View source for Notes:Poisson and Gamma distribution

::: '''These are PRELIMINARY NOTES:'''  I got somewhere and don't want to lose the work on paper.
__TOC__{{ProbMacros}}
==Initial notes==
Here we will use {{M|X\sim}}[[Poisson distribution|{{M|\text{Poi} }}]]{{MM|(\lambda)}} for {{M|\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 {{M|k\in\mathbb{N}_{\ge 1} }}, we are interested in the distribution of the ''time until'' {{M|k}} events have accumulated.
* Let {{M|T}} be the ''time'' until {{M|k}} accumulations, so:
** {{M|F(t):\eq\P{T\le t} \eq 1-\P{T>t} }}<ref group="Note">Standard [[cdf]] stuff</ref> and we can use {{M|\P{T>t} }} to be "fewer events than {{M|k}} occurred for the range of time {{M|[0,t]}}" or to be more formal / specific:
*** {{M|\P{T>t}\eq\P{\text{the number of events that occurred for the range of time }[0,t]<k} }}
***: {{M|\eq\P{\text{the number of events that occurred for the range of time }[0,t]\le k-1} }}


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


We have now discovered:
* {{M|F(t)\eq\P{T\le t}\eq 1-\P{X_t\le k-1} }}, for {{M|k\in\mathbb{N}_{\ge 1} }} remember
==Evaluation==
We now compute {{M|\P{T\le t} }}{{M|\newcommand{\d}[0]{\mathrm{d} } }}{{M|\newcommand{\ddt}[1]{\frac{\d}{\d t}\left[{#1}\middle]\right\vert_{t} } }}
* Let's start with {{M|\P{T\le t} \eq 1-\P{X_t\le k-1} }}
** {{M|\P{T\le t} }}
**: {{MM|\eq 1-\left(\sum^{k-1}_{i\eq 0}\P{X_t\eq i}\right)}} - notice the sum starts at {{M|i\eq 0}}, as {{M|k\ge 1}} we must at least have one term, the {{M|(i\eq 0)^\text{th} }} one.
**: {{MM|\eq 1-\left(\sum^{k-1}_{i\eq 0}e^{-t\lambda}\frac{(t\lambda)^i}{i!}\right)}}
**: {{MM|\eq 1-e^{-t\lambda}-e^{-t\lambda}\sum^{k-1}_{i\eq 1}\frac{(t\lambda)^i}{i!} }} - note the sum now may be zero if {{M|k\eq 1}} otherwise it will have terms.
** However we will write it as:
*** {{MM|\P{T\le t}\eq 1-e^{-\lambda t}+\sum^{k-1}_{i\eq 1}\frac{-\lambda^i}{i!}\big(t^i\cdot e^{-\lambda t}\big)}} which will help greatly with the next step
* We now must [[differentiate]] this to find the [[pdf]], {{MM|f(t)\eq F'(t)\eq \frac{\mathrm{d} }{\mathrm{d} t}\Big[F(t)\Big]\Big\vert_{t} }}
** {{MM|f(t)\eq 0-\ddt{e^{-\lambda t} }+\sum^{k-1}_{i\eq 1}\frac{-\lambda^i}{i!}\left(\ddt{t^i\cdot e^{-\lambda t} }\right) }}
*** Let us look at the components
**** First, {{M|\ddt{e^{-\lambda t} } \eq -\lambda e^{-\lambda t} }} by {{XXX|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 {{M|\ddt{t^i\cdot e^{-\lambda t} } }}
****: {{M|\eq t^i\ddt{e^{-\lambda t} }+e^{-\lambda t}\ddt{t^i} }}
****: {{M|\eq -\lambda t^i e^{-\lambda t}+ie^{-\lambda t}t^{i-1} }} - this uses the first result
****: {{M|\eq e^{-\lambda t}\big(it^{i-1}-\lambda t^i\big)}}
*** thus: {{M|f(t)}}
***: {{MM|\eq \lambda e^{-\lambda t}+\sum^{k-1}_{i\eq 1}\frac{-\lambda^i}{i!}e^{-\lambda t}(it^{i-1}-\lambda t^i) }}
==Notes==
<references group="Note"/>