Stretched exponential function

Following the usual physical interpretation, we interpret the function argument t as time, and f_β(t) is the differential distribution. The area under the curve can thus be interpreted as a mean relaxation time. One finds

$${\displaystyle \langle \tau \rangle \equiv \int _{0}^{\infty }dt\,e^{-(t/\tau _{K})^{\beta }}={\tau _{K} \over \beta }\Gamma {\left({\frac {1}{\beta }}\right)}}$$

where Γ is the gamma function. For exponential decay, ⟨τ⟩ = τ_K is recovered.

The higher moments of the stretched exponential function are[8]

$${\displaystyle \langle \tau ^{n}\rangle \equiv \int _{0}^{\infty }dt\,t^{n-1}\,e^{-(t/\tau _{K})^{\beta }}={{\tau _{K}}^{n} \over \beta }\Gamma {\left({\frac {n}{\beta }}\right)}.}$$

In physics, attempts have been made to explain stretched exponential behaviour as a linear superposition of simple exponential decays. This requires a nontrivial distribution of relaxation times, ρ(u), which is implicitly defined by

$${\displaystyle e^{-t^{\beta }}=\int _{0}^{\infty }du\,\rho (u)\,e^{-t/u}.}$$

Alternatively, a distribution

$${\displaystyle G=u\rho (u)}$$

is used.

ρ can be computed from the series expansion:[9]

$${\displaystyle \rho (u)=-{1 \over \pi u}\sum _{k=0}^{\infty }{(-1)^{k} \over k!}\sin(\pi \beta k)\Gamma (\beta k+1)u^{\beta k}}$$

For rational values of β, ρ(u) can be calculated in terms of elementary functions. But the expression is in general too complex to be useful except for the case β = 1/2 where

$${\displaystyle G(u)=u\rho (u)={1 \over 2{\sqrt {\pi }}}{\sqrt {u}}e^{-u/4}}$$

Figure 2 shows the same results plotted in both a linear and a log representation. The curves converge to a Dirac delta function peaked at u = 1 as β approaches 1, corresponding to the simple exponential function.

Figure 2. Linear and log-log plots of the stretched exponential distribution function ${\displaystyle G}$ vs ${\displaystyle t/\tau }$ for values of the stretching parameter β between 0.1 and 0.9.

The moments of the original function can be expressed as

$${\displaystyle \langle \tau ^{n}\rangle =\Gamma (n)\int _{0}^{\infty }d\tau \,t^{n}\,\rho (\tau ).}$$

The first logarithmic moment of the distribution of simple-exponential relaxation times is

$${\displaystyle \langle \ln \tau \rangle =\left(1-{1 \over \beta }\right){\rm {Eu}}+\ln \tau _{K}}$$

where Eu is the Euler constant.[10]

If the integrated distribution is a stretched exponential, the normalized probability density function is given by

$${\displaystyle p(\tau \mid \lambda ,\beta )~d\tau ={\frac {\lambda }{\Gamma (1+\beta ^{-1})}}~e^{-(\tau \lambda )^{\beta }}~d\tau }$$

Note that confusingly some authors have been known to use the name "stretched exponential" to refer to the Weibull distribution.[18]

A modified stretched exponential function

$${\displaystyle f_{\beta }(t)=e^{-t^{\beta (t)}}}$$

with a slowly t-dependent exponent β has been used for biological survival curves.[19][20]

In wireless communications, a scaled version of the stretched exponential function has been shown to appear in the Laplace Transform for the interference power ${\displaystyle I}$ when the transmitters' locations are modeled as a 2D Poisson Point Process with no exclusion region around the receiver.[21]

The Laplace Transform can be written for arbitrary fading distribution as follows:

$${\displaystyle L_{I}(s)=\exp \left(-\pi \lambda \mathbb {E} {\left[g^{\frac {2}{\eta }}\right]}\Gamma {\left(1-{\frac {2}{\eta }}\right)}s^{\frac {2}{\eta }}\right)=\exp \left(-ts^{\beta }\right)}$$

where ${\displaystyle g}$ is the power of the fading, ${\displaystyle \eta }$ is the path loss exponent, ${\displaystyle \lambda }$ is the density of the 2D Poisson Point Process, ${\displaystyle \Gamma (\cdot )}$ is the Gamma function, and ${\displaystyle \mathbb {E} [x]}$ is the expectation of the variable ${\displaystyle x}$.

The same reference also shows how to obtain the inverse Laplace Transform for the stretched exponential ${\displaystyle \exp \left(-s^{\beta }\right)}$ for higher order integer ${\displaystyle \beta =\beta _{q}\beta _{b}}$ from lower order integers ${\displaystyle \beta _{a}}$ and ${\displaystyle \beta _{b}}$.