Q) Let $f(B_t)$ be a complex Brownian motion in the complex plane. (This means that $B_t = B_t^1 + iB_t^2$ where $B_t^1$ and $B_t^2$ are independent Brownian motions.) Let $f(z):\mathbb{C} \to \mathbb{C}$ be the entire function $f(z) = e^z$. Show that $f(B_t)$ is a time-change of Brownian motion, and conclude that complex Brownian motion never hits $0$, almost surely.

At a higher level, there is a theorem that says if a continuous local martingale has quadratic variation $\uparrow\infty$, then it is a time change of Brownian motion.

$f(x+iy) = e^x\cos y+i\sin y$. Let $f_1(x+iy) = e^x\cos y$. Then:

$$\frac{\partial f_1}{\partial x} = e^x\cos y, \frac{\partial f_1}{\partial y} = -e^x\sin y, \frac{\partial^2 f_1}{\partial x^2} = e^x\cos y, \frac{\partial^2 f_1}{\partial y^2} = -e^x\cos y\implies \nabla f_1 = 0$$

where $\nabla$ is the Laplacian operator. Now, by Ito's formula it follows that:

$$f(B_t^1,B_t^2) = f_1(0,0)+\int_0^t e^{B_s^1}\cos B_s^2 dB_s^1 - \int_0^t e^{B_s^1}\sin B_s^2 dB_s^2 + \text{ terms related to }f_2$$

Thus the quadratic variation of the two integrals seen above is:

$$\int_0^t e^{2B_s^1}\cos^2 B_s^2 ds + \int_0^t e^{2B_s^1}\sin^2 B_s^2 ds = \int_0^t e^{2B_s^1} ds$$

1) So, how do I show that $\int_0^t e^{2B_s} ds \uparrow \infty$ as $t \uparrow \infty$?

2) Also, how does $f(B_t)$ is a time-change of Brownian motion imply that complex Brownian motion never hits $0$?

  • $\begingroup$ May I know where do you get this question from? Is it from a book? If yes, can you specify the book's title? $\endgroup$
    – Idonknow
    Commented Dec 24, 2019 at 0:07

1 Answer 1


For the first question, I will outline a proof of the fact that if $W_s$ is a real-valued Brownian motion then $C = \int_0^\infty e^{2W_s} ds = \infty$. This is sufficient since $\int_0^t e^{2W_s}ds$ is increasing in $t$.

The most simple first guess for how to do this is to bound $C \geq \int_0^\infty 1_{\{W_s \geq 0\}} ds$ and try to show that there almost surely exists an infinite sequence of times $t_n > 1$ such that $W(s) \geq 0$ for all $s \in [t_n - 1, t_n]$ (note that $t_n$ is a random variable here).

One can check that $T = \inf\{t > 1: W_u \geq 0 \text{ for all } u \in [t-1, t]\}$ is a stopping time for the Brownian filtration and so to see the existence of the sequence $t_n$ it suffices to show that $\mathbb{P}(T < \infty) = 1$ and apply the usual recursive argument using the strong Markov property.

Let $H_n = \inf\{t > 0: W_t = n\}$. Then, by the strong Markov property and symmetry, $$\mathbb{P}(W_t < 0 \text{ for some } t \in [H_n, H_n + 1]) = \mathbb{P}( H_n \leq 1).$$ By the reflection principle, $\mathbb{P}(H_n \leq 1) = 2 \mathbb{P}(W_1 \geq n) \to 0$ as $n \to \infty$ since $W_1$ is a standard Gaussian.

In particular, we can deduce that $$\mathbb{P}(\forall n, W_t < 0 \text{ for some } t \in [H_n, H_n + 1]) = 0$$ which in turn implies that $$\mathbb{P}(T < \infty) \geq \mathbb{P}( \exists n, W_t \geq 0 \text{ for all } t \in [H_n,H_n + 1]) = 1.$$

For your second question, if $f(B_s)$ is a time-change of the Brownian motion $\tilde{B}_s$ then the events $A = \{\exists s: f(B_s) = 0\}$ and $B = \{\exists u: \tilde{B}_u = 0\}$ are equal. Since $f(B_s) = e^{B_s}$, you know that $\mathbb{P}(A) = 0$ which implies that $\mathbb{P}(B) = 0$ which is the desired result (for the Brownian motion $\tilde{B}$).

  • $\begingroup$ Also see Nate Eldredge's excellent answer here for a really slick approach to the first problem using Kolmogorov's $0$-$1$ law. $\endgroup$ Commented Dec 23, 2019 at 21:41
  • $\begingroup$ Thanks, both arguments are great. May I know why $\{\int_{0}^{\infty}e^{B_t}<\infty\}$ in in the tail sigma-field. Is it because if we change the value of $B_t$ at finitely many places, the value of the integral does not change for any finite set? $\endgroup$
    – user621937
    Commented Dec 23, 2019 at 22:05
  • $\begingroup$ There is an argument for why this is true in the comments at the answer I link to. Essentially it's because for any fixed $s$, $\int_0^\infty e^{B_t} = \infty$ if and only if $\int_s^\infty e^{B_u - B_s} du = \infty$. $\endgroup$ Commented Dec 23, 2019 at 22:25

You must log in to answer this question.