad \square \end{array}\]

Conditional Distribution¶

Corollary 1.6.9 (Conditional PDF of Maximum Given Endpoint)

Given \(W_t = w\), the conditional density of \(M_t\) is:

\[f_{M_t | W_t}(m | w) = \frac{f_{M_t, W_t}(m, w)}{f_{W_t}(w)} = \frac{2(2m - w)}{t} \exp\left(-\frac{2m(m - w)}{t}\right)\]

for \(m \ge w\).

Proof:

\[f_{W_t}(w) = \frac{1}{\sqrt{2\pi t}} e^{-w^2/(2t)}\]

\[f_{M_t | W_t}(m | w) = \frac{f_{M_t, W_t}(m, w)}{f_{W_t}(w)} = \frac{\frac{2(2m-w)}{t\sqrt{2\pi t}} e^{-(2m-w)^2/(2t)}}{\frac{1}{\sqrt{2\pi t}} e^{-w^2/(2t)}}\]

Simplify the exponentials:

\[\frac{-(2m-w)^2/(2t) + w^2/(2t)}{1} = -\frac{2m(m-w)}{t}\]

Therefore:

\[f_{M_t | W_t}(m | w) = \frac{2(2m - w)}{t} e^{-2m(m-w)/t} \quad \square\]

Summary¶

Maximum distribution: \(\mathbb{P}(M_t \ge a) = 2\mathbb{P}(W_t \ge a) = 2\Phi(-a/\sqrt{t})\)
Joint distribution: \(\mathbb{P}(M_t \ge a, W_t \le b) = \mathbb{P}(W_t \ge 2a - b)\) for \(b < a\)
Joint density: \(f_{M_t, W_t}(m, w) = \frac{2(2m-w)}{t\sqrt{2\pi t}} e^{-(2m-w)^2/(2t)}\) for \(m > 0\), \(w \le m\)
First passage: developed in § First Passage Times via \(\{\tau_a \le t\} = \{M_t \ge a\}\).

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 \(a > 0\):

\[ \mathbb{P}(M_t \ge a) = 2\mathbb{P}(W_t \ge a) \]

Proof: Partition \(\{M_t \ge a\}\) into paths ending above and below \(a\):

\[ \mathbb{P}(M_t \ge a) = \mathbb{P}(M_t \ge a, W_t \ge a) + \mathbb{P}(M_t \ge a, W_t < a) \]

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:

\[ \mathbb{P}(M_t \ge a) = \mathbb{P}(W_t \ge a) + \mathbb{P}(W_t > a) = 2\mathbb{P}(W_t \ge a) \]

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 \(a > 0\):

\[ \mathbb{P}(M_t \le a) = 1 - \mathbb{P}(M_t \ge a) = 1 - 2\mathbb{P}(W_t \ge a) = 1 - 2\left[1 - \Phi\left(\frac{a}{\sqrt{t}}\right)\right] = 2\Phi\left(\frac{a}{\sqrt{t}}\right) - 1 \]

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 \(a > 0\)):

\[ f_{M_t}(a) = \frac{d}{da}\left[2\Phi\left(\frac{a}{\sqrt{t}}\right) - 1\right] = \frac{2}{\sqrt{t}}\,\phi\left(\frac{a}{\sqrt{t}}\right) = \frac{2}{\sqrt{2\pi t}}\exp\left(-\frac{a^2}{2t}\right) \]

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 \(\mathbb{P}(W_t \ge a) = \mathbb{P}(W_t \le -a) = \mathbb{P}(-W_t \ge a)\).

\[ \mathbb{P}(|W_t| \ge a) = \mathbb{P}(W_t \ge a) + \mathbb{P}(W_t \le -a) = 2\mathbb{P}(W_t \ge a) = 2\left[1 - \Phi\left(\frac{a}{\sqrt{t}}\right)\right] \]

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:

\[ f_{M_t, W_t}(m, w) = \frac{2(2m - w)}{t\sqrt{2\pi t}} \exp\left(-\frac{(2m - w)^2}{2t}\right) \]

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 \(w = m\), \(u = m/\sqrt{t}\):

\[ \int_{-\infty}^m \frac{2(2m-w)}{t\sqrt{2\pi t}} e^{-(2m-w)^2/(2t)}\,dw = \frac{2}{t\sqrt{2\pi t}} \int_{m/\sqrt{t}}^\infty u\sqrt{t} \cdot e^{-u^2/2} \cdot \sqrt{t}\,du \]

\[ = \frac{2}{\sqrt{2\pi}} \int_{m/\sqrt{t}}^\infty u\,e^{-u^2/2}\,du = \frac{2}{\sqrt{2\pi}}\left[-e^{-u^2/2}\right]_{m/\sqrt{t}}^\infty = \frac{2}{\sqrt{2\pi}} e^{-m^2/(2t)} \]

