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The question is: how to prove that for a particle that follows a brownian motion we have: $$ \langle X(t)^2 \rangle = 2\int_{0}^{t}d\tau\ (t-\tau)C_{v}(\tau) = 2\int_{0}^{t}d\tau_1\int_{0}^{\tau_1}C_{v}(\tau)d\tau $$ where $X(t)$ is the random position of the particle and $C_{v}(\tau) = \langle V(t_1)V(t_1+\tau) \rangle$ is the velocity autocorrelation function, with of course $\frac{dX(t)}{dt}=V(t)$.

Remark

To avoid confusion, the Brownian Motion is intended here with the physical interpretation, similar to the original one: it is a bivariate Markov process $(X(t),V(t))$ where $V(t)=\frac{dX}{dt}$ and $V$ satisfy the following Langevin equation: $\frac{dV}{dt}= A(t)$ where $A(t)$ is a white noise random process: $\langle A(t) \rangle = 0$ and $\langle A(t)A(t') \rangle = \sigma^2\delta(t-t')$. So $X(t)$ is the process that defines the position of the particle, $V(t)$ is the process that define the velocity of the particle and the Langevin equation is like the "Newton's" law of dynamics.

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Ok I solved it. We have that $$ X(t)=\int_{0}^{t}V(t_1)dt_1 $$ and so $$ \langle X(t)^2 \rangle =\int_{0}^{t}\int_{0}^{t}\langle V(t_1)V(t_2)\rangle dt_1dt_2 = \int_{0}^{t}\int_{0}^{t}C_{v}(t_2-t_1)dt_1dt_2 $$ The last equality is because the Brownian motion is a stationary process. Remember also that the velocity autocorrelation function is an even function of $t$, i.e $C_{v}(t_2-t_1)=C_{v}(t_1-t_2)$.

Now let \begin{align} \tau & = t_2-t_1 \\ T & = t_1+t_2 \end{align} So that \begin{align} t_2 & = \frac{\tau+T}{2} \\ t_1 & =\frac{T-\tau}{2} \\ \end{align} The Jacobean of this transformation is $1/2$ and the integral becomes: $$ \langle X(t)^2 \rangle =\frac{1}{2}\left[\int_{0}^{t}C_{v}(\tau)d\tau\int_{\tau}^{2t-\tau}dT+\int_{-t}^{0}C_{v}(\tau)d\tau\int_{-\tau}^{2t+\tau}dT\right] = \int_{0}^{t}(t-\tau)C_{v}(\tau)d\tau+\int_{-t}^{0}(t+\tau)C_{v}(\tau)d\tau=2\int_{0}^{t}(t-\tau)C_{v}(\tau)d\tau $$ as required. I change $\tau=-\tau$ in the second integral and I use the parity of the VAF: $C(\tau)=C(-\tau)$.

Now integrating by part we have also: $$ 2\int_{0}^{t}(t-\tau)C_{v}(\tau)d\tau = 2\int_{0}^{t}d\tau_1\int_{0}^{\tau_1}C_{v}(\tau)d\tau $$

Remark

The relation $X(t)=\int_{0}^{t}V(t_1)dt_1$ holds because the process $V(t)$, although not differentiable, is continuous (as, for instance, it satisfy the stochastic equation $\Delta V \propto \Delta t^{1/2}\Delta W $, where $\Delta W$ is an uncorrelated stationary Gaussian process with zero mean) and so it is integrable.

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    $\begingroup$ Note how the very definition of $X(t)$, from which your own answer is starting, was impossible to guess from the text of your question. Thus... $\endgroup$ – Did Jul 8 '18 at 16:49
  • $\begingroup$ Sorry I don't understand...was it not clear that $X(t)$ is the random position of a particle that follows a Brownian motion? $\endgroup$ – Alex Jul 8 '18 at 16:55
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    $\begingroup$ No it was not. Amongst other reasons, because this is not how you define X(t) in your answer. $\endgroup$ – Did Jul 8 '18 at 16:59
  • $\begingroup$ Maybe now it is better. $\endgroup$ – Alex Jul 8 '18 at 17:05
  • $\begingroup$ So now we know for sure that X(t) cannot be Brownian motion... $\endgroup$ – Did Jul 8 '18 at 17:59

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