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My lecture notes mention the following three properties of a standard Brownian motion process $\{B(t), \ t \geq 0\}$ where $B(0) = 0$:

  • $\mathbb{P}(\text{there is some zero in } (0, t)) = 1 \qquad$ for all $t > 0$;
  • $T_0 = \inf\{t > 0 : B(t) = 0\} = 0$; and
  • "In fact, a deeper analysis shows that, with probability 1, $B(t)$ has infinitely many zeros in any non-empty time interval $(0, t)$"

I suspect the reasoning behind the third point is beyond my grasp (I have forgotten any measure theory I briefly looked at, if that's relevant at all), but I would still like to intuitively understand how all three properties could be true.

In particular, if these three properties are true, how are we able to graph Brownian motion? My lecture notes show many simulated realisations of $\{B(t)\}$ and you never see the graph dipping back to $0$ in $(0, t)$ — does this mean that any graph of Brownian motion is "low-quality"? That is, if you zoom in enough you would see those infinite zeros?

Jeremy Lindsay
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1 Answers1

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About the graph question: when we look at the graph of a Brownian motion, we see it in discrete time. If you have an interval $t\in [0,T]$, usually we partition it in $\{0,\Delta t,2\Delta t,...,(N-1)\Delta t,N\Delta t = T\}$ and then we simulate the sequence $$B_{n\Delta t} \sim \mathcal{N}(B_{(n-1)\Delta t},\Delta t), \ \ n \in \{1,2..,N\}$$ which is usually accomplished with a simple recursion: $B_{n\Delta t} = B_{(n-1)\Delta t}+\sqrt{\Delta t}\varepsilon $ where $\varepsilon$ is drawn from a standard normal distribution. This simulation is exact in the sense that there is no numerical error involved caused by the construction of the recursion, but it still has gaps and we have no information about what happens in those gaps. Brownian motion being continuous, I would say yes, any graph you see is actually a 'low quality' version of the real deal and a lot of information is lost.

A question dealing with the third fact is here.

Snoop
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