Difference between revisions of "Notes:Poisson and Gamma distribution"

Revision as of 18:13, 19 January 2018 (view source) Alec (Talk | contribs) (Changing computer, so saving work)		Latest revision as of 18:58, 19 January 2018 (view source) Alec (Talk | contribs) (Saving work, want a second set of eyes before proceeding)
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	* {{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		* {{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==		==Evaluation==
−	We now compute {{M|\P{T\le t} }}	+	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} }}		* Let's start with {{M|\P{T\le t} \eq 1-\P{X_t\le k-1} }}
−	*: {{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.	+	** {{M|\P{T\le t} }}
−	*: {{MM|\eq 1-\left(\sum^{k-1}_{i\eq 0}e^{-t\lambda}\frac{(t\lambda)^i}{i!}\right)}}	+	**: {{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==		==Notes==
	<references group="Note"/>		<references group="Note"/>

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Latest revision as of 18:58, 19 January 2018

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

$\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}$$\newcommand{\d}[0]{\mathrm{d} }$$\newcommand{\ddt}[1]{\frac{\d}{\d t}\left[{#1}\middle]\right\vert_{t} }$

• Let's start with $\P{T\le t} \eq 1-\P{X_t\le k-1}$
  • $\P{T\le t}$

    $$\eq 1-\left(\sum^{k-1}_{i\eq 0}\P{X_t\eq i}\right)$$ - 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.

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

    $$\eq 1-e^{-t\lambda}-e^{-t\lambda}\sum^{k-1}_{i\eq 1}\frac{(t\lambda)^i}{i!}$$ - note the sum now may be zero if $k\eq 1$ otherwise it will have terms.

  • However we will write it as:
    • $$\P{T\le t}\eq 1-e^{-\lambda t}+\sum^{k-1}_{i\eq 1}\frac{-\lambda^i}{i!}\big(t^i\cdot e^{-\lambda t}\big)$$ which will help greatly with the next step
• We now must differentiate this to find the pdf, $$f(t)\eq F'(t)\eq \frac{\mathrm{d} }{\mathrm{d} t}\Big[F(t)\Big]\Big\vert_{t}$$
  • $$f(t)\eq 0-\ddt{e^{-\lambda t} }+\sum^{k-1}_{i\eq 1}\frac{-\lambda^i}{i!}\left(\ddt{t^i\cdot e^{-\lambda t} }\right)$$
    • Let us look at the components
      • First, $\ddt{e^{-\lambda t} } \eq -\lambda e^{-\lambda t}$ 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 $\ddt{t^i\cdot e^{-\lambda t} }$

        $\eq t^i\ddt{e^{-\lambda t} }+e^{-\lambda t}\ddt{t^i}$

        $\eq -\lambda t^i e^{-\lambda t}+ie^{-\lambda t}t^{i-1}$ - this uses the first result

        $\eq e^{-\lambda t}\big(it^{i-1}-\lambda t^i\big)$

    • thus: $f(t)$

      $$\eq \lambda e^{-\lambda t}+\sum^{k-1}_{i\eq 1}\frac{-\lambda^i}{i!}e^{-\lambda t}(it^{i-1}-\lambda t^i)$$

Notes

1. Jump up ↑ Standard cdf stuff