Difference between revisions of "Geometric distribution"

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(Overhaul of page, marked out work to do, good progress made - STILL TO DO pull infobox out and put in /infobox - SAVING WORK)
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|header10=Properties
 
|header10=Properties
 
|label10=[[Expectation]]:
 
|label10=[[Expectation]]:
|data10={{MM|\mathbb{E}[X]\eq\frac{1}{p} }}
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|data10={{MM|\mathbb{E}[X]\eq\frac{1}{p} }}<ref>See ''[[Expectation of the geometric distribution]]''</ref>
 
|label11=[[Variance]]:
 
|label11=[[Variance]]:
 
|data11={{Nowrap|{{XXX|Unknown}}<ref group="Note">Due to different conventions on the definition of geometric (for example {{M|X':\eq X-1}} for my {{M|X}} and another's {{M|X'\sim\text{Geo}(p)}}) or even differing by using {{M|1-p}} in place of {{M|p}} in the {{M|X}} and {{M|X'}} just mentioned - I cannot be sure without working it out that it's {{MM|\frac{1-p}{p^2} }} - I record this value only for a record of what was once there with the correct expectation - DO NOT USE THIS EXPRESSION</ref>}}
 
|data11={{Nowrap|{{XXX|Unknown}}<ref group="Note">Due to different conventions on the definition of geometric (for example {{M|X':\eq X-1}} for my {{M|X}} and another's {{M|X'\sim\text{Geo}(p)}}) or even differing by using {{M|1-p}} in place of {{M|p}} in the {{M|X}} and {{M|X'}} just mentioned - I cannot be sure without working it out that it's {{MM|\frac{1-p}{p^2} }} - I record this value only for a record of what was once there with the correct expectation - DO NOT USE THIS EXPRESSION</ref>}}
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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".
 
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.
+
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:
 
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 (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} }}
 
** {{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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**** {{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|\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)}}  
 
****: {{MM|\eq (1-p)^{k-1} p}} as they all have the same distribution, namely {{M|X_i\sim\text{Borv}(p)}}  
==Convention notes==
+
===Claim 2: {{M|\mathbb{P}[X\le k]\eq 1-(1-p)^k}}===
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!)
+
{{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==
Check the variance, I did part the proof, checked the [[MEI formula book]] and moved on, I didn't confirm interpretation.
+
* [[Expectation of the geometric distribution]]
 
+
* [[Variance of the geometric distribution]]
 
+
* [[Mdm of the geometric distribution]]
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
+
===Distributions===
 +
* [[Binomial distribution]]
 +
* [[Exponential distribution]]
 +
** [[Obtaining the exponential distribution from the geometric distribution]]
 
==Notes==
 
==Notes==
 
<references group="Note"/>
 
<references group="Note"/>

Revision as of 03:17, 16 January 2018

Stub grade: A*
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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 Alec (talk) 03:35, 15 January 2018 (UTC)
  • Partial expectation proof to be found at Geometric distribution2 page.
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:
TODO: Unknown
[Note 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**
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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:

  • [ilmath]\E{X}\eq\frac{1}{p} [/ilmath] for [ilmath]p\in(0,1][/ilmath] and is undefined or tentatively defined as [ilmath]+\infty[/ilmath] if [ilmath]p\eq 0[/ilmath]

To do:

  1. Variance of the geometric distribution
  2. 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?
  2. Due to different conventions on the definition of geometric (for example [ilmath]X':\eq X-1[/ilmath] for my [ilmath]X[/ilmath] and another's [ilmath]X'\sim\text{Geo}(p)[/ilmath]) or even differing by using [ilmath]1-p[/ilmath] in place of [ilmath]p[/ilmath] in the [ilmath]X[/ilmath] and [ilmath]X'[/ilmath] just mentioned - I cannot be sure without working it out that it's [math]\frac{1-p}{p^2} [/math] - I record this value only for a record of what was once there with the correct expectation - DO NOT USE THIS EXPRESSION

References

  1. See Expectation of the geometric distribution