Changing the time in Brownian motion with a function?

Question

Let $(B_t)$ be a Brownian motion on some probability space. Let $f:\mathbb R \rightarrow \mathbb R^+$ be a continuous function different from the identity. What is "$B_{f(t)}$" ? Clearly it is not a Brownian motion anymore. How can we rigorously define it and show that it exists (or not)?

What is there to know about "$B_{f(t)}$" ? Does it even make sense if $f$ is not monotone ? Can we determine its distribution ? Can we write it as $B_{f(t)}=g(B_t)$ for some function $g$ ? How can we simulate one starting from $B_t$ ?

Answer 1

I am merely a beginner on that subject, so this is by no means a comprehensive answer, but I'll still attempt to gather some elements that you may find useful :

• First off, $B_{f(t)}$ as you said will not be a Brownian Motion, but as long as $f(t)$ is non-negative, the random variable $B_{f(t)}$ is still well-defined and we can still manipulate it as usual. In other words, the family of random variables $\{Y_t := B_{f(t)}, t\in\mathbb R \}$ defines a stochastic process on the image of $\mathbb R$ by $f$. It could be reindexed as $\{Y_T := B_T, T\in f(\mathbb R)\}$, and for all $T \in f(\mathbb R)$ we would have that $Y_T$ is a Gaussian random variable with mean $0$ and variance $T$.
  Supposing that $f$ is continuous, we know that the image is some interval $f(\mathbb R) = [t_1,t_2]$ with $t_1,t_2 \ge 0$ and possibly $t_2 = \infty$. So essentially what you'd be looking at is a Brownian motion "truncated" to the interval $[t_1,t_2]$.

• As I said, $B_{f(t)}$ only needs $f$ to be non-negative for it to be well-defined. By definition of $B$, we thus have that $B_{f(t)}$ is normal, centered and of variance $f(t)$ for all $t$. It thus follows that we can compute the process' mean $\mathbb E[B_{f(t)}] = 0$ and covariance function $\mathbb E \left[B_{f(t)}B_{f(s)}\right] = f(t)\wedge f(s)$. Since the original Brownian motion is a Gaussian process, it follows that the time-changed Brownian motion also is a Gaussian process, and we can then deduce its distribution from its mean and covariance function.

• There are two well-known examples (that I can think of) of time-changing of Brownian motions which yield remarkable results :

1. If $c > 0$, then $B_{ct} \over \sqrt c$ is also a Brownian motion. That means that $B_{ct}$ is a Brownian motion "scaled" by a factor $\sqrt c$.
2. $tB_{1/t}$ is a Brownian Motion as well. This can be seen as a "time inversion" of the original Brownian motion.

Note that we cannot write something such as, e.g. , "$B_{ct} = \sqrt c B_t$" because this does not hold, they just have the same law (on the same filtration). I don't have any knowledge of cases where you could write $B_{f(t)} = g(B_t)$.

• Finally, one very important results that is quite relevant to what you're asking for is the Dubins-Schwarz theorem. It states that every continuous local martingale is a time-changed Brownian motion, or more precisely :

Dubins-Schwarz theorem : Let $M$ be a continuous local martingale with respect to a filtration $(\mathcal{F}_t)_{t\ge0}$ such that $M_0 = 0$ and $\langle M\rangle_\infty = \infty$ a.s.. Then, let for all $t \ge 0$ $$T_t=\inf\{s\geq0:\langle M\rangle_s>t\}=\langle M\rangle_t^{-1} $$ It holds that :

1. ${W={(M_{\langle M\rangle_t^{-1}})}_{t\geq0}}$ is a Brownian motion with respect to the time-stopped filtration ${{(\mathcal{F}_{T_t})}_{t\geq0}}$
2. ${{(W_{\langle M\rangle_t})}_{t\geq0}={(M_t)}_{t\geq0}}$

So for some quite specific functions $f$ (which, in fact, also depend on $\omega$), we know that $W_{\langle M\rangle_t} := W_{f(t)}$ is, in fact, the original martingale $M$ (careful with the filtrations here) ! I recommend you give a read to both Djalil Chafaï's and George Lowther's notes to learn more about these results, and time changed Brownian motions in general.

Hope that helps.