Solution

If the urn contains

g

green balls, it contains

g + 2

red balls: both update rules preserve the difference

R - G = 2

. Until absorption at

g = 0

, the green count is therefore a birth-death chain with

p_{g} = P (g \to g - 1) = \frac{g}{2 g + 2}, q_{g} = P (g \to g + 1) = \frac{g + 2}{2 g + 2} .

h (g) = \frac{1}{g + 1}

is harmonic, since

p_{g} h (g - 1) + q_{g} h (g + 1) = \frac{1}{2 ( g + 1 )} + \frac{1}{2 ( g + 1 )} = h (g) .

Let

τ_{0}

and

τ_{N}

be the hitting times of

0

and

N

. The optional sampling theorem for a supermartingale, applied to the bounded stopped martingale

h (G_{n \land τ_{0} \land τ_{N}})

, gives

h (m) = P_{m} (τ_{0} < τ_{N}) + \frac{1}{N + 1} P_{m} (τ_{N} < τ_{0}) .

Solving,

P_{m} (τ_{0} < τ_{N}) = \frac{N - m}{N ( m + 1 )} .

Letting

N \to \infty

, the events on the left increase to eventual termination. The harmonic hitting probability for the balanced-difference urn is therefore

P (the process terminates) = \frac{1}{m + 1} .

Solved by gpt-5.6-sol high.

 Ancestors (10)