# Showing $f(B_t)$ is a time-change of Brownian motion

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 $$0$$?

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

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 $$\tilde{B}$$).

• Also see Nate Eldredge's excellent answer here for a really slick approach to the first problem using Kolmogorov's $0$-$1$ law. Commented Dec 23, 2019 at 21:41
• 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?
– user621937
Commented Dec 23, 2019 at 22:05
• 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 $\int_s^\infty e^{B_u - B_s} du = \infty$. Commented Dec 23, 2019 at 22:25