View source for Geometric distribution

{{Stub page|grade=A*|msg=It's crap, look at it.
* Dire notice: "The content is iffy, and uses a weird convention where geometric is time to first failure. New page content at [[Geometric distribution2]]"
** Dire notice removed [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 03:35, 15 January 2018 (UTC)
*** Deleted [[Geometric distribution2]] as no longer needed [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 03:22, 16 January 2018 (UTC)
* {{strike|Partial expectation proof to be found at [[Geometric distribution2]] page.}}
** Turns out there was an [[Expectation of the geometric distribution]] page - I feel silly [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 03:22, 16 January 2018 (UTC)}}
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==Definition==
Consider a potentially infinite sequence of [[Borv|{{M|\text{Borv} }}]] variables, {{MSeq|X_i|i|1|n}}, each independent and identically distributed ({{iid}}) with {{M|X_i\sim}}[[Borv|{{M|\text{Borv} }}]]{{M|(p)}}, so {{M|p}} is the [[probability]] of any particular trial being a "success".

The geometric distribution models the probability that the ''first'' success occurs on the {{M|k^\text{th} }} trial, for {{M|k\in\mathbb{N}_{\ge 1} }}.

As such:
* {{M|\P{X\eq k} :\eq (1-p)^{k-1}p}} - {{link|pmf|statistics}} / {{link|pdf|statistics}} - '''''Claim 1''''' below
* {{M|\mathbb{P}[X\le k]\eq 1-(1-p)^k}} - {{link|cdf|statistics}} - '''''Claim 2''''' below
** {{M|\mathbb{P}[X\ge k]\eq (1-p)^{k-1} }} - an obvious extension.
==Convention notes==
{{Requires work|grade=A**|msg=If {{M|X\sim\text{Geo}(p)}} is defined as above then there are 3 other conventions I've seen:
# {{M|X_1\sim\text{Geo}(1-p)}} in our terminology, they would write {{M|\text{Geo}(p)}}, which measures "trials until first failure" instead of success as we do
# {{M|X_2:\eq X-1}} - the number of trials BEFORE first success
# {{M|X_3:\eq X_1-1}} - the number of trials BEFORE first failure
Document and explain [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 03:17, 16 January 2018 (UTC)}}
==Properties==
For {{M|p\in[0,1]\subseteq\mathbb{R} }} and {{M|X\sim\text{Geo}(p)}} we have the following results about the ''geometric distribution'':
* {{M|\E{X}\eq\frac{1}{p} }} for {{M|p\in(0,1]}} and is undefined or ''tentatively'' defined as {{M|+\infty}} if {{M|p\eq 0}}
** '''Proof: ''' ''[[Expectation of the geometric distribution]]''
===To do: ===
# [[Variance of the geometric distribution]]
# [[Mdm of the geometric distribution]]
==Proof of claims==
===Claim 1: {{M|\P{X\eq k}\eq (1-p)^{k-1} p }}===
{{XXX|This requires improvement, it was copy and pasted from some notes}}
* {{M|\P{X\eq k} :\eq (1-p)^{k-1}p}} - which is derived as folllows:
** {{M|\P{X\eq k} :\eq \Big(\P{X_1\eq 0}\times\Pcond{X_2\eq 0}{X_1\eq 0}\times\cdots\times \Pcond{X_{k-1}\eq 0}{X_1\eq 0\cap X_2\eq 0\cap\cdots\cap X_{k-2}\eq 0}\Big)\times\Pcond{X_k\eq 1}{X_1\eq 0\cap X_2\eq 0\cap\cdots\cap X_{k-1}\eq 0} }}
*** Using that the {{M|X_i}} are [[independent random variables]] we see:
**** {{MM|\P{X\eq k}\eq \left(\prod^{k-1}_{i\eq 1}\P{X_i\eq 0}\right)\times \P{X_k\eq 1} }}
****: {{MM|\eq (1-p)^{k-1} p}} as they all have the same distribution, namely {{M|X_i\sim\text{Borv}(p)}} 
===Claim 2: {{M|\mathbb{P}[X\le k]\eq 1-(1-p)^k}}===
{{Requires proof|grade=A**|msg=Trivial to do, direct application of ''[[Geometric series]]'' result [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 03:17, 16 January 2018 (UTC) }}
==See also==
* [[Expectation of the geometric distribution]]
* [[Variance of the geometric distribution]]
* [[Mdm of the geometric distribution]]
===Distributions===
* [[Binomial distribution]]
* [[Exponential distribution]]
** [[Obtaining the exponential distribution from the geometric distribution]]
==Notes==
<references group="Note"/>
==References==
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