The Arcsine Limit for the Last Zero

From Exact Probabilities to a Limit Law

A random walk can return to the origin many times, but its final return before a fixed time has a distribution of its own. For a simple symmetric random walk, that distribution has an exact formula involving two central binomial coefficients. The formula also explains an important feature of the limit: probabilities at interior times are of order \(1/m\), while probabilities at the very endpoints are of order \(1/\sqrt m\).

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 discrete-time walk on state space \(S=\mathbb{Z}\) by \(X_0=0\) and \(X_n=\sum_{\ell=1}^n\xi_\ell\). Its natural filtration is \(\mathcal{F}_n=\sigma(\xi_1,\ldots,\xi_n)\), with \(\mathcal{F}_0=\{\varnothing,\Omega\}\). Since the walk changes parity at every step, its zeros occur only at even times.

Write \(a_r=\binom{2r}{r}/4^r\) for \(r\ge0\). This is the probability that the walk is at zero at time \(2r\): among the \(2^{2r}\) equally likely increment sequences, exactly \(\binom{2r}{r}\) have \(r\) increments of each sign. The probability of no return during a further \(2r\) steps turns out to be the same number \(a_r\). Combining these two factors gives the exact law.

Proof. Let \(T_0^+=\inf\{n\ge1:X_n=0\}\) for a walk started at zero. A first return can occur only at an even time. For \(j\ge1\), a first return at time \(2j\) has probability \(a_{j-1}/(2j)\). To see the count, first-step symmetry gives two choices of initial sign. Conditional on the initial sign being positive, the path must take its first step up, its last step down, and stay strictly positive between them. Removing the first and last steps and lowering the intervening path by one gives a nonnegative bridge of length \(2j-2\), counted by the Catalan number \(C_{j-1}=\binom{2j-2}{j-1}/j\). The total number of first-return paths is \(2C_{j-1}\), so their probability is \(2C_{j-1}/4^j=a_{j-1}/(2j)\). Also \(a_j=a_{j-1}(2j-1)/(2j)\), hence \(a_{j-1}/(2j)=a_{j-1}-a_j\). Therefore, for \(r\ge0\), the probability of no return in steps \(1,\ldots,2r\) is $$ \mathbb{P}(T_0^+>2r)=1-\sum_{j=1}^r(a_{j-1}-a_j)=a_r, $$ where for \(r=0\) the sum is empty and the probability is \(1=a_0\). Now fix \(k\in\{0,\ldots,m\}\). The event \(L_{2m}=2k\) requires \(X_{2k}=0\) and no return to zero during the next \(2(m-k)\) steps. The increments after time \(2k\) are independent of \(\mathcal{F}_{2k}\), so these events have joint probability \(a_k a_{m-k}\). When \(k=m\), the second requirement concerns zero steps and has probability \(a_0=1\), so the same formula applies. \(\square\)

Worked Distributions

The Arcsine Limit

To describe the limit, scale the last-zero time by the observation horizon. The random variable \(L_{2m}/(2m)\) takes values \(k/m\), since \(L_{2m}=2k\). The candidate limiting distribution has density \(f(x)=1/(\pi\sqrt{x(1-x)})\) for \(0<x<1\), and zero outside that interval. This is a probability density: using \(x=\sin^2\theta\), its integral from \(0\) to \(1\) is \((2/\pi)\int_0^{\pi/2}d\theta=1\). Its distribution function on \([0,1]\) is \(F(x)=(2/\pi)\arcsin\sqrt{x}\).

The interior approximation follows from the central binomial coefficient asymptotic. Stirling’s formula, a standard asymptotic result whose proof is not given here, yields $$ a_r=\frac{\binom{2r}{r}}{4^r}\sim\frac{1}{\sqrt{\pi r}}\qquad (r\to\infty). $$ In particular, there is a finite constant \(C\) such that \(a_r\le C/\sqrt{r+1}\) for all \(r\ge0\): the asymptotic gives such a bound for all sufficiently large \(r\), and increasing \(C\) covers the finitely many remaining values. When \(k/m=x\) stays away from \(0\) and \(1\), both indices \(k\) and \(m-k\) grow, so $$ \mathbb{P}(L_{2m}=2k)=a_k a_{m-k} \sim\frac{1}{\pi\sqrt{k(m-k)}} =\frac{1}{m}\frac{1}{\pi\sqrt{x(1-x)}}. $$ Thus, at interior grid points, the probability mass is approximately the density times the grid spacing \(1/m\).

