The Arcsine Limit for Positive Occupation

Positive Occupation and Its Limit

A simple symmetric random walk repeatedly takes one step to the right or one step to the left, each with probability one half. Although its expected position remains zero, the fraction of time it spends above zero does not settle down to one half in the way a long-run average of independent coin flips might. Instead, its limiting distribution places relatively more probability near zero and one than near the middle. This distribution is called the arcsine law.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $(\xi_k)_{k\ge1}$ be independent random variables with $\mathbb{P}(\xi_k=1)=\mathbb{P}(\xi_k=-1)=\tfrac12$, and define the walk $S_0=0$, $S_n=\sum_{k=1}^n\xi_k$. This is a discrete-time process with state space $S=\mathbb{Z}$. For $n\ge1$, define its positive occupation count and fraction by

N_n=\sum_{k=1}^n\mathbf{1}_{\{S_k>0\}}, \qquad U_n=\frac{N_n}{n}.

The inequality is strict: a visit to zero is not counted as time spent positive. The sum starts at $k=1$, so the initial position $S_0=0$ is not counted either. Thus $N_n$ is an integer from $0$ to $n$, and $U_n$ lies in $[0,1]$.

Theorem (Arcsine limit for positive occupation): For the simple symmetric random walk just defined, $U_n$ converges in distribution to a random variable $U$ with density $$ f_U(x)=\frac{1}{\pi\sqrt{x(1-x)}},\qquad 0<x<1. $$ The density is zero outside $(0,1)$, and the limiting distribution has no atoms at $0$ or $1$. The path-counting proof of this limit is given in the derivation of the occupation-time arcsine law. Here we use the result to understand and calculate with its limiting distribution.

Convergence in distribution describes the limiting probabilities of intervals whose boundary has probability zero under $U$. It does not say that $U_n$ converges in probability to a fixed number. In particular, the limiting fraction remains random. The density is symmetric around $1/2$, but it is not largest there: it diverges as $x$ approaches either endpoint. This endpoint behavior is the source of the name “arcsine” and of the law’s distinctive interpretation.

From Density to Probabilities

The density gives probability by integration. For $0\le x\le1$, the cumulative distribution function is

F_U(x)=\mathbb{P}(U\le x)=\frac{2}{\pi}\arcsin\sqrt{x}.

Proof. For $0<x<1$, substitute $y=\sin^2\theta$ into the density integral. As $y$ runs from $0$ to $x$, $\theta$ runs from $0$ to $\arcsin\sqrt{x}$. Since $dy=2\sin\theta\cos\theta\,d\theta$ and $\sqrt{y(1-y)}=\sin\theta\cos\theta$ on this range,

\int_0^x\frac{dy}{\pi\sqrt{y(1-y)}} =\frac{2}{\pi}\int_0^{\arcsin\sqrt{x}}d\theta =\frac{2}{\pi}\arcsin\sqrt{x}.

At $x=0$ this expression is zero, and at $x=1$ it is $\tfrac{2}{\pi}\cdot\tfrac{\pi}{2}=1$. These endpoint values agree with the total probability and show that the density integrates to one. Since the distribution has a density, any individual point, including either endpoint, has probability zero. $\square$

For an interval inside $[0,1]$, subtract cumulative probabilities. The symmetry follows either from the density, since $f_U(x)=f_U(1-x)$, or from the identity $F_U(1-x)=1-F_U(x)$. The latter follows from $\arcsin\sqrt{1-x}=\tfrac{\pi}{2}-\arcsin\sqrt{x}$ for $0\le x\le1$.

Worked Example: A Middle-Interval Probability

Find the limiting probability that the positive occupation fraction is between $\tfrac14$ and $\tfrac34$, including either boundary.

The distribution is continuous, so including or excluding the boundary points makes no difference. Using the cumulative distribution function,

\begin{aligned} \mathbb{P}\left(\tfrac14\le U\le\tfrac34\right) &=F_U\left(\tfrac34\right)-F_U\left(\tfrac14\right)\\ &=\frac{2}{\pi}\left(\arcsin\sqrt{\tfrac34}-\arcsin\sqrt{\tfrac14}\right)\\ &=\frac{2}{\pi}\left(\arcsin\tfrac{\sqrt3}{2}-\arcsin\tfrac12\right)\\ &=\frac{2}{\pi}\left(\tfrac{\pi}{3}-\tfrac{\pi}{6}\right) =\frac13. \end{aligned}

Check: Symmetry gives $F_U(\tfrac34)=1-F_U(\tfrac14)$. The endpoint values are $F_U(\tfrac34)=\tfrac23$ and $F_U(\tfrac14)=\tfrac13$, so their difference is indeed $\tfrac13$.

Moments of the Limiting Fraction

The mean and variance give a compact summary of the limiting law, but they do not capture its endpoint-heavy shape. Both can be calculated directly from the density.

Result: If $U$ has the arcsine density above, then $\mathbb{E}[U]=\tfrac12$ and $\operatorname{Var}(U)=\tfrac18$.

Proof. In the moment integrals, make the substitution $x=\sin^2\theta$ for $0<\theta<\tfrac{\pi}{2}$. As above, $dx/\sqrt{x(1-x)}=2\,d\theta$. Therefore

\mathbb{E}[U] =\frac{2}{\pi}\int_0^{\pi/2}\sin^2\theta\,d\theta =\frac{2}{\pi}\cdot\frac{\pi}{4} =\frac12.

Here the integral is $\tfrac{\pi}{4}$ because $\sin^2\theta=\tfrac12(1-\cos 2\theta)$, whose cosine term integrates to zero between $0$ and $\tfrac{\pi}{2}$. For the second moment, use $\sin^4\theta=\tfrac18(3-4\cos 2\theta+\cos 4\theta)$. Both cosine terms integrate to zero on this interval, so

\mathbb{E}[U^2] =\frac{2}{\pi}\int_0^{\pi/2}\sin^4\theta\,d\theta =\frac{2}{\pi}\cdot\frac{3\pi}{16} =\frac38.

