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 failure 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 |
Variance: | TODO: Unknown [Note 2]
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Contents
[hide]Notes
during proof of P[X≤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 Sn 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 P[X=k]:=(1−p)k−1p along with the interpretation that allows 0
Definition
References
Notes
- <cite_references_link_accessibility_label> ↑ Do we make this distinction for cumulative distributions?
- <cite_references_link_accessibility_label> ↑ Due to different conventions on the definition of geometric (for example X′:=X−1 for my X and another's X′∼Geo(p)) or even differing by using 1−p in place of p in the X and X′ just mentioned - I cannot be sure without working it out that it's 1−pp2 - I record this value only for a record of what was once there with the correct expectation - DO NOT USE THIS EXPRESSION