Difference between revisions of "Geometric distribution"

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{{Stub page|grade=A*|msg=It's crap, look at it}}
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{{Stub page|grade=C|msg=Removed previous stub message and demoted [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 15:14, 16 January 2018 (UTC)}}
{{Dire page|msg=The content is iffy, and uses a weird convention where geometric is time to first failure. New page content at [[Geometric distribution2]]}}
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{{/Infobox}}
{{Infobox
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{{ProbMacros}}
|style=max-width:25em;
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__TOC__
|title=Geometric Distribution
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==Definition==
|above=<span style="font-size:1.5em;">{{M|X\sim\text{Geo}(p)}}</span><br/>
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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".
<span style="font-size:0.75em;">''for {{M|p}} the [[probability (event)|probability]] of each trials' success''</span>
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|subheader={{M|X\eq k}} means that the first failure occurred on the {{M|k^\text{th} }} trial, {{M|k\in\mathbb{N}_{\ge 1} }}
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|header1=Definition
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|label1=Defined over
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|data1={{M|X}} may take values in {{M|\mathbb{N}_{\ge 0}\eq\{1,2,\ldots\} }}
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|label2=[[probability mass function|p.m.f]]
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|data2={{nowrap|{{M|\mathbb{P}[X\eq k]:\eq p^{k-1}(1-p)}}}}
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|label3={{nowrap|[[cumulative density function|c.d.f]] / [[cumulative mass function|c.m.f]]<ref group="Note">Do we make this distinction for cumulative distributions?</ref>}}
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|data3={{M|\mathbb{P}[X\le k]\eq 1-p^k}}
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|label4=''[[corollary|cor:]]''
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|data4={{M|\mathbb{P}[X\ge k]\eq p^{k-1} }}<!--
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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} }}.
  
EXP and Var
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As such:
-->
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* {{M|\P{X\eq k} :\eq (1-p)^{k-1}p}} - {{link|pmf|statistics}} / {{link|pdf|statistics}} - '''''Claim 1''''' below
|header10=Properties
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* {{M|\mathbb{P}[X\le k]\eq 1-(1-p)^k}} - {{link|cdf|statistics}} - '''''Claim 2''''' below
|label10=[[Expectation]]:
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** {{M|\mathbb{P}[X\ge k]\eq (1-p)^{k-1} }} - an obvious extension.
|data10={{MM|\mathbb{E}[X]\eq\frac{1}{p} }}
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==Convention notes==
|label11=[[Variance]]:
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{{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:
|data11={{MM|\text{Var}(X)\eq\frac{1-p}{p^2} }}
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# {{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
}}
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# {{M|X_2:\eq X-1}} - the number of trials BEFORE first success
__TOC__
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# {{M|X_3:\eq X_1-1}} - the number of trials BEFORE first failure
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Document and explain [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 03:17, 16 January 2018 (UTC)}}
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==Properties==
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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'':
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* {{MM|\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}}
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** '''Proof: ''' ''[[Expectation of the geometric distribution]]''
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* {{MM|\Var{X}\eq\frac{1-p}{p^2} }} for {{M|p\in(0,1]}} and like for expectation we ''tentatively'' define is as {{M|+\infty}} for {{M|p\eq 0}}
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** '''Proof: ''' ''[[Variance of the geometric distribution]]''
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===To do: ===
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# [[Mdm of the geometric distribution]]
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==Proof of claims==
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===Claim 1: {{M|\P{X\eq k}\eq (1-p)^{k-1} p }}===
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{{XXX|This requires improvement, it was copy and pasted from some notes}}
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* {{M|\P{X\eq k} :\eq (1-p)^{k-1}p}} - which is derived as folllows:
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** {{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} }}
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*** Using that the {{M|X_i}} are [[independent random variables]] we see:
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**** {{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} }}
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****: {{MM|\eq (1-p)^{k-1} p}} as they all have the same distribution, namely {{M|X_i\sim\text{Borv}(p)}}
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===Claim 2: {{M|\mathbb{P}[X\le k]\eq 1-(1-p)^k}}===
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{{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) }}
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==See also==
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* [[Expectation of the geometric distribution]]
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* [[Variance of the geometric distribution]]
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* [[Mdm of the geometric distribution]]
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===Distributions===
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* [[Binomial distribution]]
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* [[Exponential distribution]]
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** [[Obtaining the exponential distribution from the geometric distribution]]
 
==Notes==
 
==Notes==
during proof of {{M|\mathbb{P}[X\le k]}} the result is obtained using a [[geometric series]], however one has to align the sequences (not adjust the sum to start at zero, unless you adjust the {{M|S_n}} formula too!)
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<references group="Note"/>
 
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Check the variance, I did part the proof, checked the [[MEI formula book]] and moved on, I didn't confirm interpretation.
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Make a note that my Casio calculator uses {{M|1-p}} as the parameter, giving {{M|\mathbb{P}[X\eq k]:\eq (1-p)^{k-1}p}} along with the interpretation that allows 0
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==Definition==
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==References==
 
==References==
 
<references/>
 
<references/>
==Notes==
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{{Fundamental probability distributions navbox|show}}
<references group="Note"/>
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Latest revision as of 15:14, 16 January 2018

