Notes:Poisson and Gamma distribution

These are PRELIMINARY NOTES: I got somewhere and don't want to lose the work on paper.

[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]

Initial notes

Here we will use [ilmath]X\sim[/ilmath][ilmath]\text{Poi} [/ilmath][math](\lambda)[/math] for [ilmath]\lambda\in\mathbb{R}_{>0} [/ilmath] as "events per unit time" for simplicity of conceptualising what's going on, in practice it wont matter what continuous unit is used.

We use the following:

• Let [ilmath]k\in\mathbb{N}_{\ge 1} [/ilmath], we are interested in the distribution of the time until [ilmath]k[/ilmath] events have accumulated.
• Let [ilmath]T[/ilmath] be the time until [ilmath]k[/ilmath] accumulations, so:
  • [ilmath]F(t):\eq\P{T\le t} \eq 1-\P{T>t} [/ilmath] and we can use [ilmath]\P{T>t} [/ilmath] to be "fewer events than [ilmath]k[/ilmath] occurred for the range of time [ilmath][0,t][/ilmath]" or to be more formal / specific:
    • [ilmath]\P{T>t}\eq\P{\text{the number of events that occurred for the range of time }[0,t]<k} [/ilmath]

      [ilmath]\eq\P{\text{the number of events that occurred for the range of time }[0,t]\le k-1} [/ilmath]

Lastly,

• If [ilmath]\lambda[/ilmath] is the rate of events per unit time, then for [ilmath]t[/ilmath] units of time [ilmath]t\lambda[/ilmath] is the rate of events per [ilmath]t[/ilmath]-units of time, so we define:
  • [ilmath]X_t\sim\text{Poi}(t\lambda) [/ilmath]
• And we observe:
  • [ilmath]\P{\text{the number of events that occurred for the range of time }[0,t]\le k-1} [/ilmath]

    [ilmath]\eq \P{X_t\le k-1} [/ilmath]

We have now discovered:

• [ilmath]F(t)\eq\P{T\le t}\eq 1-\P{X_t\le k-1} [/ilmath], for [ilmath]k\in\mathbb{N}_{\ge 1} [/ilmath] remember

Evaluation

We now compute [ilmath]\P{T\le t} [/ilmath]

• Let's start with [ilmath]\P{T\le t} \eq 1-\P{X_t\le k-1} [/ilmath]

  [math]\eq 1-\left(\sum^{k-1}_{i\eq 0}\P{X_t\eq i}\right)[/math] - notice the sum starts at [ilmath]i\eq 0[/ilmath], as [ilmath]k\ge 1[/ilmath] we must at least have one term, the [ilmath](i\eq 0)^\text{th} [/ilmath] one.

  [math]\eq 1-\left(\sum^{k-1}_{i\eq 0}e^{-t\lambda}\frac{(t\lambda)^i}{i!}\right)[/math]