Geometric distribution
- Dire notice: "The content is iffy, and uses a weird convention where geometric is time to first failure. New page content at Geometric distribution2"
- Partial expectation proof to be found at Geometric distribution2 page.
| 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 (1-p)^{k-1}p[/ilmath] |
| c.d.f / c.m.f[Note 1] | [ilmath]\mathbb{P}[X\le k]\eq 1-(1-p)^k[/ilmath] |
| cor: | [ilmath]\mathbb{P}[X\ge k]\eq (1-p)^{k-1} [/ilmath] |
| Properties | |
| Expectation: | [math]\mathbb{E}[X]\eq\frac{1}{p} [/math] |
| Variance: | TODO: Unknown [Note 2]
|
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.
As such:
- [ilmath]\P{X\eq k} :\eq (1-p)^{k-1}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_{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} [/ilmath]
- Using that the [ilmath]X_i[/ilmath] are independent random variables we see:
- [math]\P{X\eq k}\eq \left(\prod^{k-1}_{i\eq 1}\P{X_i\eq 0}\right)\times \P{X_k\eq 1} [/math]
- [math]\eq (1-p)^{k-1} 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^{k-1}_{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_{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} [/ilmath]
Convention notes
during proof of [ilmath]\mathbb{P}[X\le k][/ilmath] 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 [ilmath]S_n[/ilmath] 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 [ilmath]1-p[/ilmath] as the parameter, giving [ilmath]\mathbb{P}[X\eq k]:\eq (1-p)^{k-1}p[/ilmath] along with the interpretation that allows 0
Notes
- ↑ Do we make this distinction for cumulative distributions?
- ↑ Due to different conventions on the definition of geometric (for example [ilmath]X':\eq X-1[/ilmath] for my [ilmath]X[/ilmath] and another's [ilmath]X'\sim\text{Geo}(p)[/ilmath]) or even differing by using [ilmath]1-p[/ilmath] in place of [ilmath]p[/ilmath] in the [ilmath]X[/ilmath] and [ilmath]X'[/ilmath] just mentioned - I cannot be sure without working it out that it's [math]\frac{1-p}{p^2} [/math] - I record this value only for a record of what was once there with the correct expectation - DO NOT USE THIS EXPRESSION
References
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