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## ELEMENTARY COMPUTATION OF AUTCOVARIANCE FOR NONSTOCHASTIC MITTAG-LEFFLER

F. Mainardi (mainardi-2014-approximations-mittag-leffler-function) introduced the following approximation for the Mittag-Leffler function $E_{a}(-t^a)$ when $t \rightarrow 0$

$f(t) = 1/(1+Ct^a)$

with $C=\Gamma(1+a)$.  By Taylor expansion

$f(t+h) \sim f(t) - C a t^{a-1} /(1+Ct^a)^2 h + O(h^2)$

Therefore $f(t)f(t+h) = f(t)^2 ( 1 + f'(t)/f(t) h + O(h^2)$ and

$f(t) f(t+h) = f(t)^2 ( 1 - C a t^{a-1}/(1+ Ct^a) h + O(h^2))$

This is then the behavior of the ‘autocorrelation’ function satisfying $D^a x_t = -x_t$. This elementary example provides some intuition for stochastic models where the autocorrelation function can be fit by Mittag-Leffler functions.

CAPUTO DERIVATIVE

The Caputo derivative is more useful for computations because one can inherit the product rule from the usual derivative and constants are annihilated by it.

$D^\mu f(t) = \frac{1}{\Gamma(1-\mu)} \int_0^t \frac{f'(s)}{(t-s)^\mu} ds$

Suppose $D^\mu x_t = -\lambda x_t$.  Then the product rule inside the integral and integration by parts gives

$D^\mu(x_t x_{t+h}) = - 2 \lambda x_t x_{t+h} - \int_0^t (D^\mu x_s D x_{s+h} + D^\mu x_{s+h} D x_s) ds$

We can now add stochastic components $D^\mu x_t = -\lambda x_t + A dw_t$ with zero expectations and still have

$D^\mu E(x_t x_{t+h} ) = -2 \lambda E(x_t x_{t+h} ) + ER(t,h)$

where $R(t,h) = \int_0^t (D^\mu x_s D x_{s+h} + D^\mu x_{s+h} D x_s) ds$.  Assuming this is small, the Mittag-Leffler function will be a good approximation for the autocorrelation function of $x_t$.