The Arcsine Law for the Last Brownian Zero

The Last Zero Before a Fixed Time

Brownian motion crosses zero repeatedly, so its last zero before a fixed time is a random location within the time interval, not a deterministic fraction of the horizon. Its distribution is the classical arcsine law: it places comparatively high probability near the beginning and end of the interval, and less near the midpoint. The derivation connects the last zero to the probability that Brownian motion, once observed at an intermediate time, avoids zero for the rest of the interval.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space carrying a standard Brownian motion $(B_s)_{s\ge0}$, with state space $\mathbb{R}$, continuous paths, $B_0=0$, and independent Gaussian increments. Fix $t>0$, and define the last zero before time $t$ by $g_t=\sup\{s\in[0,t]:B_s=0\}$. The set in this supremum is nonempty because it contains $0$; it is closed by continuity, so its supremum is attained. We derive the law of $g_t$.

Definition: The cumulative distribution function of $g_t$ is $F_t(u)=\mathbb{P}(g_t\le u)$. For $0\le u\le t$, the event $g_t\le u$ means that there is no zero of the path in $(u,t]$. The interval is open at $u$: a zero at $u$ is permitted.

From Avoiding Zero to the Distribution Function

First recall the Brownian survival probability that will be used. If Brownian motion starts at $x>0$, the reflection principle gives the probability of hitting zero by elapsed time $v>0$ as $2\Phi(-x/\sqrt v)$, where $\Phi$ is the standard normal cumulative distribution function. Therefore the probability of not hitting zero by time $v$ is $1-2\Phi(-x/\sqrt v)=2\Phi(x/\sqrt v)-1$. For $x<0$, symmetry gives the same expression with $|x|$ in place of $x$. At $x=0$, the probability of avoiding zero throughout a positive time interval is $0$, also the value given by the formula.

Now take $0<u<t$. The increments after the deterministic time $u$ are independent of $B_u$. Conditional on $B_u=x$, the path has no zero in $(u,t]$ exactly when the Brownian motion started at $x$ avoids zero for the next $t-u$ units of time. Since $B_u$ has the same distribution as $\sqrt{u}Z$, where $Z$ is standard normal, it follows that

F_t(u)=\mathbb{E}\left[2\Phi\left(\sqrt{\frac{u}{t-u}}\,|Z|\right)-1\right].

This conditioning is at a fixed time, so only the Gaussian distribution of $B_u$, independent increments, and the reflection-principle survival formula are needed. In particular, no Markov property at a random time is being used.

Theorem (Last-zero arcsine law): For $0\le u\le t$, $$ F_t(u)=\mathbb{P}(g_t\le u)=\frac{2}{\pi}\arcsin\sqrt{\frac{u}{t}}. $$ Consequently $g_t$ has density $$ f_{g_t}(u)=\frac{1}{\pi\sqrt{u(t-u)}},\qquad 0<u<t, $$ and has no atoms at $0$, $t$, or any interior time.

Proof. For $a\ge0$, set $H(a)=\mathbb{E}[2\Phi(a|Z|)-1]$. We have $H(0)=0$. Differentiation under the expectation is valid: the absolute value of the derivative of the integrand is $2|Z|\phi(a|Z|)\le 2|Z|/\sqrt{2\pi}$, which has finite expectation. Using the standard normal density $\phi(z)=(2\pi)^{-1/2}e^{-z^2/2}$ and symmetry of $Z$, for $a\ge0$ we obtain

\begin{aligned} H'(a) &=4\int_0^\infty z\phi(z)\phi(az)\,dz\\ &=\frac{4}{2\pi}\int_0^\infty z\exp\left(-\frac{1+a^2}{2}z^2\right)\,dz\\ &=\frac{2}{\pi}\cdot\frac{1}{1+a^2}. \end{aligned}

For the last equality, substitute $y=(1+a^2)z^2/2$; then $z\,dz=dy/(1+a^2)$ and $\int_0^\infty e^{-y}\,dy=1$. Integrating from $0$ to $a$ gives $H(a)=(2/\pi)\arctan a$. In the expression for $F_t(u)$, take $a=\sqrt{u/(t-u)}$. If $\theta=\arcsin\sqrt{u/t}$, then $0<\theta<\pi/2$, $\sin\theta=\sqrt{u/t}$, and $\cos\theta=\sqrt{(t-u)/t}$. Hence $\tan\theta=a$, so $\arctan a=\arcsin\sqrt{u/t}$. This proves the stated formula for $0<u<t$. Its limits at $u=0$ and $u=t$ are $0$ and $1$, respectively, matching $F_t(0)=0$ and $F_t(t)=1$. Thus $F_t$ is continuous on $[0,t]$, which also shows there are no atoms at the endpoints or at any interior point. Differentiating the formula on $(0,t)$ yields the stated density. $\square$

The density integrates to one, as required. Indeed, with $u=t\sin^2\theta$, $0<\theta<\pi/2$, we have $du=2t\sin\theta\cos\theta\,d\theta$ and $\sqrt{u(t-u)}=t\sin\theta\cos\theta$. Therefore

\int_0^t\frac{du}{\pi\sqrt{u(t-u)}} =\frac{2}{\pi}\int_0^{\pi/2}d\theta =1.

Computing Probabilities for Time Ranges

For $0\le a<b\le t$, continuity of the distribution means that including or excluding either endpoint does not change the probability. Subtracting the cumulative distribution values gives the convenient interval formula

\mathbb{P}(a\le g_t\le b) =\frac{2}{\pi}\left(\arcsin\sqrt{\frac{b}{t}}-\arcsin\sqrt{\frac{a}{t}}\right).

