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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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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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    @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 Mike2010-10-17