A4 - Waiting time to observe the $k$-event

It is clear that if we are interested in the probability that the first count occurs in the $i$-th time interval of amplitude $\Delta T$, we recover `in principle' a geometric distribution. But since $\Delta T$ can be arbitrary small, it makes no sense in numbering the intervals. Nevertheless, thinking in terms of the $n$ Bernoulli process can be again very useful. Indeed, the probability that the first count occurs after the $x-th$ trial is equal to the probability that it never occurred in the trials from 1 to $x$:

$$P(X>x) = (1-p)^x\,.$$

In the domain of time, indicating now by $T$ the time at which the first event can occur, the probability that this variable is larger than the value $t$, the latter being $n$ times $\Delta T$, is given by

$$P(T>t) = (1-p)^n$$

$$= \left(1-r\cdot \frac{t}{n}\right)^n \xrightarrow[\ n\rightarrow\infty\ ]{}\, e^{-r\, t}\,.$$

As a complement, the cumulative distribution of $T$, from which the probability density function follows, is given by

$$F(t\,\vert\,r) \equiv P(T\le t) = 1 - P(T>t) = 1- e^{-r\, t}$$

$$f(t\,\vert\,r) \equiv \frac{\mbox{d}F(t\,\vert\,r)}{\mbox{d}t} = r\, e^{-r\, t}\,.$$

The time at which the first count is recorded is then described by an exponential distribution having expected value, standard deviation and variation coefficient equal to

$$(T) = 1/r \ [\equiv \tau]$$

$$\sigma(T) = 1/r = \tau$$

$$v = 1\,,$$

while the mode (`most probable value') is always at $T=0$, independently of $r$.

As we can see, as it is reasonable to be, the higher is the intensity of the process, the smaller is the expected time at which the first count occurs (but note that the distribution extends always rather slowly to $T\rightarrow\infty$, a mathematical property reflecting the fact that such a distribution has always a 100% standard uncertainty, that is $v=1$). Moreover, since the choice of the instant at which we start waiting from the first event is arbitrary (this is related to the so called `property of no memory' of the exponential distribution, which has an equivalent in the geometric one), we can choose it to be the instant at which a previous count occurred. Therefore, the same distribution describes the time intervals between the occurrence of subsequent counts.

Once we have got the probability distribution of $k=1$, using probability rules we can get that of $k=2$, reasoning on the fact that the associated variable is the sum of two exponentials, and so on. We shall not enter into details,³⁹but only say that we end with the Erlang distribution, given by

$$f(t\,\vert\,r,k) = \frac{r^{k}}{(k-1)!}\cdot t^{k-1}\cdot e^{-r\,t} \ \ \ \ \left\{\begin{array}{l}r>0\\ k: \mbox{integer}, \ge 1 \end{array} \right.$$

The extension of $k$ to the continuum, indicated for clarity as $c$, leads to the famous Gamma distribution (here written for our variable $t$)

$$f(t\,\vert\,r,c) = \frac{r^{c}}{\Gamma(c)}\cdot t^{c-1}\cdot e^{-r\,t} \ \ \ \left\{\begin{array}{l}r>0\\ c > 0 \end{array} \right.$$

with $r$ the `rate parameter' (and it is now clear the reason for the name) and $c$ the `shape parameter' (the special cases in which $c$ is integer help to understand its meaning), having expected value and standard deviation equal to $c/r$ and $\sqrt{c}/r$, both having the dimensions of time (this observation helps to remember their expression).

However, since in the text the symbol $r$ is assigned to the intensity of the physical process of interest, we are going to use for the Gamma distribution the standard symbols met in the literature (see e.g. [31] and [32]) applying the following replacements:

$$c \rightarrow \alpha$$

$$r \rightarrow \beta\,.$$

Using also the usual symbol $X$ for generic variable, here is a summary of the most important expressions related to the Gamma distribution (we also add the mode, easily obtained by the condition of maximum⁴⁰):

$X\sim$   Gamma$(\alpha,\beta)$:

$$f(x\,\vert\,\alpha,\beta) = \frac{\beta^{\,\alpha}}{\Gamma(\alpha)}\cdot x^{\,\alpha-1}\cdot e^{-\beta\,x} \ \ \ \left\{\begin{array}{l}\alpha>0\\ \beta > 0 \end{array} \right.$$

$$(X) = \frac{\alpha}{\beta}$$

$$(X) = \frac{\alpha}{\beta^2}$$

$$\sigma(X) = \frac{\sqrt{\alpha}}{\beta}$$

$$(X) = \left\{ \begin{array}{ll} 0 & \mbox{if}\ \alpha < 1 \\ \frac... ...if}\ \alpha \ge 1 \end{array}\right. % \frac{\alpha-1}{\beta}.$$

Here is, finally, a summary of the distributions derived from the `apparently insignificant' Bernoulli process:

For completeness, let us also remind that:

• the famous $\chi^2$ distribution is technically a Gamma, with $\alpha=\nu/2$ and $\beta=1/2$;
• most distributions appearing in this scheme, with the obvious exception of the geometric and the exponential, which have fixed shape, `tend to a Gaussian distribution' for some values of the parameters. In particular, for what concerns this paper, the Poisson distribution tends to `normality' for `large' values of $\lambda$, as well known. However, it is perhaps worth remembering that, in general, such a limit applies to the cumulative distribution, and not to the probability function, defined for the Poisson distribution only for non negative integers:
  $$F(x\,\vert\,$$Poisson$$(\lambda))\, \xrightarrow [\,\ \lq \lambda\rightarrow \infty'\,\ ]{} \,F(x\,\vert\,{\cal N}(\lambda,\sqrt{\lambda}))\,.$$