Continuous Poisson Distribution Is there a Continuous analogous of the Poisson Distribution? Under the analogous, I mean such a distribution that:

*

*It is a one-parameter distribution

*Its distribution function is similar to the Poisson one

 A: There are different types of continuous extensions of Poisson distribution.
I favor the one presented in this paper:
"Continuous counterparts
of Poisson and binomial distributions
and their properties"
by Andrii Ilienko.
They are based on integral
representations of Poisson and binomial probability distribution functions using complete and incomplete (Euler) $\Gamma$ and $\operatorname{B}$ functions.
A: I would like to complement the accepted answer by deriving again the Continuous Poisson distribution shown in the paper of the accepted answer.
The derivation I show here is more natural although slightly less formal.
We will use an experiment to connect the distributions $\text{Poisson}(\lambda)$ whose domain is $k$, and $\text{Erlang}(k)$ whose domain is $\lambda$. Since the former is discrete and the latter is continuous, let's compare cumulative distribution functions. Shortening less and greater than with .lt and .gt, let $\text{Poisson}(\lambda)\text{.lt}(k)$ and $\text{Erlang}(k)\text{.gt}(\lambda)$ denote the probability that $X < k$ for $X\sim\text{Poisson}(\lambda)$ , and that $X > \lambda$ for $X\sim\text{Erlang}(k)$, respectively.
The experiment is a sequence of events that occur at $t_1, t_2, t_3, ...$ (in seconds), where $t_1 \sim \text{Exp}(1)$, $t_2 \sim t_1 + \text{Exp}(1)$, $t_3 \sim t_2 + \text{Exp}(1), ...$. In other words, $t_k \sim \text{Erlang}(k,1) = \text{Gamma}(k,1)$. This experiment is known as a Poisson process and to be more precise, I'm differing from the standard of a rate of $1/\lambda$ during one second by considering instead a fixed rate of 1 during $\lambda$ seconds. But these two points of view are equivalent.
Notice the following. $\text{Poisson}(\lambda)\text{.lt}(k)$ measures the probability that less than $k$ events occur during the first $\lambda$ seconds. Equivalently, this is the probability that the $k$'th event occurs after $\lambda$ seconds.
Therefore, this is the probability of $t_k>\lambda$, which coincides with $\text{Erlang}(k)\text{.gt}(\lambda)$.
Thus
$$ \text{Poisson}(\lambda)\text{.lt}(k) = \text{Erlang}(k)\text{.gt}(\lambda) $$
Let us now extend $\text{Poisson}(\lambda)$ into a $\text{Continuous-Poisson}(\lambda)$.
We know $\text{Gamma}(k)$ extends $\text{Erlang}(k)$ to non-integer values of $k$. This can informally still be thought of as waiting for the $k$'th event, as if events were fractioned in abstract manner by the $\text{Gamma}$ distribution. As a consequence, the Continuous-Poisson distribution can be naturally characterized with:
$$ \text{Continuous-Poisson}(\lambda)\text{.lt}(k) = \text{Gamma}(k)\text{.gt}(\lambda) $$
It's straightforward to get the CDF of our $\text{Continuous-Poisson}(\lambda)$ by plugging in the CDF of the Gamma distribution from Wikipedia as
$$ \text{Continuous-Poisson}(\lambda)\text{.cdf}(k) := 1 - \frac{\gamma(k,\lambda)}{\Gamma(k)} = \frac{\Gamma(k,\lambda)}{\Gamma(k)} $$
where $\gamma$ and $\Gamma$ are the lower and upper incomplete gamma functions. This coincides with the formula in the paper of the accepted answer. Regarding the PDF, we would need to compute the derivative of the CDF with respect to $k$, which looks difficult. The paper contains some formulas, although they contain integrals inside.
In terms of meaning, our CDF will measure the probability of having at most $k$ events until $\lambda$ seconds, where $k$ is possibly non-integer. Fractional events should be understood in the same abstract sense used by the Gamma distribution (a definition that mathematically makes sense, and maybe only mathematically).
