Process for a Gamma distribution with non integer shape parameter I am sampling the distribution of lifetimes of computers participating in massive volunteer computing initiatives (BOINC projects). While a phenomenological Weibull distribution makes a good description, I find that a better fit happens with a Gamma distribution with a shape parameter $k < 1$
Now, while a Gamma distribution with integer shape parameter is produced as the waiting time for the $k$-th event of a Poisson process, I do not know of a similar, process-based, description of the meaning of a Gamma distribution for a non integer parameter, nor to say for  non integer and less than unity. Is there one?
 A: This is just a hint (actually yours is a question I planned to ask here myself sooner or later):
What's the model for the gamma distribution with non-integer parameter $\lambda$? 
Since possibly the most relevant aspect is the semigroup law w.r.to convolution, my feeling is that it should describe properly waiting times for reaching given amount $\lambda$ of mass, arriving continuously and randomly under independence assumptions, like e.g. rain, falls of powder on the soil. I'm very interested in further more precise answers. 
A: Here is some interesting, though natural, view of the problem (including a small but quite elegant result at the end). Denote by ${\rm Gamma}(k,\lambda)$ the gamma distribution with density function $f(x;k,\lambda) = \lambda ^k x^{k - 1} e^{ - \lambda x}/ \Gamma (k)$, $x>0$. Here $k$ and $\lambda$ are arbitrary positive constants. Clearly, the problem reduces to the case $0<k<1$. Indeed, if $k$ is non-integer and greater than $1$, then a ${\rm Gamma}(k,\lambda)$ random variable can be viewed as a sum of independent ${\rm Gamma}(\left\lfloor k \right\rfloor,\lambda)$ and ${\rm Gamma}(k - \left\lfloor k \right\rfloor,\lambda)$ random variables. Now, let's consider a gamma process $X = \lbrace X_t : t \geq 0\rbrace$ such that $X_1 \sim {\rm exponential}(\lambda)$. That is, $X$ is a process with stationary independent increments, starting at $0$ and having right-continuous with left limits sample paths (i.e., a L\'evy process), such that $X_t - X_s \sim {\rm Gamma}(t-s,\lambda)$, for any $0 \leq s < t$. The process $X$ is a pure jump process, having positive jumps only, and it holds $X_t  = \sum\nolimits_{0 < s \le t} {\Delta X_s } $, where $\Delta X_s = X_s - X_{s-}$. The number of jumps is a.s. infinite countable in any time interval, no matter how small (yet, $\Delta X_t =0$ a.s., for any fixed $t>0$). The process $X$ is characterized by its jump measure (L\'evy measure) given by $\nu({\rm d}x)=x^{-1}e^{-\lambda x}\,{\rm d}x$: given a time interval $[t_1,t_2]$, the number of jumps of $X$ whose sizes lie in $[a,b] \subset (0,\infty)$ is Poisson distributed with mean $(t_2-t_1)\int_a^b {x^{ - 1} e^{ - \lambda x } \,{\rm d}x}$ (the key concept here is Poisson random measure). Hence, roughly speaking, the process usually increases by tiny jumps (whose sum is negligible in practice). 
Now, consider the process $X$ on the time interval $[0,1]$. On the one hand, $X_1$ is the (infinite) sum of jumps of $X$ up to time $t=1$. On the other hand, $X_1$ is distributed as the waiting time until the first occurrence in a Poisson process with rate $\lambda$. This is already quite interesting. Returning to the original problem, we can view a ${\rm Gamma}(k,\lambda)$ random variable, $0<k<1$ fixed, as the random variable $X_k$, i.e., the sum of jumps of $X$ up to time $k$ (note that $\lim _{k \uparrow 1} X_k  = X_1 $, a.s.). So, in some respect, this already gives a process-based description of the meaning of a Gamma distribution for a non-integer parameter. But much more can be said, using the fact that if $X$ and $Y$ are independent ${\rm Gamma}(\alpha,\lambda)$ and ${\rm Gamma}(\beta,\lambda)$ rv's, respectively, then the ratio $X/(X+Y)$ has a ${\rm Beta}(\alpha,\beta)$ distribution. Namely, the ratio $R = X_k / X_1$, $0<k<1$, is distributed as a ${\rm Beta}(k,1-k)$ random variable, implying that $R$ has density function $f_{R} (x;k) = x^{k-1}(1-x)^{-k}\sin(\pi k)/\pi$, $0<x<1$, where we have used $\Gamma(k)\Gamma(1-k) = \pi/\sin(\pi k)$. In particular, $f_R (x;1/2) = 1/(\pi \sqrt {x(1 - x)} )$, $0<x<1$.
