# integral of exponential of Brownian motion

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.

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 $-1$.

• Is this so slick because of experience or is this a more general argument in disguise? +1.
– snar
Commented Apr 24, 2015 at 18:43
• @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. Commented Apr 24, 2015 at 19:45
• 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 Commented Mar 27, 2019 at 10:27
• Why $A$ is a tail event ? Commented Jun 2, 2019 at 11:09
• @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. 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.

• Yes, I have this intuition. But I don't know how to justify this rigorously.
– erik
Commented Apr 24, 2015 at 17:12
• @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... Commented Apr 24, 2015 at 17:14
• 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. Commented Apr 24, 2015 at 17:28
• @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? Commented Apr 24, 2015 at 17:31