Difference between revisions of "Geometric distribution"
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− | {{Stub page|grade= | + | {{Stub page|grade=C|msg=Removed previous stub message and demoted [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 15:14, 16 January 2018 (UTC)}} |
− | + | {{/Infobox}} | |
− | + | {{ProbMacros}} | |
− | + | __TOC__ | |
− | | | + | ==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". | |
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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} }}. | ||
− | + | 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 \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== | ==Notes== | ||
− | + | <references group="Note"/> | |
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==References== | ==References== | ||
<references/> | <references/> | ||
− | + | {{Fundamental probability distributions navbox|show}} | |
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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] |
Contents
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
The message provided is:
- [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
- [ilmath]X_2:\eq X-1[/ilmath] - the number of trials BEFORE first success
- [ilmath]X_3:\eq X_1-1[/ilmath] - the number of trials BEFORE first failure
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:
Proof of claims
Claim 1: [ilmath]\P{X\eq k}\eq (1-p)^{k-1} p [/ilmath]
- [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]
- [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]
- Using that the [ilmath]X_i[/ilmath] are independent random variables we see:
- [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]
Claim 2: [ilmath]\mathbb{P}[X\le k]\eq 1-(1-p)^k[/ilmath]
The message provided is:
See also
- Expectation of the geometric distribution
- Variance of the geometric distribution
- Mdm of the geometric distribution
Distributions
Notes
- ↑ Do we make this distinction for cumulative distributions?
References
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