Outer integral (over \(m\)):

\[ \int_0^\infty \frac{2}{\sqrt{2\pi}} e^{-m^2/(2t)}\,dm = \frac{2}{\sqrt{2\pi}} \cdot \sqrt{\frac{\pi t}{2}} \cdot \frac{1}{\sqrt{t}} \cdot \sqrt{t} = \frac{2}{\sqrt{2\pi}} \cdot \sqrt{\frac{\pi t}{2}} = 1 \]

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 (cross-references)¶

Recall (see § First Passage Times): for \(a > 0\), \(\tau_a := \inf\{t \ge 0 : W_t = a\}\) satisfies \(\mathbb{P}(\tau_a < \infty) = 1\), \(\mathbb{E}[\tau_a] = \infty\), and has Lévy density \(f_{\tau_a}(t) = \frac{a}{\sqrt{2\pi t^3}}e^{-a^2/(2t)}\). Detailed exercises (recurrence, moments, normalization, Laplace transform) live in that section.

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 5

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 6

Using \(f_{M_t|W_t}(m|w) = \frac{2(2m-w)}{t} e^{-2m(m-w)/t}\) for \(m \ge \max(w, 0)\):

\[ \mathbb{E}[M_T | W_T = w] = \int_{\max(w,0)}^\infty m \cdot \frac{2(2m-w)}{T} e^{-2m(m-w)/T}\,dm \]

For \(w \ge 0\), substitute \(v = m - w/2\) (so \(2m - w = 2v\) and \(m = v + w/2\)), and let \(c = 2/T\):

\[ = \int_0^\infty (v + w/2) \cdot \frac{4v}{T} e^{-cv(v+w/2) \cdot 2}\,dv \]

This integral does not simplify to a closed form in elementary functions for general \(w\). However, for \(w = 0\):

\[ \mathbb{E}[M_T | W_T = 0] = \int_0^\infty m \cdot \frac{4m}{T} e^{-2m^2/T}\,dm = \frac{4}{T}\int_0^\infty m^2 e^{-2m^2/T}\,dm \]

Using \(\int_0^\infty x^2 e^{-\beta x^2}\,dx = \frac{\sqrt{\pi}}{4\beta^{3/2}}\) with \(\beta = 2/T\):

\[ = \frac{4}{T} \cdot \frac{\sqrt{\pi}}{4(2/T)^{3/2}} = \frac{4}{T} \cdot \frac{\sqrt{\pi}T^{3/2}}{4 \cdot 2\sqrt{2}} = \frac{\sqrt{\pi T}}{2\sqrt{2}} = \sqrt{\frac{\pi T}{8}} \]

Applications¶

Recall (see § First Passage Times): Laplace transform \(\mathbb{E}[e^{-\alpha\tau_a}] = e^{-a\sqrt{2\alpha}}\) and the verification \(\int_0^\infty f_{\tau_a}(t)\,dt = 1\) are exercises in that section.

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 7

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:

\[ \max_{0 \le s \le T} S_s \ge B \iff \max_{0 \le s \le T} W_s \ge \frac{\log(B/S_0) - (r - \sigma^2/2)T}{\sigma} \]

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:

\[ a = \frac{\log(B/S_0)}{\sigma} \]

The activation probability is:

\[ \mathbb{P}\left(\max_{0 \le s \le T} W_s \ge a\right) = 2\Phi\left(-\frac{a}{\sqrt{T}}\right) = 2\Phi\left(-\frac{\log(B/S_0)}{\sigma\sqrt{T}}\right) \]

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 8

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 \(f_{M_T, W_T}(m, w)\):

\[ \mathbb{P}(DD_T > d) = \int_0^\infty \int_{-\infty}^{m-d} f_{M_T, W_T}(m, w)\,dw\,dm \]

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 \(v \ge m + d\)):

\[ = \int_0^\infty \int_{m+d}^\infty \frac{2v}{T\sqrt{2\pi T}} e^{-v^2/(2T)}\,dv\,dm \]

The inner integral evaluates as:

\[ \int_{m+d}^\infty \frac{2v}{T\sqrt{2\pi T}} e^{-v^2/(2T)}\,dv = \frac{2}{\sqrt{2\pi T}} e^{-(m+d)^2/(2T)} \]

Therefore:

\[ \mathbb{P}(DD_T > d) = \frac{2}{\sqrt{2\pi T}} \int_0^\infty e^{-(m+d)^2/(2T)}\,dm \]

Substitute \(u = (m + d)/\sqrt{T}\):

\[ = \frac{2}{\sqrt{2\pi}} \int_{d/\sqrt{T}}^\infty e^{-u^2/2}\,du = 2\left[1 - \Phi\left(\frac{d}{\sqrt{T}}\right)\right] = 2\Phi\left(-\frac{d}{\sqrt{T}}\right) \]

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)
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 - Barrier Options)