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Simple Random Walks on the Integers · Tutorial 295 of 994

The Time of the Maximum and Its Arcsine Limit

Learn how to define the time of a random-walk maximum, interpret its arcsine scaling limit, and distinguish finite-sample tie conventions.

Intermediate 10 min read

What You'll Learn

  • Define the earliest and latest times attaining a walk’s running maximum
  • Compute finite-sample maximum-time distributions by enumerating paths
  • State the normalized arcsine limit and its Brownian-motion basis
  • Prove why earliest and latest maximizers have the same scaling limit
  • Identify how ties affect exact finite-sample probabilities

Which Time Is “the” Time of the Maximum?

A walk may reach its highest level more than once. In a finite sample, asking for “the time of the maximum” therefore requires a tie convention. We will use the earliest time at which the running maximum is attained. This choice makes the statistic unambiguous, but it affects finite-sample probabilities: counting the latest maximizing time can give a different distribution. After dividing time by the observation horizon, however, both conventions have the same arcsine limit.

Let \((\Omega,\mathcal{F},\mathbb{P})\) carry independent random variables \(\xi_1,\xi_2,\ldots\), each equal to \(1\) or \(-1\) with probability \(1/2\). Define the simple symmetric random walk by \(S_0=0\) and \(S_k=\sum_{r=1}^k \xi_r\) for \(k\ge1\). Its state space is \(\mathbb{Z}\), and its time index is discrete, \(k\ge0\). For a fixed integer \(n\ge1\), let \(M_n=\max_{0\le k\le n}S_k\).

Definition: The earliest time of the maximum through time \(n\) is $$ T_n^{\max}=\min\{k\in\{0,1,\ldots,n\}:S_k=M_n\}. $$ The latest time of the maximum is \(L_n^{\max}=\max\{k\in\{0,1,\ldots,n\}:S_k=M_n\}\). The set being minimized or maximized is nonempty because a maximum over finitely many values is attained.

The normalized time \(T_n^{\max}/n\) lies in \([0,1]\). Its limit law is the arcsine distribution, whose cumulative distribution function is \(F(x)=\frac{2}{\pi}\arcsin\sqrt{x}\) for \(0\le x\le1\), and whose density on \(0<x<1\) is \(f(x)=\frac{1}{\pi\sqrt{x(1-x)}}\). The density is unbounded near both endpoints, so the limiting maximum time is more likely to lie near the beginning or end than near the middle.

Theorem (Arcsine limit for the time of the maximum): For the simple symmetric random walk just defined, $$ \frac{T_n^{\max}}{n}\ \xrightarrow[n\to\infty]{\ d\ }\ A, \qquad \mathbb{P}(A\le x)=\frac{2}{\pi}\arcsin\sqrt{x}\quad(0\le x\le1). $$ The same convergence holds with \(L_n^{\max}\) in place of \(T_n^{\max}\). Here convergence in distribution means convergence of the cumulative distribution functions at every continuity point of the limiting distribution.

The proof below reduces the random-walk statement to standard results about Brownian motion. The Brownian invariance principle (Donsker’s theorem), almost-sure uniqueness of the time at which Brownian motion attains its maximum on \([0,1]\), and the Brownian argmax arcsine law are standard results; their proofs are beyond this tutorial. We state their precise uses so that the reduction from the walk to Brownian motion is explicit.

Worked Example: Finding the Earliest Maximum on One Path

Suppose the five increments observed are \(+1,-1,+1,+1,-1\). The successive walk values, including time zero, are \(S_0=0\), \(S_1=1\), \(S_2=0\), \(S_3=1\), \(S_4=2\), and \(S_5=1\). Thus \(M_5=2\), attained only at time \(4\), so \(T_5^{\max}=L_5^{\max}=4\) and the normalized time is \(4/5\).

Check: Comparing the six values \(0,1,0,1,2,1\), the largest is \(2\); its index is \(4\). Dividing that index by the horizon gives \(4/5\), as required by the definition.

