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Assume we have a stochastic process $$ dX_t = V_tdt $$ $$ dV_t = -KX_tdt -MV_tdt +dB_t. $$ with $X_t \in \mathbb{R}^n$ representing position and $V_t$ velocity. This is a linear elastic model described by the equation $$ dZ_t = AZ_t dt + dB_t $$ where $A$ is a $2n \times 2n$ matrix which has $0$ and $I$ on top $n\times 2n$ part of $A$ and $-K$, $-M$ on bottom $n \times 2n$ part. Here $M$ represents friction terms. I want to understand how scaling of $M$ by a uniform constant effects the normalized time delayed cross correlations: $$ C(\epsilon) = \lim_{t\rightarrow \infty} \frac{\mathbb{E}(Z(t+\epsilon)Z^T(t))}{\sqrt{\mathbb{E}(Z(t+\epsilon)^2)\mathbb{E}(Z(t)^2)}} $$

This matrix $A$ usually has some $0$ eigenvalues and in such cases calculating the correlation matrix above one needs to quotient them out otherwise the limit as $t$ goes to infinity does not converge. But otherwise you more or less treat it like the regular case where you disregard the zero eigenvalue in any summation over eigenvalues that appear. Now I did some computational experiments and I see that if I scale $M$ by some large constant $\mu$ then the decay rate of $C(\epsilon)$ slows down considerably while if I decrease friction by scaling $M$ by a small constant the decay rate increases up to a point where if it is too small then you again get a slow decay rate. Of course the correlation matrix does not converge if the friction is $0$ so I can test very small values but beyond a certain value it might not be reasonable. So my question is:

1-Is this the kind of behavior one should expect,

2-And if so does any body have a better explanation for this physical intuition or mathematical that is better than some numerical simulations.

3-No matter what the values are $C(\epsilon)$ decays to $0$ as $\epsilon$ goes to $\infty$ which kind of makes sense from a physical point of view but is it somehow possible to devise systems where the time delayed cross correlations dont go to $0$?

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    $\begingroup$ -1. Not clear. I recommend that you make your question easier to read by separating your text into more paragraphs and by highlighting what you are asking, because this is lost in a wall of text without even a question mark as a clue. $\endgroup$ – sammy gerbil Feb 26 '17 at 20:00
  • $\begingroup$ I agree, I don't actually see what the question is here, it is just a bunch of statements about the process & your results. $\endgroup$ – Kyle Kanos Feb 28 '17 at 11:16
  • $\begingroup$ @KyleKanos: These are not really importants results but some simple simulations to get a feeling about the relation between time delayed correlation and friction. But in the end I was kind of confused. In any case based on your suggestion, I improved the question. $\endgroup$ – Sina Feb 28 '17 at 23:10
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We have

\begin{align} \mathrm{d}\mathbf{x}_t &= \mathbf{v}_t \mathrm{d}t \\ \mathrm{d}\mathbf{v}_t &= -K \mathbf{x}_t \mathrm{d}t - M \mathbf{v}_t\mathrm{d}t + \sigma_v\mathrm{d}\mathbf{B}_t \end{align}

where I assume $M$, $\sigma$ are diagonal, $K$ not so, and the Brownian processes are uncorrelated.

Defining $\mathbf{z}_t = (\mathbf{x}_t, \mathbf{v}_t)$, $$\mathrm{d}\mathbf{z}_t = \begin{pmatrix}0 & I \\ -K & -M \end{pmatrix} \mathbf{z}_t \mathrm{d}t + \sigma\mathrm{d}\mathbf{W}_t$$

Where the Brownian process bit changed notation as it now has length $2n$, and the first matrix we define as $A$.

The solution is now (Ornstein-Uhlenbeck), so you can just use this directly instead of running simulations: $$\mathbf{z}_t = e^{tA}\mathbf{z}_0 + \int_0^t e^{-(s-t)A}\sigma\mathrm{d}\mathbf{W}_s$$

We want to compute the autocorrelation, $$C(\varepsilon) = \frac{\operatorname{cov}(\mathbf{z}_t, \mathbf{z}_{t+\varepsilon})}{\sqrt{\operatorname{var}(\mathbf{z}_t)\operatorname{var}(\mathbf{z}_{t+\varepsilon})}}$$

Now (Ito isometry): $\operatorname{var}(\mathbf{z}_t) = \int_0^t \left(e^{-(s-t)A}\sigma\right)^2\mathrm{d}s$, and with some additional mathematical assuptions ($A$ invertible), $\operatorname{var}(\mathbf{z}_t) = \frac{1}{2}\sigma^2A^{-1}(e^{2tA}-I)$. Similarly $\operatorname{cov}(\mathbf{z}_t, \mathbf{z}_{t+\varepsilon}) = e^{\varepsilon A}\operatorname{var}(\mathbf{z}_t)$.

With some physics intuition from O-U processes, let's say that the variance is bounded (because some part of the network holds the entire thing close to a fixed coordinate, and the friction dissipates the rest; you could say that the system as a whole is free to move, but then you'd make a similar argument restricting all parts of the system to the center of mass; if the system is really two separate systems, we would just separate those into two and treat each in turn).

Put another way: $\lim_{t\to\infty} e^{2tA} = 0$ (i.e. $\lim_{t\to\infty} \operatorname{var}(\mathbf{z}_t) = - \frac{1}{2}\sigma^2A^{-1}$), so we see that $\lim_{\varepsilon\to\infty} C(\varepsilon) = 0.$

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  • $\begingroup$ I am not completely sure if the variance is $A^{-1}$ (you can take pseudoinverse if $A$ is not invertible). That is only true when $A$ is symmetric otherwise it is related to pseudoinverse of $A\oplus A$ where $\oplus$ is Kronecker summation. In particular if you the relevant computions assuming $A$ is diagonalizable, you get terms like $\frac{1}{\lambda_i+\lambda_j}$ where $\lambda_i$ are eigenvalues. $\endgroup$ – Sina Mar 6 '17 at 13:57

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