Geometric distribution
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Geometric Distribution | |
X∼Geo(p) for p the probability of each trials' success | |
X=k means that the first success occurred on the kth trial, k∈N≥1 | |
Definition | |
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Defined over | X may take values in N≥1={1,2,…} |
p.m.f | P[X=k]:=(1−p)k−1p |
c.d.f / c.m.f[Note 1] | P[X≤k]=1−(1−p)k |
cor: | P[X≥k]=(1−p)k−1 |
Properties | |
Expectation: | E[X]=1p[1] |
Variance: | Var(X)=1−pp2[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} }
Contents
[hide]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
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If X\sim\text{Geo}(p) is defined as above then there are 3 other conventions I've seen:
- 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
- X_2:\eq X-1 - the number of trials BEFORE first success
- X_3:\eq X_1-1 - the number of trials BEFORE first failure
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Properties
For p\in[0,1]\subseteq\mathbb{R} and X\sim\text{Geo}(p) we have the following results about the geometric distribution:
- \E{X}\eq\frac{1}{p} for p\in(0,1] and is undefined or tentatively defined as +\infty if p\eq 0
- \Var{X}\eq\frac{1-p}{p^2} for p\in(0,1] and like for expectation we tentatively define is as +\infty for p\eq 0
To do:
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)
- \P{X\eq k}\eq \left(\prod^{k-1}_{i\eq 1}\P{X_i\eq 0}\right)\times \P{X_k\eq 1}
- Using that the X_i are independent random variables we see:
- \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}
Claim 2: \mathbb{P}[X\le k]\eq 1-(1-p)^k
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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
Notes
- Jump up ↑ Do we make this distinction for cumulative distributions?
References
- Jump up ↑ See Expectation of the geometric distribution
- Jump up ↑ See Variance of the geometric distribution
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