Calculation of Autocorrelation Functions¶

First, suppose we have calculate the left eigenvectors and eigenvalues:

\[T^T \phi_i = \lambda \phi_i\]

Suppose that \(\pi\) is the equilibrium population. Then, we can normalize the eigenvectors such that:

\[\phi_i^T \pi^{-1} \phi_j = \delta_{ij}\]

Above, we denote \(\pi^{-1}\) to be a diagonal matrix with elements \(\pi_i^{-1}\).

The autocorrelation function of the observable \(f_i\) can be denoted:

\[E(f(z_t) f(z_0)) = \sum_{i,j} f_i P(z_0 = i) f_j P(z_t = j | z_0 = i) = \sum_{i,j} f_i f_j \pi_i T_{ij} =\]

We know that

\[T_{ab}(t) = \sum_k \lambda_k(t) (\psi_k)_a (\phi_k)_b = \sum_k \lambda_k(t) (\pi_a)^{-1} (\phi_k)_a (\phi_k)_b\]

Thus,

\[E(f(z_t) f(z_0)) = \sum_{i,j,k} f_i f_j \lambda_k(t) (\phi_k)_i (\phi_k)_j = \sum_k \lambda_k(t) s_k^2\]

Where

\[s_k = \sum_i f_i (\phi_k)_i\]

Finally, note that \(\lambda_i(\infty) = \delta_{i0}\), so the long-timescale behavior is simply:

\[E(f(z_\infty) f(z_0)) = s_0^2\]

For most applications, one is interested in the zero-centered ACF, so we simply skip the \(k = 0\) term in the summation.