Difference between revisions of "Geometric distribution"

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* Dire notice: "The content is iffy, and uses a weird convention where geometric is time to first failure. New page content at [[Geometric distribution2]]"
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** Dire notice removed [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 03:35, 15 January 2018 (UTC)
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*** Deleted [[Geometric distribution2]] as no longer needed [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 03:22, 16 January 2018 (UTC)
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* {{strike|Partial expectation proof to be found at [[Geometric distribution2]] page.}}
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** Turns out there was an [[Expectation of the geometric distribution]] page - I feel silly [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 03:22, 16 January 2018 (UTC)}}
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Latest revision as of 15:14, 16 January 2018

Stub grade: C
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Removed previous stub message and demoted Alec (talk) 15:14, 16 January 2018 (UTC)
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{\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

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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:

  • [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:

  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]

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?

References

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