Difference between revisions of "Geometric distribution"
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| − | {{Stub page|grade=A*|msg=It's crap, look at it | + | {{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) | ||
| + | * Partial expectation proof to be found at [[Geometric distribution2]] page.}} | ||
{{Infobox | {{Infobox | ||
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|above=<span style="font-size:1.5em;">{{M|X\sim\text{Geo}(p)}}</span><br/> | |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> | <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 | + | |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 | |header1=Definition | ||
|label1=Defined over | |label1=Defined over | ||
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|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>}} | ||
| − | }} | + | }}{{ProbMacros}} |
__TOC__ | __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". | ||
| + | |||
| + | The geometric distribution models the probability that the ''first'' success occurs on the {{M|k^\text{th} }} trial. | ||
| + | |||
| + | As such: | ||
| + | * {{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)}} | ||
| + | ==Convention 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!) | 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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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 | 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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| − | |||
| − | |||
==Notes== | ==Notes== | ||
<references group="Note"/> | <references group="Note"/> | ||
| + | ==References== | ||
| + | <references/> | ||
| + | {{Fundamental probability distributions navbox|show}} | ||
Revision as of 03:35, 15 January 2018
Stub grade: A*
This page is a stub
This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
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"
- 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] |
| 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.
As such:
- [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]
Convention notes
during proof of [ilmath]\mathbb{P}[X\le k][/ilmath] 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 [ilmath]S_n[/ilmath] formula too!)
Check the variance, I did part the proof, checked the MEI formula book and moved on, I didn't confirm interpretation.
Make a note that my Casio calculator uses [ilmath]1-p[/ilmath] as the parameter, giving [ilmath]\mathbb{P}[X\eq k]:\eq (1-p)^{k-1}p[/ilmath] along with the interpretation that allows 0
Notes
- ↑ Do we make this distinction for cumulative distributions?
- ↑ 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
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