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{\bf Question}

The Ehrenfest model of diffusion consists of a Markov chain with
states labeled 0, 1, 2, ..., d and one-step transition
probabilities \begin{eqnarray*} P_{j,j-1} &=&
\frac{j}{d},\hspace{.6in} j = 1, 2,\ldots, d  , \\ P_{j, j+1} &=&
1 - \frac{j}{d}, \hspace{.3in} j = 0, 1,\ldots, d-1
.\end{eqnarray*}
\begin{description}
\item[(i)] Classify the states as periodic or aperiodic.
\item[(ii)] If initially the states 0 and 2 are likely to be
occupied, each with probability $\frac{1}{2}$, find the
probability distribution over the states after two steps.
\item[(iii)] