Difference between revisions of "Geometric distribution"
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Latest revision as of 15:14, 16 January 2018
Stub grade: C
This page is a stub
This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
| Geometric Distribution | |
| X∼Geo(p) for p the probability of each trials' success | |
| X=k means that the first success occurred on the kth trial, k∈N≥1 | |
| Definition | |
|---|---|
| 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 [1]
|
| Variance: | Var(X)=1−pp2 [2]
|
Contents
[hide]Definition
Consider a potentially infinite sequence of Borv variables, (Xi)ni=1, each independent and identically distributed (i.i.d) with Xi∼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 kth trial, for k∈N≥1.
As such:
- P[X=k]:=(1−p)k−1p - pmf / pdf - Claim 1 below
- P[X≤k]=1−(1−p)k - cdf - Claim 2 below
- P[X≥k]=(1−p)k−1 - an obvious extension.
Convention notes
Grade: A**
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Some aspect of this page is incomplete and work is required to finish it
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If X∼Geo(p) is defined as above then there are 3 other conventions I've seen:
- X1∼Geo(1−p) in our terminology, they would write Geo(p), which measures "trials until first failure" instead of success as we do
- X2:=X−1 - the number of trials BEFORE first success
- X3:=X1−1 - the number of trials BEFORE first failure
Warning:That grade doesn't exist!
Properties
For p∈[0,1]⊆R and X∼Geo(p) we have the following results about the geometric distribution:
- E[X]=1pfor p∈(0,1] and is undefined or tentatively defined as +∞ if p=0
- Var(X)=1−pp2for p∈(0,1] and like for expectation we tentatively define is as +∞ for p=0
To do:
Proof of claims
Claim 1: P[X=k]=(1−p)k−1p
TODO: This requires improvement, it was copy and pasted from some notes
- P[X=k]:=(1−p)k−1p - which is derived as folllows:
- P[X=k]:=(P[X1=0]×P[X2=0 | X1=0]×⋯×P[Xk−1=0 | X1=0∩X2=0∩⋯∩Xk−2=0])×P[Xk=1 | X1=0∩X2=0∩⋯∩Xk−1=0]
- Using that the Xi are independent random variables we see:
- P[X=k]=(k−1∏i=1P[Xi=0])×P[Xk=1]
- =(1−p)k−1pas they all have the same distribution, namely Xi∼Borv(p)
- =(1−p)k−1p
- P[X=k]=(k−1∏i=1P[Xi=0])×P[Xk=1]
- Using that the Xi are independent random variables we see:
- P[X=k]:=(P[X1=0]×P[X2=0 | X1=0]×⋯×P[Xk−1=0 | X1=0∩X2=0∩⋯∩Xk−2=0])×P[Xk=1 | X1=0∩X2=0∩⋯∩Xk−1=0]
Claim 2: P[X≤k]=1−(1−p)k
Grade: A**
This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
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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
Notes
- Jump up ↑ Do we make this distinction for cumulative distributions?
References
- Jump up ↑ See Expectation of the geometric distribution
- Jump up ↑ See Variance of the geometric distribution
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