Shot-noise Fano factor

A variability measure of the times of uniform events based on a shot-noise process is proposed and studied. The measure is inspired by the Fano factor, which we generalize by considering the time-weighted influence of the events given by a shot-noise response function. The sequence of events is assumed to be an equilibrium renewal process, and based on this assumption we present formulas describing the behavior of the variability measure. The formulas are derived for a general response function, restricted only by some natural conditions, but the main focus is given to the shot noise with exponential decrease. The proposed measure is analyzed and compared with the Fano factor.

Figure 1

Data construction based on events that occurred in $m$ time intervals of a length $t>0$. (a) The events are counted and their numbers $n_1,...,n_m$ obtained. (b) To every event a value is assigned using a weighting (exponential) function, and these values are summed. The individual sums are denoted as $s_1,...,s_m$.

Figure 2

(a) Illustration of times of events that occurred after time zero. (b) Illustration of the counting process $N\left(t\right)$. (c) Illustration of the shot-noise process $S(t,\tau )$ with an exponential response function.

Figure 3

Illustration of the behavior of $G_E(t,\tau )$ and $F\left(t\right)$ for gamma (a),(c), IG (b),(d), Weibull (e), and log-normal (f) probability distributions of IEIs with mean 1 and with various values of $C_v^2$ (values that correspond to the lines from bottom to top: $0.25,0.5,1,2,4$). The vertical axes are logarithmically scaled. (a),(b) Solid lines: $G_E(t,\tau )$ in dependence on $t$ for $\tau =1$. Dash-dotted lines: $G\left(t\right)$ in dependence on $t$. (c)–(f) Solid lines: $G_E\left(\tau \right)$ in dependence on $\tau$. Dash-dotted lines: $F\left(\tau \right)/2$ in dependence on $\tau$. We can see that the behavior in dependence on $t$ of the variability measures differs; however, at the steady state, $G_E\left(\tau \right)$ is approximately $G\left(\tau \right)/2$.