Worked Example: All Paths Through Time Two

There are \(2^2=4\) equally likely increment sequences, each with probability \(1/4\). Listing the walk values identifies the earliest maximum for each path:

  • \(++\): values \(0,1,2\); the maximum is \(2\), first attained at time \(2\).
  • \(+-\): values \(0,1,0\); the maximum is \(1\), first attained at time \(1\).
  • \(-+\): values \(0,-1,0\); the maximum is \(0\), first attained at time \(0\).
  • \(--\): values \(0,-1,-2\); the maximum is \(0\), first attained at time \(0\).

Therefore \(\mathbb{P}(T_2^{\max}=0)=2/4=1/2\), \(\mathbb{P}(T_2^{\max}=1)=1/4\), and \(\mathbb{P}(T_2^{\max}=2)=1/4\). The probabilities sum to \(1/2+1/4+1/4=1\).

For the latest convention, the path \(-+\) has its maximum at both times \(0\) and \(2\), so its latest maximizing time is \(2\), not \(0\). The latest-time probabilities are consequently \(1/4\) at time \(0\), \(1/4\) at time \(1\), and \(2/4=1/2\) at time \(2\); these also sum to \(1\). Thus the two conventions give different finite-sample distributions.

Why Both Tie Conventions Have the Same Limit

We first give the deterministic fact that makes the limiting argument work. Consider continuous functions on the compact interval \([0,1]\). A maximizer of a function \(f\) is a point \(t\) with \(f(t)\ge f(s)\) for every \(s\in[0,1]\).

Lemma (Stability of a unique maximizer): Suppose continuous functions \(f_m\) converge uniformly to a continuous function \(f\) on \([0,1]\), and \(f\) has a unique maximizer \(t_*\). If \(t_m\) is any maximizer of \(f_m\), then \(t_m\to t_*\). In particular, the result applies whether \(t_m\) is chosen as the earliest or the latest maximizer.

Proof. Take any subsequence of \((t_m)\). Since \([0,1]\) is compact, that subsequence has a further subsequence converging to some \(t\in[0,1]\). Along this further subsequence, uniform convergence gives \(|f_m(t_m)-f(t_m)|\to0\). Continuity of \(f\) and \(t_m\to t\) give \(f(t_m)\to f(t)\). For every fixed \(s\in[0,1]\), the maximizing property gives \(f_m(t_m)\ge f_m(s)\). Uniform convergence gives \(f_m(s)\to f(s)\), so taking limits yields \(f(t)\ge f(s)\). This holds for every \(s\), meaning \(t\) is a maximizer of \(f\). Uniqueness forces \(t=t_*\). Every convergent further subsequence of \((t_m)\) therefore has limit \(t_*\); compactness then implies the original sequence converges to \(t_*\). The argument used no particular rule for choosing a maximizer. \(\square\)

For each \(n\), linearly interpolate the values \(S_k/\sqrt{n}\) at times \(k/n\), \(0\le k\le n\), to obtain a continuous function \(U_n\) on \([0,1]\). The simple symmetric walk has independent increments with mean \(0\) and variance \(1\). Donsker’s theorem states that \(U_n\) converges in distribution, as a random continuous function, to standard Brownian motion \(B_t\), \(0\le t\le1\). The interpolation has no flat segments because each increment is \(1\) or \(-1\); its maximum is attained at one or more grid points. Its earliest maximizing time is exactly \(T_n^{\max}/n\).

Proof of the theorem using the stated standard results. Brownian motion has, with probability one, a unique maximizing time \(\tau_*=\operatorname*{arg\,max}_{0\le t\le1}B_t\). The stability lemma shows why maximizer locations pass to the limit under uniform convergence whenever the limiting function has a unique maximizer. Applied to the random-function convergence in Donsker’s theorem, this gives convergence in distribution of the earliest maximizing times to \(\tau_*\). The same lemma applies to the latest maximizing times, so they converge to \(\tau_*\) as well. The standard Brownian argmax arcsine law states that \(\mathbb{P}(\tau_*\le x)=\frac{2}{\pi}\arcsin\sqrt{x}\) for \(0\le x\le1\). Substitution of this distribution identifies both limiting laws and proves the asserted convergences. The Brownian uniqueness and argmax-law results are the deep inputs here; they have been named rather than claimed as proved. \(\square\)

The discrete-time tie convention matters because a walk can return to its current maximum. In the limit, the Brownian path has a unique maximizing time almost surely. The lemma explains the distinction: when the limiting maximizer is unique, any finite-sample choice among maximizers approaches that same time. This conclusion concerns the normalized times; it does not say that earliest and latest maximizing times have the same finite-sample distribution.