Stub grade: C
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Removed previous stub message and demoted Alec (talk) 15:14, 16 January 2018 (UTC)
Geometric Distribution
[ilmath]X\sim\text{Geo}(p)[/ilmath]

for [ilmath]p[/ilmath] the probability of each trials' success

[ilmath]X\eq k[/ilmath] means that the first success occurred on the [ilmath]k^\text{th} [/ilmath] trial, [ilmath]k\in\mathbb{N}_{\ge 1} [/ilmath]
Definition
Defined over [ilmath]X[/ilmath] may take values in [ilmath]\mathbb{N}_{\ge 1}\eq\{1,2,\ldots\} [/ilmath]
p.m.f [ilmath]\mathbb{P}[X\eq k]:\eq (1-p)^{k-1}p[/ilmath]
c.d.f / c.m.f[Note 1] [ilmath]\mathbb{P}[X\le k]\eq 1-(1-p)^k[/ilmath]
cor: [ilmath]\mathbb{P}[X\ge k]\eq (1-p)^{k-1} [/ilmath]
Properties
Expectation: [math]\mathbb{E}[X]\eq\frac{1}{p} [/math][1]
Variance: [math]\text{Var}(X)\eq\frac{1-p}{p^2} [/math][2]
[ilmath]\newcommand{\P}[2][]{\mathbb{P}#1{\left[{#2}\right]} } \newcommand{\Pcond}[3][]{\mathbb{P}#1{\left[{#2}\!\ \middle\vert\!\ {#3}\right]} } \newcommand{\Plcond}[3][]{\Pcond[#1]{#2}{#3} } \newcommand{\Prcond}[3][]{\Pcond[#1]{#2}{#3} }[/ilmath]
[ilmath]\newcommand{\E}[1]{ {\mathbb{E}{\left[{#1}\right]} } } [/ilmath][ilmath]\newcommand{\Mdm}[1]{\text{Mdm}{\left({#1}\right) } } [/ilmath][ilmath]\newcommand{\Var}[1]{\text{Var}{\left({#1}\right) } } [/ilmath][ilmath]\newcommand{\ncr}[2]{ \vphantom{C}^{#1}\!C_{#2} } [/ilmath]

Definition

Consider a potentially infinite sequence of [ilmath]\text{Borv} [/ilmath] variables, [ilmath] ({ X_i })_{ i = 1 }^{ n } [/ilmath], each independent and identically distributed (i.i.d) with [ilmath]X_i\sim[/ilmath][ilmath]\text{Borv} [/ilmath][ilmath](p)[/ilmath], so [ilmath]p[/ilmath] is the probability of any particular trial being a "success".

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

As such:

  • [ilmath]\P{X\eq k} :\eq (1-p)^{k-1}p[/ilmath] - pmf / pdf - Claim 1 below
  • [ilmath]\mathbb{P}[X\le k]\eq 1-(1-p)^k[/ilmath] - cdf - Claim 2 below
    • [ilmath]\mathbb{P}[X\ge k]\eq (1-p)^{k-1} [/ilmath] - an obvious extension.

Convention notes

Grade: A**
This page requires some work to be carried out
Some aspect of this page is incomplete and work is required to finish it
The message provided is:
If [ilmath]X\sim\text{Geo}(p)[/ilmath] is defined as above then there are 3 other conventions I've seen:
  1. [ilmath]X_1\sim\text{Geo}(1-p)[/ilmath] in our terminology, they would write [ilmath]\text{Geo}(p)[/ilmath], which measures "trials until first failure" instead of success as we do
  2. [ilmath]X_2:\eq X-1[/ilmath] - the number of trials BEFORE first success
  3. [ilmath]X_3:\eq X_1-1[/ilmath] - the number of trials BEFORE first failure
Document and explain Alec (talk) 03:17, 16 January 2018 (UTC)

Warning:That grade doesn't exist!

Properties

For [ilmath]p\in[0,1]\subseteq\mathbb{R} [/ilmath] and [ilmath]X\sim\text{Geo}(p)[/ilmath] we have the following results about the geometric distribution:

  • [math]\E{X}\eq\frac{1}{p} [/math] for [ilmath]p\in(0,1][/ilmath] and is undefined or tentatively defined as [ilmath]+\infty[/ilmath] if [ilmath]p\eq 0[/ilmath]
  • [math]\Var{X}\eq\frac{1-p}{p^2} [/math] for [ilmath]p\in(0,1][/ilmath] and like for expectation we tentatively define is as [ilmath]+\infty[/ilmath] for [ilmath]p\eq 0[/ilmath]

To do:

  1. Mdm of the geometric distribution

Proof of claims

Claim 1: [ilmath]\P{X\eq k}\eq (1-p)^{k-1} p [/ilmath]

TODO: This requires improvement, it was copy and pasted from some notes
  • [ilmath]\P{X\eq k} :\eq (1-p)^{k-1}p[/ilmath] - which is derived as folllows:
    • [ilmath]\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} [/ilmath]
      • Using that the [ilmath]X_i[/ilmath] are independent random variables we see:
        • [math]\P{X\eq k}\eq \left(\prod^{k-1}_{i\eq 1}\P{X_i\eq 0}\right)\times \P{X_k\eq 1} [/math]
          [math]\eq (1-p)^{k-1} p[/math] as they all have the same distribution, namely [ilmath]X_i\sim\text{Borv}(p)[/ilmath]

Claim 2: [ilmath]\mathbb{P}[X\le k]\eq 1-(1-p)^k[/ilmath]

Grade: A**
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Trivial to do, direct application of Geometric series result Alec (talk) 03:17, 16 January 2018 (UTC)

See also

Distributions

Notes

  1. Do we make this distinction for cumulative distributions?

References

  1. See Expectation of the geometric distribution
  2. See Variance of the geometric distribution