Geometric Distribution

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 Poisson
Often, P(x|p) = (1-p)x-1 . p,   integer x≥1, μ=1/p, μ≥1, but
here, P(x|p) = (1-p)x . p,   integer x≥0, μ=(1/p)-1, μ≥0, p=1/(μ+1), 1-p=μ/(μ+1).
 
In μ-space:
p = 1/(μ+1),   so
P(x|μ) = (1 - 1/(μ+1))x / (μ+1)
= (μ / (μ+1))x / (μ+1)
 
Given n data, x1, ..., xn, the likelihood
= P(x1, ..., xn | μ) = (μ / (μ+1))∑xi / (μ+1)n
 
neg log likelihood
L = (∑xi).(log(μ+1) - log μ) + n.log(μ+1)
1st derivative
d L / d μ = (∑xi).(1/(μ+1) - 1/μ) + n/(μ+1)
If we equate this to zero,
(∑xi).μ - (∑xi).(μ+1) + n.μ = 0,
μmaxLH = (∑xi) / n.
2nd derivative
d2 L / d μ2 = (∑xi).(1/μ2 - 1/(μ+1)2) - n/(μ+1)2
Note that E ∑xi = n.μ.
which has expectation, i.e., Fisher information, Fμ
= n.μ.(1/μ2 - 1/(μ+1)2) - n/(μ+1)2
= n/μ - n.μ/(μ+1)2 - n/(μ+1)2
= n.(1/μ - 1/(μ+1))
= n / (μ (μ+1))
 
Assume prior,   h μ = (1/A).e-μ/A,   which has mean A.
 
The two-part message length, m
= - log(h μ) + L + (1/2)log Fμ + (-log 12 + 1)/2
= log A + μ/A - (∑xi).log(μ/(μ+1)) + n.log(μ+1) + (1/2)log n - (1/2)logμ - (1/2)log(μ+1) + (-log 12 + 1)/2
 
To estimate μ, differentiate m with respect to μ
d m / d μ
= 1/A + (∑xi).{1/(μ+1) - 1/μ} + n/(μ+1) - 1/(2μ) - 1/(2(μ+1))
= 1/A + (1/(μ+1)).{∑xi + n - 1/2} - (1/μ).{∑xi + 1/2}
equate to zero, multiply by μ(μ+1)
0 = μ(μ+1)/A + μ{∑xi + n - 1/2} - (μ+1){∑xi + 1/2}
= μ2/A + μ{1/A + n - 1} - 1/2 - ∑xi
(Note that if A is "very large",   μMML = (∑xi + 1/2) / (n - 1).)
The quadratic has solutions
μMML = (1 - n - 1/A ±√{n2 + 1/A2 + 1 + 2n/A - 2/A - 2n + 2/A + 4(∑xi)/A}) / (2/A)
  
= (1 - n - 1/A ±√{n2 + 1/A2 + 1 + 2n/A - 2n + 4(∑xi)/A}) / (2/A)
only the "+" solution is admissible.
A=, data=,
op=
(Also see Poisson.)
-- L.A., July 2007.

Thanks to Daniel Schmidt and Enes Makalic.

See [IP 1.2] for an implementation.

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