Notes:Poisson and Gamma distribution
From Maths
- These are PRELIMINARY NOTES: I got somewhere and don't want to lose the work on paper.
Contents
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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}
- \P{T>t}\eq\P{\text{the number of events that occurred for the range of time }[0,t]<k}
- 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:
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}
- \P{\text{the number of events that occurred for the range of time }[0,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)