4.7 Ehrenfest model of diffusion and its invariant distribution

(From Grimmett and Stirzaker) Two containers A and B are placed adjacent to each other and gas is allowed to pass through a small aperture joining them. A total of $m$ gas molecules is distributed between the containers. We assume that at each time-step one molecule, picked uniformly at random from the $m$ available, passes through this aperture.

Let $X_n$ be the number of molecules in container A after $n$ units of time have passed. Clearly $\{X_n\}$ is a Markov chain with transition matrix

$$P_{i,i-1}=\frac{i}{m},\mathrm{\hspace{0.17em}\hspace{0.17em}1}\le i\le m{P}_{i,i+1}=1-\frac{i}{m},\mathrm{\hspace{0.17em}\hspace{0.17em}0}\le i\le m-1.$$

i.e.

(Check that the formula makes sense for the end points!) Another interpretation for this model is as follows. Consider an urn containing $m$ balls each of which is either of colour A or colour B. A ball is drawn from the urn at random and is replaced by a ball of the other colour. Let $X_n$ denote the number of balls of colour A after the $n$-th draw.

We could find the stationary distribution by solving $\pi P=\pi$; this involves solving a rather unfriendly second order difference equation. Alternatively it seems sensible that when at equilibrium someone observing a video of the system would be unable to tell whether or not it was being shown backwards, and so the system should satisfy detailed balance equations; Theorem 4.4.10 confirms this.

$${\pi }_i{P}_{i,i+1}={\pi }_{i+1}{P}_{i+1,i},\text{i.e.}{\pi }_i\left(1-\frac{i}{m}\right)={\pi }_{i+1}\frac{i+1}{m},\mathrm{\hspace{0.17em}\hspace{0.17em}0}\le i\le m-1,$$

or equivalently

$${\pi }_{i+1}=\frac{m-i}{i+1}{\pi }_i,\mathrm{\hspace{0.17em}\hspace{0.17em}0}\le i\le m-1.$$

Thus

	${\pi }_1$	$=$	${\displaystyle \frac{m}{1}}{\pi }_0$
	${\pi }_2$	$=$	${\displaystyle \frac{m-1}{2}}{\pi }_1={\displaystyle \frac{m(m-1)}{2\times 1}}{\pi }_0$
	${\pi }_3$	$=$	${\displaystyle \frac{m-2}{3}}{\pi }_2={\displaystyle \frac{m(m-1)(m-2)}{3\times 2\times 1}}{\pi }_0.$

In general

$${\pi }_i=\frac{m!}{i!(m-i)!}{\pi }_0.$$

So

$$1=\sum _{i=0}^m{\pi }_i={\pi }_0\sum _{i=0}^m\frac{m!}{i!(m-i)!}={\pi }_0{(1+1)}^m.$$

Therefore ${\pi }_0=1/2^m$ and

$${\pi }_i=\frac{1}{2^m}\left(\begin{array}{c}\hfill m\hfill \\ \hfill i\hfill \end{array}\right),\mathrm{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}0}\le i\le m,$$

which is the binomial probability with parameters $m$ and $1/2$.

This is the chance if the objects were allocated at random we would end up with $i$ in A and $m-i$ in B.

In a large system at equilibrium it is very likely that $X\approx m/2$ since this is the mean of the binomial distribution.

The next topic is of interest on its own and is related to the existence of the stationary distribution.