Poisson mixture

$\newcommand{\P}[2][]{\mathbb{P}#1{\left[{#2}\right]} } \newcommand{\Pcond}[3][]{\mathbb{P}#1{\left[{#2}\!\ \middle\vert\!\ {#3}\right]} } \newcommand{\Plcond}[3][]{\Pcond[#1]{#2}{#3} } \newcommand{\Prcond}[3][]{\Pcond[#1]{#2}{#3} }$

$\newcommand{\E}[1]{ {\mathbb{E}{\left[{#1}\right]} } }$$\newcommand{\Mdm}[1]{\text{Mdm}{\left({#1}\right) } }$$\newcommand{\Var}[1]{\text{Var}{\left({#1}\right) } }$$\newcommand{\ncr}[2]{ \vphantom{C}^{#1}\!C_{#2} }$

Stub grade: A**

This page is a stub

This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:

Warning:The notation and terminology used here was developed by me for a project, it has not been researched yet.

Before we start I must point out something which may without deeper thought appear to be a Poisson mixture, recall that:

• The addition of two Poisson distributions is itself a Poisson distribution, and
• TODO: Link here
  If we have a Poisson distribution and each of its events being noticed i.i.d and BORV(p) then the observed process is also a Poisson distribution
  • That is, let $X\sim\text{Poi}(\lambda)$ and let $(X_i)_{i\in\mathbb{N} }$ be the events detected by this "process". Let $(D_i)_{i\in\mathbb{N} }$ be the detection of each event, i.i.d and $\forall i\in\mathbb{N}\big[D_i\sim\text{Borv}(p)\big]$ then $Y\sim\text{Poi}(p\lambda)$ is the distribution of the $(D_i)_{i\in\mathbb{N} }$ events.
    • This is not phrased very well.

Definition

Let $(\lambda_i)_{i\eq 1}^n\subseteq\mathbb{R}_{\ge 0}$ be given, and let $\big(X_i\sim\text{Poi}(\lambda_i)\big)_{i\eq 1}^n$ be a sequence of $\mathbb{N}$-valued Poisson distributed random variables, then

Let $\mathcal{C}$ be an $N$-valued random variable where $N:\eq\{1,2,3,\ldots,n\}\subseteq\mathbb{N}$¸ next

• Define $X$ to be an $\mathbb{N}$-valued random variable as follows:
  • for any $k\in\mathbb{N}$, define: $$\P{X\eq k}:\eq \sum^n_{i\eq 1}\Big(\P{\mathcal{C}\eq i}\cdot\Pcond{X_i \eq k}{\mathcal{C}\eq i}\Big)$$ - which is just a standard composition
    TODO: Is this law of total probability?
    • Notice: $$\P{X\eq k}\eq\sum^n_{i\eq 1}\Big(\P{\mathcal{C}\eq i}\cdot\P{X_i\eq m}\Big)$$
      TODO: Justification?

      $$\eq \sum^n_{i\eq 1}\left(\P{\mathcal{C}\eq i}\cdot\frac{e^{-\lambda_i}\cdot \lambda_i^k}{k!} \right)$$

      $$\eq \frac{1}{k!}\sum^n_{i\eq 1}\Big(e^{-\lambda_i}\cdot\P{\mathcal{C}\eq i}\cdot \lambda_i^k\Big)$$

The idea is that we first let $(x_i)_{i\eq 1}^n$ be a sample from the $(X_i)_{i\eq 1}^n$ random variables. Let $c$ be a sample of $\mathcal{C}$, then $x$, our sample of $X$ is:

• $x:\eq x_c$

Special case: $n\eq 2$

Let $\P{\mathcal{C}\eq 1}:\eq p$, thus $\P{\mathcal{C}\eq 2}\eq 1-p$, then:

• $$\P{X\eq k}\eq\frac{1}{k!}\Big(p e^{-\lambda_1} \lambda_1^k+(1-p)e^{-\lambda_2}\lambda_2^k\Big)$$

Finding the parameters would be an optimisation problem (minimise error with $3$ parameters here, $\lambda_1,\ \lambda_2\in\mathbb{R}_{\ge 0}$ and $p\in[0,1]\subseteq\mathbb{R}$