By the definition of variance, $\operatorname{Var}(U)=\mathbb{E}[U^2]-(\mathbb{E}[U])^2=\tfrac38-(\tfrac12)^2=\tfrac18$. $\square$

The mean of one half is consistent with the symmetry between positive and negative locations. The variance is positive, so the limiting fraction does not collapse onto its mean. In fact, the density’s shape shows that fractions nearer the endpoints are more likely than fractions in a same-width interval around the center.

Worked Example: Comparing Endpoint and Central Probabilities

Compare the limiting probability of being in either outer quarter, $[0,\tfrac14]\cup[\tfrac34,1]$, with the probability of being in the central half, $[\tfrac14,\tfrac34]$.

The first outer-quarter probability is $F_U(\tfrac14)=\tfrac{2}{\pi}\arcsin(\tfrac12)=\tfrac13$. By symmetry, the probability in $[\tfrac34,1]$ is also $\tfrac13$. These two intervals are disjoint, so their combined probability is $\tfrac13+\tfrac13=\tfrac23$. The central-half probability, calculated in the previous example, is $\tfrac13$. Thus the two outer quarters together have twice the probability of the central half, even though both regions have total length $\tfrac12$.

Check: The probabilities add to $\tfrac23+\tfrac13=1$, as the outer and central regions cover $[0,1]$ and share only boundary points. Those boundary points have probability zero under a density.

Endpoint Concentration Is Not an Endpoint Atom

The density tends to infinity near both endpoints: as $x$ decreases to zero, $f_U(x)$ behaves like $1/(\pi\sqrt{x})$; by symmetry the same behavior occurs near one. This divergence is integrable. It means that short intervals next to the endpoints have more probability than equally long intervals in the middle, not that the limiting variable equals an endpoint with positive probability.

For example, for $0<\varepsilon<\tfrac12$, symmetry and the cumulative distribution function give

\mathbb{P}\big(U\le\varepsilon\ \text{or}\ U\ge1-\varepsilon\big) =\frac{4}{\pi}\arcsin\sqrt{\varepsilon}.

As $\varepsilon$ decreases to zero, this probability also decreases to zero because $\arcsin\sqrt{\varepsilon}$ decreases to zero. Thus there is no point mass at either endpoint. The term “endpoint concentration” describes the relative weight close to the endpoints, not mass located exactly at them.

Worked Example: Probability Near the Endpoints

Find the limiting probability that the occupation fraction is at most $\tfrac14$ or at least $\tfrac34$, using strict inequality at neither boundary.

The lower-tail probability is $F_U(\tfrac14)=\tfrac{2}{\pi}\arcsin(\tfrac12)=\tfrac13$. By symmetry, the upper-tail probability is also $\tfrac13$. The events are disjoint because $\tfrac14<\tfrac34$, so the requested probability is $\tfrac13+\tfrac13=\tfrac23$.

Check: The complement is $\tfrac14<U<\tfrac34$, whose probability is $\tfrac13$ by continuity and the middle-interval calculation. The two probabilities sum to one. Changing either endpoint from included to excluded does not affect the answer, since $U$ has no atoms.

Why the Symmetric-Walk Assumption Matters

The arcsine limit is specific to the zero-drift simple symmetric walk. For a biased walk with independent steps satisfying $\mathbb{P}(\xi_k=1)=p$ and $\mathbb{P}(\xi_k=-1)=1-p$, where $p\ne\tfrac12$, the mean step is $2p-1$. The strong law of large numbers gives $S_n/n\to2p-1$ almost surely. If $p>\tfrac12$, the limit is positive, so eventually $S_n>0$ almost surely and $N_n/n\to1$ almost surely. If $p<\tfrac12$, eventually $S_n<0$ almost surely and $N_n/n\to0$ almost surely. For instance, at $p=\tfrac34$, the limiting positive occupation fraction is one, not an arcsine-distributed variable. This illustrates why zero drift and symmetry are essential hypotheses of the stated limit.

Common mistake. A tempting but incorrect argument is to say that symmetry makes each $S_k$ equally likely to be positive or negative, and therefore $N_n$ behaves like a sum of independent Bernoulli variables with success probability one half. Symmetry does give $\mathbb{P}(S_k>0)=\mathbb{P}(S_k<0)$, but the two events do not exhaust the possibilities when $S_k=0$, and the indicators at different times are not independent. For example, at $n=2$, the four step sequences $++,+-,-+,--$ each have probability $\tfrac14$. Their positive-occupation counts are respectively $2,1,0,0$, since a visit to zero is excluded. Thus $U_2$ has probabilities $\tfrac14,\tfrac14,\tfrac12$ at $1,\tfrac12,0$, respectively—not the distribution of the average of two independent fair Bernoulli variables. The limit theorem concerns the dependent occupation indicators of the walk, and its arcsine distribution reflects that dependence.

Check Your Understanding

Use the definition with strict positivity and the arcsine cumulative distribution function to answer these questions.

For the positive occupation count $N_n$, which times are included, and does a time with $S_k=0$ contribute?
Calculate $\mathbb{P}(U\le\tfrac1{16})$ from the cumulative distribution function.
Explain why a density that diverges at zero does not imply $\mathbb{P}(U=0)>0$.
Use the symmetry of the arcsine density to find $\mathbb{P}(U\ge\tfrac34)$.
For a walk with $p=\tfrac34$, what does the strong law imply about $N_n/n$, and why does this differ from the symmetric-walk limit?