# Geometric distribution

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Removed previous stub message and demoted Alec (talk) 15:14, 16 January 2018 (UTC)
Definition 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] [ilmath]X[/ilmath] may take values in [ilmath]\mathbb{N}_{\ge 1}\eq\{1,2,\ldots\} [/ilmath] [ilmath]\mathbb{P}[X\eq k]:\eq (1-p)^{k-1}p[/ilmath] [ilmath]\mathbb{P}[X\le k]\eq 1-(1-p)^k[/ilmath] [ilmath]\mathbb{P}[X\ge k]\eq (1-p)^{k-1} [/ilmath] $\mathbb{E}[X]\eq\frac{1}{p}$[1] $\text{Var}(X)\eq\frac{1-p}{p^2}$[2]

## 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)

## 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:

• $\E{X}\eq\frac{1}{p}$ for [ilmath]p\in(0,1][/ilmath] and is undefined or tentatively defined as [ilmath]+\infty[/ilmath] if [ilmath]p\eq 0[/ilmath]
• $\Var{X}\eq\frac{1-p}{p^2}$ 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:
• $\P{X\eq k}\eq \left(\prod^{k-1}_{i\eq 1}\P{X_i\eq 0}\right)\times \P{X_k\eq 1}$
$\eq (1-p)^{k-1} p$ 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]

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Trivial to do, direct application of Geometric series result Alec (talk) 03:17, 16 January 2018 (UTC)