Geometric distribution
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| 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 failure 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]
|
Contents
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
Definition
References
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
