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Law of the Brownian maximum

Statement

Assume the Axiom of Choice, let $B$ be a standard Brownian motion Brownian motion. Use the everywhere-continuous, zero-start representative fixed in Brownian reflection principle: replace the path by zero outside a measurable probability-one event of continuity and zero start, retaining the notation $B$ . Let $t > 0$ and $M_{t} : = \sup_{0 \leq s \leq t} B_{s}$ . This is a finite nonnegative random variable: continuity gives boundedness on $[0, t]$ and identifies its supremum with the supremum over the countable dense set $(Q \cap [0, t]) \cup {t}$ . With $Φ$ the standard normal distribution function, $Φ (x) = N (0, 1) ((- \infty, x])$ Standard normal and normal laws Cumulative distribution function of a real random variable, one has for every $x \geq 0$ $P (M_{t} \leq x) = 2 Φ ⁣ (\frac{x}{\sqrt{t}}) - 1.$ Consequently $M_{t}$ has the same law as $∣ B_{t} ∣$ , and on $x > 0$ the law of $M_{t}$ has the density $f (x) = \sqrt{\frac{2}{π t}} \exp ⁣ (- \frac{x^{2}}{2 t}) .$

Facts & Assumptions

Given: AC, a standard Brownian motion $B$ in the stated everywhere-continuous zero-start representative, $t > 0$ and $x \geq 0$ .

[F1]

$P (M_{t} \geq a) = 2 P (B_{t} \geq a)$ for every $a > 0$ , and $M_{t} \geq a ⟺ \sup_{[0, t]} B \geq a$ . Brownian reflection principle

[F2]

The law of $B_{t}$ is $N (0, t)$ , the law of $\sqrt{t} Z$ for a standard normal $Z$ , whose density is $φ (y) = e^{- y^{2} / 2} / \sqrt{2 π}$ ; hence $P (a \leq B_{t} \leq b) = \int_{a}^{b} φ (y / \sqrt{t}) t^{- 1 / 2} d y$ for $a < b$ and $P (B_{t} = c) = 0$ . The Brownian kernels form a semigroup Standard normal and normal laws

[F3]

Substitution for continuous integrands on compact intervals, with the Riemann–Lebesgue equality (under countable choice, supplied by AC), and monotone convergence for increasing nonnegative functions, in particular indicators. Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral Monotone convergence for the integral

[F4]

A probability measure on $R$ is determined by its distribution function on the intervals $(- \infty, x]$ ; this uses countable choice, which AC supplies. Probability laws correspond to distribution functions The Axiom of Countable Choice ( ${A C}_{ω}$ ) The Axiom of Choice

Proof

technique · direct

1.1

For $x \geq 0$ and positive integers $n$ , put $a_{n} = x + 1 / n > 0$ . The events ${M_{t} \geq a_{n}}$ increase to ${M_{t} > x}$ , and ${B_{t} \geq a_{n}}$ increase to ${B_{t} > x}$ . Applying [F3] to their indicators and [F1] at each $a_{n}$ gives $P (M_{t} > x) = 2 P (B_{t} > x)$ . (To use an index starting at zero, replace $n$ by $n + 1$ .) By [F2], $P (B_{t} > x) = 1 - Φ (x / \sqrt{t})$ . Taking complements yields the asserted formula for every $x \geq 0$ , including $x = 0$ since the even normal density has mass one and no atom, so $Φ (0) = 1 / 2$ .

F1F2F3given

1.2

Define $G (x) : = \int_{0}^{x} f (y) d y$ for $x \geq 0$ with $f (y) = \sqrt{2 / (π t)} e^{- y^{2} / (2 t)}$ . The substitution $y = \sqrt{t} u$ , applied to the continuous integrand on $[0, x]$ , gives $G (x) = 2 (Φ (x / \sqrt{t}) - Φ (0)) = 2 Φ (x / \sqrt{t}) - 1$ for every $x > 0$ : indeed $f (\sqrt{t} u) \sqrt{t} = 2 φ (u)$ . For $x \geq 0$ , the bound $0 \leq G (x) \leq \sqrt{2 / (π t)} x$ gives $G (0^{+}) = 0$ , and the same computation with the upper limit tending to $+ \infty$ , together with $\lim_{u \to \infty} Φ (u) = 1$ , gives $\int_{0}^{\infty} f = 1$ .

F2F3

2.1

For $x \geq 0$ , $P (∣ B_{t} ∣ \leq x) = P (- x \leq B_{t} \leq x) = Φ (x / \sqrt{t}) - Φ (- x / \sqrt{t}) = 2 Φ (x / \sqrt{t}) - 1$ by the symmetry $Φ (- u) = 1 - Φ (u)$ of the standard normal law, which follows from the symmetry of its density $φ$ ; for $x < 0$ both $P (M_{t} \leq x)$ and $P (∣ B_{t} ∣ \leq x)$ vanish. Since the two distribution functions agree on all of $R$ , [F4] identifies the laws, so $M_{t}$ and $∣ B_{t} ∣$ have the same law.

F2F4step 1.1

2.2

The measure with density $f$ on $(0, \infty)$ , extended by zero on $(- \infty, 0]$ , is a probability measure whose distribution function at $x \geq 0$ is $G (x) = 2 Φ (x / \sqrt{t}) - 1$ and at $x < 0$ is $0$ ; by [F4] it therefore equals the law of $M_{t}$ . Hence the law of $M_{t}$ has the density $f$ on $x > 0$ and no atom at $0$ .

F4step 1.1step 1.2

3.1

The cases $x = 0$ and $t > 0$ are included in steps 1.1 and 2.2; the strict-tail identity was obtained by increasing indicator limits in step 1.1, and atomlessness of the maximum follows from its density in step 2.2. Countable choice in the Riemann–Lebesgue bridge and [F4] is supplied by the assumed AC.

F2F4givenstep 1.2∎

Source notes

Lawler, Proposition 2.7.2, supplies the reflection-based maximum formula. The strict tail is derived here by increasing indicator limits, including the endpoint zero. The density is identified through compact-interval substitution, the Riemann–Lebesgue bridge, and equality of distribution functions.

Depends on

Used by

Distribution of a one-sided Brownian hitting time Corollary
Strong Markov fails at a nonstopping random time Counterexample
Maximum crossing before a fixed time Example
Brownian law of the iterated logarithm at infinity Theorem
The Brownian zero set has no isolated points Theorem
The last Brownian zero has the arcsine law Theorem
Two-sided Brownian exit probability Theorem

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Sources

Gregory F. Lawler, Stochastic Calculus: An Introduction with Applications, Proposition 2.7.2 (standard reference, not scraped)
Rick Durrett, Probability: Theory and Examples, fifth edition, Section 7.4 (standard reference, not scraped)