Geometric distribution

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Geometric Distribution
XGeo(p)

for p the probability of each trials' success

X=k means that the first success occurred on the kth trial, kN1
Definition
Defined over X may take values in N1={1,2,}
p.m.f P[X=k]:=(1p)k1p
c.d.f / c.m.f[Note 1] P[Xk]=1(1p)k
cor: P[Xk]=(1p)k1
Properties
Expectation: E[X]=1p[1]
Variance: Var(X)=1pp2[2]
\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} }
\newcommand{\E}[1]{ {\mathbb{E}{\left[{#1}\right]} } } \newcommand{\Mdm}[1]{\text{Mdm}{\left({#1}\right) } } \newcommand{\Var}[1]{\text{Var}{\left({#1}\right) } } \newcommand{\ncr}[2]{ \vphantom{C}^{#1}\!C_{#2} }

Definition

Consider a potentially infinite sequence of \text{Borv} variables, ({ X_i })_{ i = 1 }^{ n } , each independent and identically distributed (i.i.d) with X_i\sim\text{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 k^\text{th} trial, for k\in\mathbb{N}_{\ge 1} .

As such:

  • \P{X\eq k} :\eq (1-p)^{k-1}p - pmf / pdf - Claim 1 below
  • \mathbb{P}[X\le k]\eq 1-(1-p)^k - cdf - Claim 2 below
    • \mathbb{P}[X\ge k]\eq (1-p)^{k-1} - an obvious extension.

Convention notes

Grade: A**
This page requires some work to be carried out
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If X\sim\text{Geo}(p) is defined as above then there are 3 other conventions I've seen:
  1. X_1\sim\text{Geo}(1-p) in our terminology, they would write \text{Geo}(p), which measures "trials until first failure" instead of success as we do
  2. X_2:\eq X-1 - the number of trials BEFORE first success
  3. X_3:\eq X_1-1 - 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 p\in[0,1]\subseteq\mathbb{R} and X\sim\text{Geo}(p) we have the following results about the geometric distribution:

To do:

  1. Mdm of the geometric distribution

Proof of claims

Claim 1: \P{X\eq k}\eq (1-p)^{k-1} p

TODO: This requires improvement, it was copy and pasted from some notes
  • \P{X\eq k} :\eq (1-p)^{k-1}p - which is derived as folllows:
    • \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 X_i 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 X_i\sim\text{Borv}(p)

Claim 2: \mathbb{P}[X\le k]\eq 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

Distributions

Notes

  1. <cite_references_link_accessibility_label> Do we make this distinction for cumulative distributions?

References

  1. <cite_references_link_accessibility_label> See Expectation of the geometric distribution
  2. <cite_references_link_accessibility_label> See Variance of the geometric distribution