# 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.

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