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Brownian Motion

  1. Feb 10, 2010 #1
    Introductory texts on Wiener Process often introduce the topic by dividing the time into small time steps [tex]h=\Delta t[/tex].
    Then the value of the process [tex]W_j[/tex] at time j, [tex]t_j[/tex] is given by adding up many independent and normally distributed increments:


    The Wiener process [tex]W(t)[/tex] is generated in the limit as the step size [tex]h \rightarrow 0[/tex].

    My question is why is the step size [tex]\sqrt{h}Z_j[/tex] and not some other scale of h, such as [tex]h\cdot Z_j[/tex] or [tex]h^2 \cdot Z_j[/tex], say?
  2. jcsd
  3. Feb 10, 2010 #2
    This scale is the only one where the variance converges to a finite non-zero value.
  4. Feb 11, 2010 #3


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    I think of it as follows. Let Zj be standard normal, N(0,1) for all j. Let Wt be a Wiener (Brownian) process constructed, perhaps, through a Brownian bridge. Wt - Ws is distributed normally with mean 0 and variance |t - s|.

    Wt(j) - Wt(j-1) = Zj[itex]\sqrt{t(j) - t(j-1)}[/itex]. Then Var[Wt(j) - Wt(j-1)] = Var[Zj]|t(j) - t(j-1)| = |t(j) - t(j-1)|.

    Had I replaced [itex]\sqrt{t(j) - t(j-1)}[/itex] with anything else then I would not have obtained the correct variance for a Wiener process.
    Last edited: Feb 11, 2010
  5. Feb 12, 2010 #4
    Yes, I see that that upon choosing the scale as [tex]\sqrt{h}[/tex], the distribution of [tex]W(t)[/tex] will not depend on the time step [tex]h[/tex].

    However, notice that the value of Weiner process itself will diverge as we take smaller and smaller time steps. Why?

    Below I've done the calculations and I'm almost certain they are correct. (I use the law of large numbers to go from the fourth to the fifth equation)
    W(t)=lim_{n \rightarrow \infty} \sum_{k=0}^n \sqrt{\frac{t}{n}} Z_k [/tex]
    W(t)=\sqrt{t}\cdot lim_{n \rightarrow \infty}\frac{1}{\sqrt{n}}\sum_{k=0}^n Z_k[/tex]
    W(t)=\sqrt{t}\cdot lim_{n \rightarrow \infty}\frac{\sqrt{n}}{n}\sum_{k=0}^n Z_k[/tex]
    W(t)=\sqrt{t}\cdot \left(lim_{n \rightarrow \infty}\sqrt{n}\right) \cdot \left(lim_{n \rightarrow \infty}\frac{1}{n}\sum_{k=0}^n Z_k\right)[/tex]
    W(t)=\sqrt{t}\cdot \left(lim_{n \rightarrow \infty}\sqrt{n}\right) \left(\mu_z\right)[/tex]
    W(t)=\sqrt{t}\cdot \mu_z \cdot lim_{n \rightarrow \infty}\sqrt{n} \rightarrow \infty[/tex]

    where above [tex]W_0=0, t=n\cdot h[/tex] and [tex]\mu_z[/tex] is the mean of the [tex]Z_j[/tex]
  6. Feb 12, 2010 #5
    Here [tex]\mu_z=0[/tex]. To add a drift term (eg in a stochastic DE) it must have size dt not [tex]\sqrt{dt}[/tex].
  7. Feb 12, 2010 #6
    The gaussian increments should have a zero mean, as mentioned above. So in line four your are splitting a limit into one part that approaches infinity and one part that approaches 0. That will of course not lead you to the answer.

    Anyway, when you perform this limiting procedure, the right hand side will converge towards a random variable, namely the Wiener process at time t, so it will not approach any real number.

    In order to show the equivalence between the two on the level of distributions, you can check the different moments of the distribution of the random variables on each side. In particular the second moment about 0 should be t, since this is one of the defining properties of the Wiener process:

    Left hand side, by def. of Wiener process

    E[W(t)W(t)] = t

    Then check the right hand side, before taking the limit. We get

    \frac{t}{n}\sum_{i,j}E[Z_i Z_j] = \frac{t}{n}\sum_{i}E[Z_i Z_i]
    = \frac{t}{n}\sum_{i} 1 = t

    where I have used independence of increments:

    E[Z_i Z_j] = 0\quad\mathrm{when}\quad i\neq j

    and by definition:

    E[Z_iZ_i] = 1

    So the second moment of the right hand side has the correct value, and is even independent of n.

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