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.

Initial notes

Here we will use XPoi(λ)

for λR>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 kN1, 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):=P[Tt]=1P[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]=P[the number of events that occurred for the range of time [0,t]<k]
        =P[the number of events that occurred for the range of time [0,t]k1]


Lastly,

  • If λ is the rate of events per unit time, then for t units of time tλ is the rate of events per t-units of time, so we define:
    • XtPoi(tλ)
  • And we observe:
    • P[the number of events that occurred for the range of time [0,t]k1]
      =P[Xtk1]


We have now discovered:

  • F(t)=P[Tt]=1P[Xtk1], for kN1 remember

Evaluation

We now compute P[Tt]

  • Let's start with P[Tt]=1P[Xtk1]
    • P[Tt]
      =1(k1i=0P[Xt=i])
      - notice the sum starts at i=0, as k1 we must at least have one term, the (i=0)th one.
      =1(k1i=0etλ(tλ)ii!)
      =1etλetλk1i=1(tλ)ii!
      - note the sum now may be zero if k=1 otherwise it will have terms.
    • However we will write it as:
      • P[Tt]=1eλt+k1i=1λii!(tieλt)
        which will help greatly with the next step
  • We now must differentiate this to find the pdf, f(t)=F(t)=ddt[F(t)]|t
    • f(t)=0ddt[eλt]|t+k1i=1λii!(ddt[tieλt]|t)
      • Let us look at the components
        • First, ddt[eλt]|t=λeλt by
          TODO: link thing here that talks about differentiation of e raised to a power of a function of the variable
          - involves chain rule
        • Next we apply the product rule to ddt[tieλt]|t
          =tiddt[eλt]|t+eλtddt[ti]|t
          =λtieλt+ieλtti1 - this uses the first result
          =eλt(iti1λti)
      • thus: f(t)
        =λeλt+k1i=1λii!eλt(iti1λti)

Notes

  1. Jump up Standard cdf stuff