Proof. We verify convergence by testing against any bounded continuous function \(g\) on \([0,1]\), which is the definition of convergence in distribution on this interval. Fix \(0<\delta<1/2\). For indices with \(\delta\le k/m\le1-\delta\), both \(k\) and \(m-k\) are at least \(\delta m\). The asymptotic for \(a_r\) therefore holds uniformly over these indices as \(m\to\infty\): for any prescribed error tolerance, the defining asymptotic bound holds for every \(r\) above a fixed threshold, and \(\delta m\) eventually exceeds that threshold. Consequently the contribution from these indices to \(\mathbb{E}[g(L_{2m}/(2m))]\) converges, by Riemann sums with mesh \(1/m\), to $$ \int_\delta^{1-\delta}\frac{g(x)}{\pi\sqrt{x(1-x)}}\,dx. $$ It remains to control both endpoint ranges. For \(k\le\delta m\), using \(m-k\ge m/2\) and the bound on \(a_r\) gives $$ \sum_{k=0}^{\lfloor\delta m\rfloor}a_k a_{m-k} \le \frac{C\sqrt{2}}{\sqrt m}\sum_{k=0}^{\lfloor\delta m\rfloor}\frac{C}{\sqrt{k+1}} \le \frac{2\sqrt{2}C^2}{\sqrt m}\sqrt{\lfloor\delta m\rfloor+1}. $$ The last inequality follows by comparing the sum of \((k+1)^{-1/2}\) with its integral, or by summing the bound \(\sum_{j=1}^N j^{-1/2}\le2\sqrt N\). Its limit superior is at most \(2\sqrt{2}C^2\sqrt{\delta}\). By symmetry, the same bound controls the range \(k\ge(1-\delta)m\). Multiplying these bounds by \(\sup_{x\in[0,1]}|g(x)|\) controls the absolute endpoint contributions to the expectation. The limiting density has endpoint integrals tending to zero as \(\delta\downarrow0\), since \(1/\sqrt{x(1-x)}\) is integrable at both endpoints. First letting \(m\to\infty\), then letting \(\delta\downarrow0\), proves $$ \mathbb{E}\!\left[g\!\left(\frac{L_{2m}}{2m}\right)\right] \longrightarrow \int_0^1\frac{g(x)}{\pi\sqrt{x(1-x)}}\,dx. $$ This is the asserted convergence in distribution. \(\square\)

Why the Endpoints Need Separate Attention

The interior approximation is a local statement: it applies when \(k\) and \(m-k\) both grow proportionally to \(m\). At the left endpoint, the exact mass is \(\mathbb{P}(L_{2m}=0)=a_m\), which is asymptotic to \(1/\sqrt{\pi m}\). At a fixed interior location \(k/m\approx x\), the mass is instead of order \(1/m\). There is no contradiction with the limit density: the endpoint is a single grid point whose mass tends to zero, while an interval containing many grid points accumulates mass according to the integrable density. The endpoint estimate in the proof controls a whole shrinking region, not just the probability at \(k=0\).

Common mistake. It is incorrect to say that \(\mathbb{P}(L_{2m}=2k)=a_k\), using only the probability of being at zero at time \(2k\). That counts paths that may return to zero later. For \(m=2\), this error assigns probabilities \(1,1/2,3/8\) to \(L_4=0,2,4\), respectively; their sum is \(1+1/2+3/8=15/8\), so they cannot form a probability distribution. The correction is to include the probability of no later return: multiply by \(a_{m-k}\). The resulting values \(3/8,1/4,3/8\) sum to \(1\).