Why Brownian First-Passage Times Have Infinite Mean

The First-Passage Time and Its Tail

Let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ be a filtered probability space carrying a standard Brownian motion $B_t$, adapted to $(\mathcal{F}_t)$. Fix a level $a>0$, and define the first-passage time $\tau_a=\inf\{t\ge0:B_t=a\}$, with $\inf\varnothing=\infty$. The question is not whether the level can be reached at all, but how long it takes on average. The answer is striking: $\tau_a$ is finite almost surely, yet its expectation is infinite.

For $t>0$, continuity of Brownian paths gives $\{\tau_a\le t\}=\{\sup_{0\le s\le t}B_s\ge a\}$. The reflection principle for Brownian motion states that, for $a>0$, the probability on the right equals $2\mathbb{P}(B_t\ge a)$. Since $B_t$ has distribution $N(0,t)$, write $\Phi$ for the standard normal cumulative distribution function to obtain the distribution and survival tail below. The tail concerns the strict event $\{\tau_a>t\}$; the distribution has no atom at any $t>0$, so replacing $\le t$ by $<t$ in the distribution formula would not change its value.

Result: For $a>0$ and $t>0$, $$ \mathbb{P}(\tau_a\le t)=2\left(1-\Phi\left(\frac{a}{\sqrt{t}}\right)\right), \qquad \mathbb{P}(\tau_a>t)=2\Phi\left(\frac{a}{\sqrt{t}}\right)-1. $$

As $t$ increases to infinity, $a/\sqrt{t}$ decreases to zero, and the survival probability tends to zero. Thus $\mathbb{P}(\tau_a=\infty)=0$. This almost-sure finiteness does not imply a finite expectation: rare passages at very large times can dominate the average.

Worked Example: Tail and Density at the Level's Natural Time Scale

Take $a=2$ and $t=4=a^2$. Substitution into the tail formula gives

\mathbb{P}(\tau_2>4)=2\Phi\left(\frac{2}{\sqrt{4}}\right)-1 =2\Phi(1)-1\approx 0.6827.

Differentiating the distribution function, as shown below, gives the density at $t=4$: $$ f_2(4)=\frac{2}{\sqrt{2\pi}\,4^{3/2}}\exp\left(-\frac{4}{2\cdot4}\right) =\frac{e^{-1/2}}{4\sqrt{2\pi}}\approx 0.0605. $$

Check: the tail formula at $t=a^2$ always reduces to $2\Phi(1)-1$, because $a/\sqrt{a^2}=1$ for $a>0$. For the density, $4^{3/2}=8$, so the original expression is $2e^{-1/2}/(8\sqrt{2\pi})$, agreeing with the simplified value.

Density and Its Long-Time Behavior

The standard normal density is $\phi(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$, and $\Phi'(x)=\phi(x)$. Differentiating the distribution function for $t>0$, using the chain rule on $a/\sqrt{t}$, yields the density.

Definition: The first-passage time $\tau_a$ has density $$ f_a(t)=\frac{a}{\sqrt{2\pi}\,t^{3/2}} \exp\left(-\frac{a^2}{2t}\right),\qquad t>0. $$

Indeed, if $F(t)=2(1-\Phi(a t^{-1/2}))$, then $$ F'(t)=-2\phi(a t^{-1/2})\left(-\frac{a}{2}t^{-3/2}\right) =\frac{a}{\sqrt{2\pi}\,t^{3/2}}e^{-a^2/(2t)}. $$ The density is nonnegative. Since $F(t)$ tends to zero as $t$ decreases to zero and to one as $t$ increases to infinity, integrating $F'$ over $(0,\infty)$ gives total mass one. These endpoint limits also show that the distribution has no mass at zero or at infinity.

For large $t$, the tail decays like $t^{-1/2}$, not exponentially. To see its exact leading order, use $\Phi(z)-\tfrac12=\int_0^z\phi(u)\,du$ for $z>0$. Continuity of $\phi$ at zero implies $\lim_{z\downarrow0}z^{-1}\int_0^z\phi(u)\,du=\phi(0)=1/\sqrt{2\pi}$. Consequently,

\lim_{t\to\infty}\sqrt{t}\,\mathbb{P}(\tau_a>t) =\lim_{t\to\infty}2a\, \frac{\Phi(a/\sqrt{t})-\tfrac12}{a/\sqrt{t}} =a\sqrt{\frac{2}{\pi}}.

