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I am trying to understand why Brownian Motion is not differentiable.

Here we are tossing a coin n-times.

So we form a sample space $ \Omega $.

Each of the paths are represented by ω = ω1 ω2 ω3…. ωn

In all likely situation, sample space will include the following two paths

  • outcomes of the tosses are either purely Heads-H or purely Tails Then

these two paths can be written as o (omega) ω =HHHHHHHHHHHHHHHHHHHHHH…n times o (omega) ω = TTTTTTTTTTTTTTTTTTTTTTTTTT..ntimes

In my understanding both the above cases the paths/Brownian motion is a straight line. In other words, not a zig-zag

In my view, the above two omegas are differentiable as both are straight lines. So two of the random walks/paths in the Omega $ \omega $ are differentiable.

I have come across in several texts and videos that Brownian motion is not differentiable. I agree that except for the above two paths remaining are not differentiable.

The clarification I seek is whether Brownian Motion is differentiable at each time period • For the above two paths at every/any point • for the remaining paths in the omega (sample space other than the above two) This might sound naïve, why is it important to be differentiable at every step.

Coming from a non-math background, I have built my skills to this stage with the help of PhD student. But there is always the possibility my fundamentals may not be perfect or complete.

Kindly guide/help me.

Thank you

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    $\begingroup$ You have left off how your coin-flipping relates to Brownian motion. If it's as a result of Donsker's invariance principle, then you've left off the time scaling that would make this path a vertical line. If it's as some result of sampling, then it has nothing to do with differentiability. ----- As a final point, when we say something like "Brownian motion is nowhere differentiable", we technically mean "Almost every path of Brownian motion is nowhere differentiable". That is, the set of paths on which Brownian motion is differentiable somewhere has probability zero. $\endgroup$ Commented Jun 29, 2022 at 20:58
  • $\begingroup$ A random walk (i.e. a sequence of heads and tails) is not a Brownian motion. A Brownian motion is a stochastic process which is continuous in time, which can be realized as the limit of random walks. Even if you regard a straight line as a Brownian motion (i.e. the path traced out by a specific particle moving according to a Brownian motion), the probability that a particle will trace out such a path is zero (under suitable constructions of "probability"), hence one might rephrase the statement as "the path of a Brownian motion is nowhere differentiable with probability 1". $\endgroup$
    – Xander Henderson
    Commented Jun 29, 2022 at 20:59
  • $\begingroup$ A suggestion: check the defition of differentiability from a formal point of view. The argument "the above two omegas are differentiable as both are straight lines", makes little sense. $\endgroup$
    – Math-fun
    Commented Jun 29, 2022 at 21:03
  • $\begingroup$ @Math-fun I don't think one really needs to check the definition of differentiability to conclude a straight line is differentiable $\endgroup$ Commented Jun 29, 2022 at 21:58
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    $\begingroup$ @user6247850 maybe I don't under the concept in this framework myself. Thank you very much for the hint. What does it mean that a "certain $\omega$", as is referred to by OP, is a straight line which is the differentiable [with respect to what]? $\endgroup$
    – Math-fun
    Commented Jun 30, 2022 at 6:08

1 Answer 1

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The proof of a.s. nondifferentiability of Brownian motion is explained in Theorem 1.30, page 21 of the book [1]. The theorem was first proved by Paley, Wiener and Zygmund in [PWZ33], but the proof in the book is due to Dvoretzky, Erdos and Kakutani [DEK61].

[1] Mörters, Peter, and Yuval Peres. Brownian motion. Vol. 30. Cambridge University Press, 2010. https://yuvalperes.com/brownian-motion/

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