random walk on a torus [riddle]

Galgate, Lancastershire, July 19, 2011The Riddler of this week(-end) has a simple riddle to propose, namely given a random walk on the {1,2,…,N} torus with a ⅓ probability of death, what is the probability of death occurring at the starting point?

The question is close to William Feller’s famous Chapter III on random walks. With his equally famous reflection principle. Conditioning on the time n of death, which as we all know is definitely absorbing (!), the event of interest is a passage at zero, or any multiple of N (omitting the torus cancellation), at time n-1 (since death occurs the next time). For a passage in zero, this does not happen if n is even (since n-1 is odd) and else it is a Binomial event with probability

{n \choose \frac{n-1}{2}} 2^{-n}

For a passage in kN, with k different from zero, kN+n must be odd and the probability is then

{n \choose \frac{n-1+kN}{2}} 2^{-n}

which leads to a global probability of

\sum_{n=0}^\infty \dfrac{2^n}{3^{n+1}} \sum_{k=-\lfloor (n-1)/N \rfloor}^{\lfloor (n+1)/N \rfloor} {n \choose \frac{n-1+kN}{2}} 2^{-n}


\sum_{n=0}^\infty \dfrac{1}{3^{n+1}} \sum_{k=-\lfloor (n-1)/N \rfloor}^{\lfloor (n+1)/N \rfloor} {n \choose \frac{n-1+kN}{2}}

Since this formula is rather unwieldy I looked for another approach in a métro ride [to downtown Paris to enjoy a drink with Stephen Stiegler]. An easier one is to allocate to each point on the torus a probability p[i] to die at position 1 and to solve the system of equations that is associated with it. For instance, when N=3, the system of equations is reduced to

p_0=1/3+2/3 p_1, \quad p_1=1/3 p_0 + 1/3 p_1

which leads to a probability of ½ to die at position 0 when leaving from 0. When letting N grows to infinity, the torus structure no longer matters and the probability of dying at position 0 implies returning in position 0, which is a special case of the above combinatoric formula, namely

\sum_{m=0}^\infty \dfrac{1}{3^{2m+1}}  {2m \choose m}

which happens to be equal to

\dfrac{1}{3}\,\dfrac{1}{\sqrt{1-4/9}}=\dfrac{1}{\sqrt{5}}\approx 0.4472

as can be [unnecessarily] checked by a direct R simulation. This √5 is actually the most surprising part of the exercise!

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.