Geometric distribution
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 (1p)^{k1}p[/ilmath] 
c.d.f / c.m.f^{[Note 1]}  [ilmath]\mathbb{P}[X\le k]\eq 1(1p)^k[/ilmath] 
cor:  [ilmath]\mathbb{P}[X\ge k]\eq (1p)^{k1} [/ilmath] 
Properties  
Expectation:  [math]\mathbb{E}[X]\eq\frac{1}{p} [/math]^{[1]} 
Variance:  [math]\text{Var}(X)\eq\frac{1p}{p^2} [/math]^{[2]} 
Contents
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 (1p)^{k1}p[/ilmath]  pmf / pdf  Claim 1 below
 [ilmath]\mathbb{P}[X\le k]\eq 1(1p)^k[/ilmath]  cdf  Claim 2 below
 [ilmath]\mathbb{P}[X\ge k]\eq (1p)^{k1} [/ilmath]  an obvious extension.
Convention notes
The message provided is:
 [ilmath]X_1\sim\text{Geo}(1p)[/ilmath] in our terminology, they would write [ilmath]\text{Geo}(p)[/ilmath], which measures "trials until first failure" instead of success as we do
 [ilmath]X_2:\eq X1[/ilmath]  the number of trials BEFORE first success
 [ilmath]X_3:\eq X_11[/ilmath]  the number of trials BEFORE first failure
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{1p}{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:
Proof of claims
Claim 1: [ilmath]\P{X\eq k}\eq (1p)^{k1} p [/ilmath]
 [ilmath]\P{X\eq k} :\eq (1p)^{k1}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_{k1}\eq 0}{X_1\eq 0\cap X_2\eq 0\cap\cdots\cap X_{k2}\eq 0}\Big)\times\Pcond{X_k\eq 1}{X_1\eq 0\cap X_2\eq 0\cap\cdots\cap X_{k1}\eq 0} [/ilmath]
 Using that the [ilmath]X_i[/ilmath] are independent random variables we see:
 [math]\P{X\eq k}\eq \left(\prod^{k1}_{i\eq 1}\P{X_i\eq 0}\right)\times \P{X_k\eq 1} [/math]
 [math]\eq (1p)^{k1} 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^{k1}_{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_{k1}\eq 0}{X_1\eq 0\cap X_2\eq 0\cap\cdots\cap X_{k2}\eq 0}\Big)\times\Pcond{X_k\eq 1}{X_1\eq 0\cap X_2\eq 0\cap\cdots\cap X_{k1}\eq 0} [/ilmath]
Claim 2: [ilmath]\mathbb{P}[X\le k]\eq 1(1p)^k[/ilmath]
The message provided is:
See also
 Expectation of the geometric distribution
 Variance of the geometric distribution
 Mdm of the geometric distribution
Distributions
Notes
 ↑ Do we make this distinction for cumulative distributions?
References
