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According to an article I'm studying ("Time series, self-similarity and network traffic by Mark Crovella) the expectation of the square of a time-scaled Brownian motion process $E[ B(ct)^2 ]$ where $c$ is the time scaling is equal to $ct$.

I'd appreciate help proving this; i.e.

$E[ B(ct)^2 ] = c t$

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1 Answer 1

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This follows from the fact that $B(t)$ has a Gaussian$(0,t)$ distribution. Therefore, $B(ct)$ has a Gaussian$(0,ct)$ distribution. Thus $E[B(ct)^2] = Var[B(ct)] + (E[B(ct)])^2 = ct + 0 = ct$.

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Thanks but, I don't understand why $B(ct)$ has the Gaussian distribution $(0, ct)$. Can this be proved? I can show that a random walk has variance $t$ and is therefore "tends" to the Gaussian $(0,t)$ when $t$ becomes large. On second thoughts, perhaps is it merely sufficient to say $x = ct$ and therefore $Var[B(x)] = x = ct$? (PS: I'm awarding you the points, but please correct me if I'm wrong.) –  Olumide Oct 17 '10 at 8:45
    
Re: "I don't understand why B(ct) has the Gaussian distribution (0,ct). Can this be proved?" This is part of the definition of the Brownian process. –  Rasmus Oct 17 '10 at 12:12
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Yes, it is sufficient to say $x = ct$ and then proceed as you have done. –  Mike Spivey Oct 17 '10 at 13:05
    
@Rasmus I've been able to prove that the variance of a random walk of $n$ steps each of length $\epsilon$ is $n \epsilon^2$. And I suppose that if $n = \frac{x}{\epsilon}$, where $x$ is the distance of the "drunk" from the starting point. The variance becomes $x \epsilon$ not $x$ as I hoped -- unless I assume $\epsilon = 1$ -- which makes me wonder if the formulation of the random walk requires unit step lengths. (PS: on second thoughts, the assumption $n = \frac{x}{\epsilon}$ may be incorrect.) Update: Thanks Mike –  Olumide Oct 17 '10 at 13:07

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