Worked Example: The First and Last Quarters

For a fixed $t>0$, find the probabilities that $g_t$ lies in the first quarter, the middle half, and the final quarter of $[0,t]$. The ranges are $[0,t/4]$, $[t/4,3t/4]$, and $[3t/4,t]$, respectively.

For the first quarter, $\sqrt{(t/4)/t}=1/2$, so

\mathbb{P}(0\le g_t\le t/4) =F_t(t/4) =\frac{2}{\pi}\arcsin(1/2) =\frac{2}{\pi}\cdot\frac{\pi}{6} =\frac13.

For the upper endpoint of the middle half, $\sqrt{(3t/4)/t}=\sqrt{3}/2$, and $\arcsin(\sqrt{3}/2)=\pi/3$. Thus

\mathbb{P}(t/4\le g_t\le3t/4) =F_t(3t/4)-F_t(t/4) =\frac{2}{\pi}\left(\frac{\pi}{3}-\frac{\pi}{6}\right) =\frac13.

For the final quarter, $F_t(t)=1$, so $\mathbb{P}(3t/4\le g_t\le t)=1-F_t(3t/4)=1-2/3=1/3$. As a check, the three disjoint intervals obtained by assigning their shared boundaries to just one interval cover $[0,t]$; their probabilities sum to $1/3+1/3+1/3=1$. Shared endpoints have probability zero.

Worked Example: A Range on a Nine-Unit Time Horizon

Take $t=9$. Find the probability that $g_9$ lies between time $1$ and time $9/4$, including both endpoints.

At $u=9/4$, the ratio $u/t=(9/4)/9=1/4$, so $F_9(9/4)=(2/\pi)\arcsin(1/2)=1/3$. At $u=1$, the ratio $u/t=1/9$, so $F_9(1)=(2/\pi)\arcsin(1/3)$. Subtracting gives

\mathbb{P}(1\le g_9\le9/4) =\frac13-\frac{2}{\pi}\arcsin\left(\frac13\right).

As a check, the density gives the same answer: $\int_1^{9/4}du/(\pi\sqrt{u(9-u)})$. Under $u=9\sin^2\theta$, the lower endpoint corresponds to $\theta=\arcsin(1/3)$, and the upper endpoint to $\theta=\arcsin(1/2)=\pi/6$. The integral becomes $(2/\pi)(\pi/6-\arcsin(1/3))$, which is exactly the expression above.

Worked Example: A Symmetric Interior Range

For any $t>0$, compute the probability that $g_t$ lies in $[t/9,4t/9]$, and check the answer by integrating the density.

At the lower endpoint, $\sqrt{(t/9)/t}=1/3$. At the upper endpoint, $\sqrt{(4t/9)/t}=2/3$. Therefore the cumulative distribution formula gives

\mathbb{P}(t/9\le g_t\le4t/9) =\frac{2}{\pi}\left[ \arcsin\left(\frac23\right)-\arcsin\left(\frac13\right) \right].

For the density check, use $u=t\sin^2\theta$. The two endpoints correspond to $\theta=\arcsin(1/3)$ and $\theta=\arcsin(2/3)$, both in $(0,\pi/2)$. Since $du/\sqrt{u(t-u)}=2\,d\theta$, the integral of the density over this interval is $\frac{2}{\pi}[\arcsin(2/3)-\arcsin(1/3)]$, agreeing with the calculation from the distribution function.

Shape of the Law and a Common Mistake

The density diverges to $+\infty$ as $u$ approaches either endpoint $0$ or $t$, but these divergences are integrable, as the normalization calculation shows. Its shape is not monotone across the entire interval. Differentiating for $0<u<t$ gives

f'_{g_t}(u)=\frac{2u-t}{2\pi[u(t-u)]^{3/2}}.

The denominator is positive. The numerator is negative for $0<u<t/2$ and positive for $t/2<u<t$. Thus the density decreases on $(0,t/2)$, increases on $(t/2,t)$, and is smallest at the midpoint, where $f_{g_t}(t/2)=2/(\pi t)$. The distribution is symmetric about $t/2$: the density has the same value at $u$ and $t-u$. This symmetry does not make the distribution uniform.

Common mistake. Treating $g_t$ as uniform would give $\mathbb{P}(g_t\le t/4)=1/4$. The correct calculation is $\mathbb{P}(g_t\le t/4)=(2/\pi)\arcsin\sqrt{1/4}=(2/\pi)(\pi/6)=1/3$. The last zero is more likely to fall in the first quarter than a uniform location would be. Likewise, the density does not decrease as $u$ approaches $t$: it increases on the second half and diverges at that endpoint.

The arcsine law is useful whenever a Brownian path is summarized by when it last visits a level, rather than by its value at the final time. The key steps here were specific: express a last-zero event as avoidance of zero after a deterministic time, condition on the position at that time, and use a survival probability obtained from reflection. The resulting density gives exact probabilities for any specified portion of the time horizon.

Check Your Understanding

Use the distribution function or density, and state whether endpoints affect each probability.

For $t>0$, compute $\mathbb{P}(g_t\le t/16)$.
Find $\mathbb{P}(t/2\le g_t\le t)$, and explain how symmetry provides a check.
For $t=4$, calculate the probability that $g_4$ lies in $[1,3]$.
Why does the density tending to infinity at both endpoints not prevent it from integrating to one?
State precisely on which subintervals the density decreases and increases.

The Arcsine Law for the Last Brownian Zero

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