Geometric distribution

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It's crap, look at it.

• 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, 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**

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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:

• $\E{X}\eq\frac{1}{p}$ for $p\in(0,1]$ and is undefined or tentatively defined as $+\infty$ if $p\eq 0$
  • Proof: Expectation of the geometric distribution

To do:

1. Variance of the geometric distribution
2. 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

• Expectation of the geometric distribution
• Variance of the geometric distribution
• Mdm of the geometric distribution

Distributions

• Binomial distribution
• Exponential distribution
  • Obtaining the exponential distribution from the geometric distribution