Poisson mixture

[ilmath]\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} }[/ilmath]

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

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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 [ilmath]X\sim\text{Poi}(\lambda)[/ilmath] and let [ilmath](X_i)_{i\in\mathbb{N} } [/ilmath] be the events detected by this "process". Let [ilmath](D_i)_{i\in\mathbb{N} } [/ilmath] be the detection of each event, i.i.d and [ilmath]\forall i\in\mathbb{N}\big[D_i\sim\text{Borv}(p)\big][/ilmath] then [ilmath]Y\sim\text{Poi}(p\lambda)[/ilmath] is the distribution of the [ilmath](D_i)_{i\in\mathbb{N} } [/ilmath] events.
    • This is not phrased very well.

Definition

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

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

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

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

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

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

• [ilmath]x:\eq x_c[/ilmath]

Special case: [ilmath]n\eq 2[/ilmath]

Let [ilmath]\P{\mathcal{C}\eq 1}:\eq p[/ilmath], thus [ilmath]\P{\mathcal{C}\eq 2}\eq 1-p[/ilmath], then:

• [math]\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)[/math]

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