Notes:Poisson and Gamma distribution

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

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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:
  • $F(t):\eq\P{T\le t} \eq 1-\P{T>t}$ 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.

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