Geometric distribution

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• 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
$X\sim\text{Geo}(p)$ for $p$ the probability of each trials' success
$X\eq k$ means that the first success occurred on the $k^\text{th}$ trial, $k\in\mathbb{N}_{\ge 1}$
Definition
Defined over	$X$ may take values in $\mathbb{N}_{\ge 1}\eq\{1,2,\ldots\}$
p.m.f	$\mathbb{P}[X\eq k]:\eq (1-p)^{k-1}p$
c.d.f / c.m.f	$\mathbb{P}[X\le k]\eq 1-(1-p)^k$
cor:	$\mathbb{P}[X\ge k]\eq (1-p)^{k-1}$
Properties
Expectation:	$$\mathbb{E}[X]\eq\frac{1}{p}$$
Variance:	TODO: Unknown

$\newcommand{\P}[2][]{\mathbb{P}#1{\left[{#2}\right]} } \newcommand{\Pcond}[3][]{\mathbb{P}#1{\left[{#2}\!\ \middle\vert\!\ {#3}\right]} } \newcommand{\Plcond}[3][]{\Pcond[#1]{#2}{#3} } \newcommand{\Prcond}[3][]{\Pcond[#1]{#2}{#3} }$

$\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.

As such:

• $\P{X\eq k} :\eq (1-p)^{k-1}p$ - 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 $X_i\sim\text{Borv}(p)$

Convention notes

during proof of $\mathbb{P}[X\le k]$ the result is obtained using a geometric series, however one has to align the sequences (not adjust the sum to start at zero, unless you adjust the $S_n$ formula too!)

Check the variance, I did part the proof, checked the MEI formula book and moved on, I didn't confirm interpretation.

Make a note that my Casio calculator uses $1-p$ as the parameter, giving $\mathbb{P}[X\eq k]:\eq (1-p)^{k-1}p$ along with the interpretation that allows 0