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, ..., x_n, the likelihood

= P(x₁, ..., x_n | μ) = (μ / (μ+1))^∑x_i / (μ+1)ⁿ

 

neg log likelihood

L = (∑x_i).(log(μ+1) - log μ) + n.log(μ+1)

1st derivative

d L / d μ = (∑x_i).(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 = (∑x_i).(1/μ² - 1/(μ+1)²) - n/(μ+1)²

Note that E ∑xi = n.μ.

which has expectation, i.e., Fisher information, F_μ

= 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 F_μ + (-log 12 + 1)/2

= log A + μ/A - (∑x_i).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 + (∑x_i).{1/(μ+1) - 1/μ} + n/(μ+1) - 1/(2μ) - 1/(2(μ+1))

= 1/A + (1/(μ+1)).{∑x_i + n - 1/2} - (1/μ).{∑x_i + 1/2}

equate to zero, multiply by μ(μ+1)

0 = μ(μ+1)/A + μ{∑x_i + n - 1/2} - (μ+1){∑x_i + 1/2}

= μ²/A + μ{1/A + n - 1} - 1/2 - ∑x_i

(Note that if A is "very large",   μ_MML = (∑x_i + 1/2) / (n - 1).)

The quadratic has solutions

μ_MML = (1 - n - 1/A ±√{n² + 1/A² + 1 + 2n/A - 2/A - 2n + 2/A + 4(∑x_i)/A}) / (2/A)

= (1 - n - 1/A ±√{n² + 1/A² + 1 + 2n/A - 2n + 4(∑x_i)/A}) / (2/A)

only the "+" solution is admissible.

Thanks to Daniel Schmidt and Enes Makalic.

See [IP 1.2] for an implementation.