Notes:Poisson and Gamma distribution

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

Initial notes

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


Lastly,

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


We have now discovered:

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

Evaluation

We now compute \P{T\le t}

  • Let's start with \P{T\le t} \eq 1-\P{X_t\le k-1}
    \eq 1-\left(\sum^{k-1}_{i\eq 0}\P{X_t\eq i}\right) - notice the sum starts at i\eq 0, as k\ge 1 we must at least have one term, the (i\eq 0)^\text{th} one.
    \eq 1-\left(\sum^{k-1}_{i\eq 0}e^{-t\lambda}\frac{(t\lambda)^i}{i!}\right)

Notes

  1. Jump up Standard cdf stuff