Difference between revisions of "Geometric distribution"

Revision as of 03:35, 15 January 2018 (view source) Alec (Talk | contribs) (ERROR IN INFO-BOX fixed, moved page about, so forth.) ← Older edit		Latest revision as of 15:14, 16 January 2018 (view source) Alec (Talk | contribs) m (Removing stub message as those issues have been addressed.)
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−	{{Stub page|grade=A*|msg=It's crap, look at it.	+	{{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 notice: "The content is iffy, and uses a weird convention where geometric is time to first failure. New page content at [[Geometric distribution2]]"	+	{{/Infobox}}
−	** Dire notice removed [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 03:35, 15 January 2018 (UTC)	+	{{ProbMacros}}
−	* Partial expectation proof to be found at [[Geometric distribution2]] page.}}	+
−	{{Infobox	+
−	|style=max-width:25em;	+
−	|title=Geometric Distribution	+
−	|above=<span style="font-size:1.5em;">{{M|X\sim\text{Geo}(p)}}</span><br/>	+
−	<span style="font-size:0.75em;">''for {{M|p}} the [[probability (event)|probability]] of each trials' success''</span>	+
−	|subheader={{M|X\eq k}} means that the first success occurred on the {{M|k^\text{th} }} trial, {{M|k\in\mathbb{N}_{\ge 1} }}	+
−	|header1=Definition	+
−	|label1=Defined over	+
−	|data1={{M|X}} may take values in {{M|\mathbb{N}_{\ge 1}\eq\{1,2,\ldots\} }}	+
−	|label2=[[probability mass function|p.m.f]]	+
−	|data2={{nowrap|{{M|\mathbb{P}[X\eq k]:\eq (1-p)^{k-1}p}}}}	+
−	|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>}}	+
−	|data3={{M|\mathbb{P}[X\le k]\eq 1-(1-p)^k}}	+
−	|label4=''[[corollary|cor:]]''	+
−	|data4={{M|\mathbb{P}[X\ge k]\eq (1-p)^{k-1} }}<!--	+
−		+
−		+
−	EXP and Var	+
−	-->	+
−	|header10=Properties	+
−	|label10=[[Expectation]]:	+
−	|data10={{MM|\mathbb{E}[X]\eq\frac{1}{p} }}	+
−	|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>}}	+
−	}}{{ProbMacros}}	+
	__TOC__		__TOC__
	==Definition==		==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".		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'':
		+	* {{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}}
		+	** '''Proof: ''' ''[[Expectation of the geometric distribution]]''
		+	* {{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}}
		+	** '''Proof: ''' ''[[Variance of the geometric distribution]]''
		+	===To do: ===
		+	# [[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"/>

---

Latest revision as of 15:14, 16 January 2018

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{\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

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]
  • Proof: Expectation of the geometric distribution
• [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]
  • Proof: Variance of the geometric distribution

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]

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

1. ↑ Do we make this distinction for cumulative distributions?

References

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

v • d • e Fundamental Probability Distributions This navbox is being designed - it is not yet stable! - see Hierarchy of fundamental probability distributions' Fundamental Primitives [ilmath]\text{BORV} [/ilmath], [ilmath]\text{MORV} [/ilmath], Uniform distribution, Rectangular distribution (AKA: Continuous uniform distribution) Fundamental distributions Binomial distribution, Geometric distribution, Exponential distribution, Multinominal distribution Fundamental derived distributions Poisson distribution (exponential distribution - is related to the Poisson distribution), Negative binomial distribution Further derived distributions Poisson race, Poisson accumulation Notable distributions Normal distribution (Student-t's distribution) Other distributions Gamma distribution