In particular, the tail is bounded above and below by positive constant multiples of $t^{-1/2}$ for all sufficiently large $t$. This comparison is the key to the moment threshold.

Exactly Which Positive Moments Are Finite?

Theorem: Fix $a>0$. For every $p>0$, $\mathbb{E}[\tau_a^p]<\infty$ if and only if $p<\tfrac12$. In particular, the first moment is infinite. For $0<p<\tfrac12$, $$ \mathbb{E}[\tau_a^p] =\frac{a^{2p}}{2^p\sqrt{\pi}}\, \Gamma\left(\frac12-p\right), $$ where $\Gamma(q)=\int_0^\infty u^{q-1}e^{-u}\,du$ for $q>0$.

Proof. Since $\tau_a\ge0$, the identity $x^p=\int_0^\infty p t^{p-1}\mathbf{1}_{\{x>t\}}\,dt$ holds for $x\in[0,\infty]$ and $p>0$. Applying Tonelli's theorem to this nonnegative integrand gives $$ \mathbb{E}[\tau_a^p] =p\int_0^\infty t^{p-1}\mathbb{P}(\tau_a>t)\,dt. $$ The part over $0<t\le1$ is finite, because the tail is at most one and $\int_0^1p t^{p-1}\,dt=1$. For large $t$, the tail is bounded above and below by positive constant multiples of $t^{-1/2}$, as proved above. Therefore the integral over large $t$ is finite exactly when $\int_1^\infty t^{p-3/2}\,dt$ is finite. A power integral over $[1,\infty)$ converges exactly when its exponent is less than $-1$, which here is equivalent to $p<\tfrac12$. This proves both directions of the finiteness criterion, including divergence at $p=\tfrac12$. In particular, $p=1$ gives $\mathbb{E}[\tau_a]=\infty$.

It remains to compute the value when $0<p<\tfrac12$. Use the density and substitute $u=a^2/(2t)$, so $t=a^2/(2u)$ and $dt=-a^2/(2u^2)\,du$. Reversing the integration limits gives $$ \mathbb{E}[\tau_a^p] =\frac{a}{\sqrt{2\pi}}\int_0^\infty t^{p-3/2}e^{-a^2/(2t)}\,dt =\frac{a^{2p}}{2^p\sqrt{\pi}} \int_0^\infty u^{-p-1/2}e^{-u}\,du. $$ The last integral is $\Gamma(\tfrac12-p)$, whose defining integral is finite precisely for $p<\tfrac12$. The calculation agrees with the tail criterion and completes the proof. $\square$

The scaling is also visible directly in the distribution: for $t>0$, $\mathbb{P}(\tau_a>a^2t)=2\Phi(1/\sqrt{t})-1=\mathbb{P}(\tau_1>t)$. Thus $\tau_a$ has the same distribution as $a^2\tau_1$. A larger level stretches passage times by the square of the level, but it does not change which moments are finite.

Worked Example: A Finite Fractional Moment

Compute the quarter moment for a passage to level $a=4$. Since $p=\tfrac14<\tfrac12$, the theorem applies: $$ \mathbb{E}[\tau_4^{1/4}] =\frac{4^{1/2}}{2^{1/4}\sqrt{\pi}}\Gamma\left(\frac14\right) =\frac{2}{2^{1/4}\sqrt{\pi}}\Gamma\left(\frac14\right). $$ This is finite because the gamma integral has exponent $u^{-3/4}$ near zero, which is integrable, and exponential decay at infinity.

Check: scaling gives $\tau_4$ the same distribution as $16\tau_1$, so its quarter moment must be $16^{1/4}\mathbb{E}[\tau_1^{1/4}]=2\mathbb{E}[\tau_1^{1/4}]$. Substituting $a=1$ into the moment formula gives $\mathbb{E}[\tau_1^{1/4}]=2^{-1/4}\Gamma(1/4)/\sqrt{\pi}$, and multiplying by $2$ reproduces the result.

