I am currently reading a proof that uses the following fact without proof:

If $B$ is a scalar standard Brownian motion, then $\int_0^\infty e^{B_s} \,ds = + \infty$ a.s..

How can we justify this fact? I don't see how this follows from any property of Brownian motion.


2 Answers 2


Let $A$ be the event that $\int_0^\infty e^{B_t} \,dt < +\infty$. By the Kolomorgov 0-1 law, $P(A) = 0$ or $1$.

Now let $B$ be the event that $\int_0^\infty e^{-B_t}\,dt < +\infty$. By symmetry, $P(A) = P(B)$. Moreover, on $A \cap B$ we have $$\int_0^\infty (e^{B_t} + e^{-B_t}) \,dt < +\infty$$ which is absurd since $e^x + e^{-x} \ge 1$ for all $x$. So $A \cap B = \emptyset$. This makes it impossible that $P(A) = P(B) = 1$, so we must have $P(A) = 0$.

Another approach (more involved) is to use recurrence and the strong Markov property to show that almost surely, there are infinitely many disjoint intervals of length 1 on which $B_t$ stays above $-1$.

  • $\begingroup$ Is this so slick because of experience or is this a more general argument in disguise? +1. $\endgroup$
    – snar
    Commented Apr 24, 2015 at 18:43
  • $\begingroup$ @snarski: Well, the Kolomogorov 0-1 law is a powerful tool, so maybe that's the "more general argument". And it's not unusual to take advantage of symmetry when using it. But it's not specifically a special case of any particular general fact, that I know of. $\endgroup$ Commented Apr 24, 2015 at 19:45
  • $\begingroup$ Nice approach! (+1). Would you mind looking at my question which is about the second approach you have mentioned? math.stackexchange.com/q/3164293/349501 $\endgroup$
    – Shashi
    Commented Mar 27, 2019 at 10:27
  • $\begingroup$ Why $A$ is a tail event ? $\endgroup$
    – user657324
    Commented Jun 2, 2019 at 11:09
  • $\begingroup$ @user657324: For arbitrary $s > 0$, write $\int_0^\infty e^{B_t} dt = \int_0^s e^{B_t}\,dt + e^{B_s} \int_s^\infty e^{B_t - B_s}\,dt$. Now the first integral is certainly finite, so this random variable is finite iff $I_s := \int_s^\infty e^{B_t - B_s}\,dt$ is finite, and $I_s$ is independent of $\sigma(B_t : 0 \le t \le s)$, hence so is $A$. So it's a tail event. If this is not the version of Kolmogorov 0-1 you like, then complete the proof by hand: since $s$ was arbitrary, $A$ is independent of $\sigma(B_t : t \ge 0)$ and also in this $\sigma$-field. $\endgroup$ Commented Jun 2, 2019 at 13:55

(Not a proof, but here's some intuition at least)

Standard Brownian motion has mean 0 and variance t for $0 \le t \lt \infty$ Thus, on average the integral becomes... $$\int_0^{\infty} e^0 \ ds=\int_0^{\infty} 1 \ ds$$ Which clearly diverges. What you should aim for proving is that Brownian motion crosses between negative and positive values in such a fashion that as the limit of the integral increases to infinity, the number of reversals also increases to infinity with a mean period between oscillations that isn't 0.

  • $\begingroup$ Yes, I have this intuition. But I don't know how to justify this rigorously. $\endgroup$
    – erik
    Commented Apr 24, 2015 at 17:12
  • $\begingroup$ @Erik I gave a method of attack. Prove that the motion crosses the x axis an infinite number of times. I can't guarantee that's true, but it would work... $\endgroup$
    – Zach466920
    Commented Apr 24, 2015 at 17:14
  • $\begingroup$ You need more than that. This would tell you that $\limsup_{t \to +\infty} e^{B_t} \ge 1$ almost surely, but that doesn't imply that the integral is infinite. You have to exclude the possibility that there could be infinitely many spikes whose total area is finite. $\endgroup$ Commented Apr 24, 2015 at 17:28
  • $\begingroup$ @NateEldredge To fix that I guess you'd have to go about proving that the oscillations have finite period non-zero period. Would that do the trick? $\endgroup$
    – Zach466920
    Commented Apr 24, 2015 at 17:31

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