Worked Example: Reading Probabilities from the Limit

Let \(A\) have the limiting arcsine distribution. The probability that the normalized maximum time lies between \(1/4\) and \(3/4\), including the left endpoint and excluding the right endpoint, is \(F(3/4)-F(1/4)\). The limiting distribution is continuous, so including or excluding either endpoint gives the same probability.

$$ F\!\left(\frac34\right) =\frac{2}{\pi}\arcsin\!\left(\frac{\sqrt3}{2}\right) =\frac{2}{\pi}\cdot\frac{\pi}{3} =\frac23, \qquad F\!\left(\frac14\right) =\frac{2}{\pi}\arcsin\!\left(\frac12\right) =\frac{2}{\pi}\cdot\frac{\pi}{6} =\frac13. $$

Hence \(\mathbb{P}(1/4\le A<3/4)=2/3-1/3=1/3\). The density gives an independent check: using \(x=\sin^2\theta\), so \(dx=2\sin\theta\cos\theta\,d\theta\), the integral of \(1/(\pi\sqrt{x(1-x)})\) from \(1/4\) to \(3/4\) becomes \(\frac{2}{\pi}\int_{\pi/6}^{\pi/3}d\theta=\frac{2}{\pi}(\pi/3-\pi/6)=1/3\).

Conditions and a Common Pitfall

The arcsine conclusion is specific to the symmetric, mean-zero, finite-variance walk and the large-time scaling described above. It is not a rule for every process that happens to have a maximum. For example, if every increment is instead deterministically \(+1\), then \(S_k=k\) and the unique maximum through time \(n\) occurs at time \(n\). Thus \(T_n^{\max}/n=1\) for every \(n\), which does not converge to the arcsine distribution. This example drops the symmetric random-walk assumption and shows why the hypotheses cannot be discarded.

Common mistake: treating a tied maximum as if it had only one occurrence. For the two-step path \(-+\), the values are \(0,-1,0\). Its maximum is \(0\), attained at both \(0\) and \(2\). Counting this path only when the maximum occurs at time \(2\) would incorrectly give \(\mathbb{P}(T_2^{\max}=0)=1/4\), counting only the path \(--\). Under the stated earliest-time convention, \(-+\) also contributes at time \(0\), so the correct probability is \(2/4=1/2\). The correction is to identify all maximizing times first, then apply the specified minimum or maximum rule.

For finite \(n\), \(T_n^{\max}/n\) takes values only in the grid \(\{0,1/n,\ldots,1\}\), and ties can shift probability between grid points depending on the convention. The arcsine law describes the limiting distribution as the grid becomes fine; it is not an exact finite-\(n\) formula. When comparing finite-sample calculations, state the convention explicitly and count paths according to it.

Check Your Understanding

Use the definitions and limiting result to answer these questions.

  1. For the path with values \(0,1,0,1,2,1\), find \(M_5\), \(T_5^{\max}\), and \(T_5^{\max}/5\).
  2. For the two-step path \(-+\), list all maximizing times and give the earliest and latest maximizing times.
  3. Evaluate the limiting probability \(\mathbb{P}(A\le 1/2)\) from the arcsine cumulative distribution function.
  4. Why does almost-sure uniqueness of the Brownian maximizing time make the choice between earliest and latest maximizers irrelevant in the normalized limit?
  5. What happens to the normalized time of the maximum if every increment is deterministically \(+1\), and which random-walk assumption has been dropped?