# Does the Poisson distribution work for non-integer exponents?

The question regards the Poisson distribution function as given by:

$$\frac{x^k e^{-x}}{k!}$$

The distribution's domain (x) goes from 0 to $\infty$, and $k \in \mathbb{N_0}$

I tried the distribution as the following function:

$$\frac{x^r e^{-x}}{\Gamma(r + 1)}$$

To my surprise, the integral

$$\int_0^\infty \frac{x^{r}e^{-x}}{\Gamma(r + 1)} dx = 1$$

At least, that's what I deduce from my tests of 0.5, 1.5, 1.2, and 7.9. I have done so non-algebraically, only numerically.

So my question is; is this a valid form of distribution? And should we perhaps replace the one currently on wikipedia stating the requirement of k to be a natural number?

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You have discovered the Gamma Distribution. Much used. –  André Nicolas Jul 6 '14 at 18:08
@AndréNicolas I did not know of this. I question the usage of the $\theta$ variable here though. –  Bourgond Aries Jul 6 '14 at 18:11
The gamma distribution is more general, since it has the extra parameter $\theta$. When we pick $\theta=1$, your distribution emerges. –  I like Serena Jul 6 '14 at 18:13
Someone should submit it as answer so I can accept it. I shall now study the gamma distribution. –  Bourgond Aries Jul 6 '14 at 18:15
In case you haven't noticed: To verify the integral "algebraically" instead of numerically, multiply both sides by $\Gamma(r+1)$, and note that the left-hand side is the definition of the Gamma function. –  angryavian Jul 6 '14 at 18:15

"The distribution's domain" is a phrase I would not have understood if you hadn't provided some context. The distribution's support is the set of values the random variable can take, or in the case of continuous distributions, the closure of that set.[Later edit: see PS below.] For the Poisson distribution whose probability mass function is $$k\mapsto\frac{x^k e^{-x}}{k!}$$ the support is given by $$k\in\{0,1,2,3,\ldots\},$$ the set of non-negative integers.

The set of possible values of $x$ is $[0,\infty)=\{x : x\ge 0\}$. That is the parameter space, not for a distribution, but for a family of distributions.

You leave us to infer, rather than explicitly saying, that $0.5$, $1.5$, $1.2$, and $7.9$ are values of $r$ rather than of $x$.

The fact that $$\int_0^\infty \frac{x^r e^{-x}}{\Gamma(r + 1)} dx = 1$$ is often taken to be the definition of the Gamma function, although the form in which it is most often stated is $$\Gamma(s) = \int_0^\infty x^{s-1} e^{-x}\,dx,$$ so the $s$ here is $r+1$. A continuous distribution called the Gamma distribution has probability density function $$x\mapsto\frac{x^{s-1} e^{-x}}{\Gamma(s)} \text{ for }x>0,$$ and if one rescales it, putting $x/\lambda$ in place of $x$, one has the density $$x\mapsto\frac{(x/\lambda)^{s-1} e^{-x/\lambda}}{\Gamma(s)}\cdot\frac 1 \lambda \text{ for } x>0$$ and that is also considered a "Gamma distribution". The extra $1/\lambda$ at the end comes from the chain rule. I like to write it is $$\frac{(x/\lambda)^{s-1} e^{-x/\lambda}}{\Gamma(s)}\cdot\frac {dx} \lambda\text{ for }x>0.\tag 1$$

A connection between the Gamma distribution and the Poisson distribution is this: suppose the number of "occurrences" during a particular time interval has a Poisson distribution whose expected value is $\lambda$ times the amount of time, and that the numbers of "occurrences" in different time intervals are independent if the time intervals don't overlap. In that case, the time you wait until the $s$th occurence has the Gamma distribution $(1)$.

PS: I was not too precise about what "support" means". For distributions for which the set of values is topologically discrete, the support is indeed the set of values that can be attained. Thus for the Poisson distribution it is $\{0,1,2,3,\ldots\}$. But technically a "discrete distribution" is one for which all the probability is in point masses, and that need not be discrete in the topological sense: it could, for example, be the set of all rational numbers. So here's a definition: The support of a distribution is the smallest closed set whose complement has probability $0$. For the Poisson distribution, the set of possible values is $\{0,1,2,3,\ldots\}$, and that's already a closed set, so we don't need to take its closure and we get no complications or subtleties.

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You have discovered the Gamma Distribution, actually an inessentially special case of it. The distribution has many uses.

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