Geometric distribution

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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 Alec (talk) 03:35, 15 January 2018 (UTC)
• 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	[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

[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

Grade: A**

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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:

• [ilmath]\E{X}\eq\frac{1}{p} [/ilmath] 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

To do:

1. Variance of the geometric distribution
2. 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**

This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.

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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

• Binomial distribution
• Exponential distribution
  • Obtaining the exponential distribution from the geometric distribution