I am trying to write the product of two stretched exponentials as another stretched exponential function. For example,

$ f(t) = \exp[-(t/\tau_1)^{\beta_1}]$ and $g(t) = \exp[-(t/\tau_2)^{\beta_2}]$

Now, can a function $h(t) = f(t)g(t)$ be also described with a stretched exponential like,

$h(t) = \exp[-(t/\tau)^{\beta}]$

where $\beta$ is related to $\beta_1, \beta_2$ (likewise for $\tau$).

An attempt when one of them is exponential, $\beta_2 = 1$

One can write stretched exponential as a distribution of single exponentials as,

$$ f(t) = \exp[-(t/\tau_1)^{\beta_1}] = \int_0^\infty ds\; P(s,\beta_1)\exp\{-s\frac{t}{\tau_1}\} $$

$$ g(t) = \exp[-t/\tau_2] = \int_0^\infty ds' \; \delta(s'-1)\exp[-s'\frac{t}{\tau_2}] $$

then $h(t)$ can be written as a double integral,

$$ h(t) = \int_0^\infty ds\int_0^\infty ds' \; P(s,\beta_1) \delta(s'-1) \exp\bigg[-s'\frac{t}{\tau_2}-s\frac{t}{\tau_1}\bigg] $$

defining a new set of variables,

$$ \frac{s}{\tau_1} = \frac{a+b}{2\tau}; \qquad \qquad \frac{s'}{\tau_2} = \frac{a-b}{2\tau} $$

with few algebraic manipulations, one obtains $$ h(t) = - \int_0^\infty da \;P\bigg[\frac{a}{2}\bigg(1+\frac{\tau_1}{\tau}\bigg)-\frac{\tau_1}{\tau_2},\beta_1\bigg] \exp[-a \frac{t}{\tau}] $$

which equivalent to saying $h(t)$ is also a stretched exponential with different parameters.

But if I did few more transformations, I could also end up with the following expression that, $$ h(t) = -\int_0^\infty da' \;P\bigg[a'-\frac{\tau_1}{\tau_2},\beta_1\bigg] \exp[-a' \frac{t}{\tau_1}] $$

which just means that the probability distribution is shifted by a constant $(\tau_1/\tau_2)$ but with same stretching exponent $\beta$. Can this also be considered as a stretched exponential?


Taking logarithm of everything your question becomes:

If $ \ln f(t) = -(t/\tau_1)^{\beta_1}$ and $\ln g(t) = -(t/\tau_2)^{\beta_2} $, can the function $\ln h(t) = \ln f(t)+ \ln g(t)$ be also described with a log of stretched exponential, i.e. $\ln h(t) = -(t/\tau)^{\beta}$.

This is false, since the sum $-(t/\tau_1)^{\beta_1} -(t/\tau_2)^{\beta_2} $ in question is not homogeneous in $t$ unless $\beta_1=\beta_2$.

Further explanation about homogeneity: A function is homogeneous of degree $\beta$ if $f(kt)=k^{\beta}f(t)$ for any $k>0$. Clearly logarithm of stretched exponential, being $-(t/\tau)^{\beta}$ is homogeneous of degree $\beta$.

However, I claim that unless $\beta_1=\beta_2$ the function $u(t)=-(t/\tau_1)^{\beta_1} -(t/\tau_2)^{\beta_2} $ is not homogeneous of any degree, essentially because it has wrong scaling behavior, which is easiest to see for very large or very small $t$ (and so not being homogeneous it can not be logarithm of a stretched exponential). Here is a formal proof:

We will actually show that $u(t)$ restricted to positive inputs $t>0$ is not homogeneous. In fact on positive $t$ we have for any non-zero homogeneous function $f(t)$ the formula

$$f(t)=f(t\times 1)=t^\beta f(1)=Ct^\beta$$

So $f(t)/t^\beta=C$ is constant, and in particular

$$\lim_{t\to 0+} f(t)/t^\beta = \lim_{t\to +\infty} f(t)/t^\beta=C$$

which is finite and must be non-zero for non-zero $f(t)$.

Without loss of generality suppose $\beta_1>\beta_2$. Write $$ u(t)/t^{\beta}= C_1 t^{\beta_1-\beta}+C_2t^{\beta_2-\beta}$$

with negative $C_1$ and $C_2$.

Since $\lim_{t\to +\infty} t^a$ is finite and nonzero only if $a=0$, $\lim_{t\to +\infty} u(t)/t^{\beta}$ is finite and non-zero if and only if $\beta =\beta_1$. Similarly, $\lim_{t\to 0+} u(t)/t^{\beta}$ is finite and non-zero if and only if $\beta =\beta_2$. Since we can not have $\beta$ equal to both of them, $u(t)$ is not homogeneous.

  • $\begingroup$ It's raised to the power of $\beta$ outside the - sign. But if the sum is not homogenous, is it impossible to write it as an exponent of any real values of $\tau, \beta$. $\endgroup$ – user35952 Apr 3 '18 at 9:38
  • $\begingroup$ Is it possible to solve for $\beta = \beta(\beta_1,\beta_2,\tau_1,\tau_2)$ and $\tau = \tau(\beta_1,\beta_2,\tau_1,\tau_2)$? $\endgroup$ – user35952 Apr 3 '18 at 11:58
  • $\begingroup$ I have corrected the sign; no, it is not possible to solve for $\beta$, no $\beta$ will work -- see the more detailed explanation in the updated answer. $\endgroup$ – Max Apr 3 '18 at 17:15

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