Worked Example: The Half Moment Is Already Infinite

For $a=1$ and $p=\tfrac12$, the tail-integral formula becomes $$ \mathbb{E}[\tau_1^{1/2}] =\frac12\int_0^\infty t^{-1/2}\mathbb{P}(\tau_1>t)\,dt. $$ The tail limit gives $\sqrt{t}\,\mathbb{P}(\tau_1>t)\to\sqrt{2/\pi}>0$. Hence, for all sufficiently large $t$, the integrand is bounded below by a positive constant times $t^{-1}$. Since $\int_1^\infty t^{-1}\,dt=\infty$, the half moment diverges.

Check: the density calculation gives the same conclusion: the integrand for this moment is $t^{1/2}f_1(t)=\frac{1}{\sqrt{2\pi}}t^{-1}e^{-1/(2t)}$. For $t\ge1$, $e^{-1/(2t)}\ge e^{-1/2}$, so its integral dominates a positive constant times $\int_1^\infty t^{-1}\,dt$, which diverges.

Truncation and a Common Pitfall

Although the untruncated mean is infinite, every truncated value $\min(\tau_a,R)$, with $R>0$, is bounded and has finite expectation. For a nonnegative random variable, writing $\min(\tau_a,R)=\int_0^R\mathbf{1}_{\{\tau_a>t\}}\,dt$ and applying Tonelli gives $$ \mathbb{E}[\min(\tau_a,R)] =\int_0^R\mathbb{P}(\tau_a>t)\,dt. $$ As $R$ grows, these expectations grow without bound. For example, with $a=1$ and $t\ge1$, set $z=1/\sqrt{t}\le1$. The normal density $\phi$ is decreasing on $[0,1]$, so $$ \mathbb{P}(\tau_1>t)=2\int_0^{1/\sqrt{t}}\phi(u)\,du \ge \frac{2\phi(1)}{\sqrt{t}}. $$

Worked Example: A Lower Bound for a Truncated Mean

For $R\ge1$, integrate the preceding bound over $1\le t\le R$: $$ \mathbb{E}[\min(\tau_1,R)] \ge \int_1^R\frac{2\phi(1)}{\sqrt{t}}\,dt =4\phi(1)(\sqrt{R}-1). $$ At $R=4$, this gives the explicit lower bound $4\phi(1)=4e^{-1/2}/\sqrt{2\pi}$ for $\mathbb{E}[\min(\tau_1,4)]$. As $R$ increases, the bound tends to infinity, so the truncated expectations cannot stay bounded.

Check: integrating $t^{-1/2}$ from $1$ to $4$ gives $2(\sqrt{4}-\sqrt{1})=2$. Multiplication by the bound's coefficient $2\phi(1)$ gives $4\phi(1)$, as stated. The bound is zero at $R=1$, consistent with integrating over an interval of zero length.

Common mistake: The density contains an exponential factor, so it is tempting to claim that the mean integral is made finite by exponential decay. The incorrect calculation treats $e^{-a^2/(2t)}$ as though it decays to zero as $t\to\infty$. In fact, it tends to one. For $t\ge a^2$, $e^{-a^2/(2t)}\ge e^{-1/2}$, and therefore $$ t f_a(t)=\frac{a}{\sqrt{2\pi}}t^{-1/2}e^{-a^2/(2t)} \ge \frac{a e^{-1/2}}{\sqrt{2\pi}}t^{-1/2}. $$ The integral of the right-hand side over $[a^2,\infty)$ diverges. The tail formula gives the same warning: a $t^{-1/2}$ survival tail is too heavy for a finite first moment.

Check Your Understanding

Use the tail, density, and moment criterion to answer each question. In moment questions, assume $a>0$ and $p>0$.

Write $\mathbb{P}(\tau_a>t)$ and $f_a(t)$ for $t>0$. Which standard Brownian result gives the tail?
For which values of $p$ is $\mathbb{E}[\tau_a^p]$ finite? Explain why the endpoint $p=\tfrac12$ is excluded.
Use scaling to relate the distributions of $\tau_3$ and $\tau_1$, and state the resulting relation between their finite $p$-th moments.
Explain why $\mathbb{P}(\tau_a=\infty)=0$ does not imply that $\mathbb{E}[\tau_a]$ is finite.
For $a=1$, give a lower bound proportional to $\sqrt{R}-1$ for $\mathbb{E}[\min(\tau_1,R)]$ when $R\ge1$.