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
Stub grade: C
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:
| Geometric Distribution | |
| X∼Geo(p) for p the probability of each trials' success | |
| X=k means that the first success occurred on the kth trial, k∈N≥1 | |
| Definition | |
|---|---|
| Defined over | X may take values in N≥1={1,2,…} |
| p.m.f | P[X=k]:=(1−p)k−1p |
| c.d.f / c.m.f[Note 1] | P[X≤k]=1−(1−p)k |
| cor: | P[X≥k]=(1−p)k−1 |
| Properties | |
| Expectation: | E[X]=1p [1]
|
| Variance: | Var(X)=1−pp2 [2]
|
Contents
[hide]Definition
Consider a potentially infinite sequence of Borv variables, (Xi)ni=1, each independent and identically distributed (i.i.d) with Xi∼Borv(p), so p is the probability of any particular trial being a "success".
The geometric distribution models the probability that the first success occurs on the kth trial, for k∈N≥1.
As such:
- P[X=k]:=(1−p)k−1p - pmf / pdf - Claim 1 below
- P[X≤k]=1−(1−p)k - cdf - Claim 2 below
- P[X≥k]=(1−p)k−1 - 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
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Warning:That grade doesn't exist!
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If X∼Geo(p) is defined as above then there are 3 other conventions I've seen:
- X1∼Geo(1−p) in our terminology, they would write Geo(p), which measures "trials until first failure" instead of success as we do
- X2:=X−1 - the number of trials BEFORE first success
- X3:=X1−1 - the number of trials BEFORE first failure
Warning:That grade doesn't exist!
Properties
For p∈[0,1]⊆R and X∼Geo(p) we have the following results about the geometric distribution:
- E[X]=1pfor p∈(0,1] and is undefined or tentatively defined as +∞ if p=0
- Var(X)=1−pp2for p∈(0,1] and like for expectation we tentatively define is as +∞ for p=0
To do:
Proof of claims
Claim 1: P[X=k]=(1−p)k−1p
TODO: This requires improvement, it was copy and pasted from some notes
- P[X=k]:=(1−p)k−1p - which is derived as folllows:
- P[X=k]:=(P[X1=0]×P[X2=0 | X1=0]×⋯×P[Xk−1=0 | X1=0∩X2=0∩⋯∩Xk−2=0])×P[Xk=1 | X1=0∩X2=0∩⋯∩Xk−1=0]
- Using that the Xi are independent random variables we see:
- P[X=k]=(k−1∏i=1P[Xi=0])×P[Xk=1]
- =(1−p)k−1pas they all have the same distribution, namely Xi∼Borv(p)
- =(1−p)k−1p
- P[X=k]=(k−1∏i=1P[Xi=0])×P[Xk=1]
- Using that the Xi are independent random variables we see:
- P[X=k]:=(P[X1=0]×P[X2=0 | X1=0]×⋯×P[Xk−1=0 | X1=0∩X2=0∩⋯∩Xk−2=0])×P[Xk=1 | X1=0∩X2=0∩⋯∩Xk−1=0]
Claim 2: P[X≤k]=1−(1−p)k
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
- Expectation of the geometric distribution
- Variance of the geometric distribution
- Mdm of the geometric distribution
Distributions
Notes
- Jump up ↑ Do we make this distinction for cumulative distributions?
References
- Jump up ↑ See Expectation of the geometric distribution
- Jump up ↑ See Variance of the geometric distribution
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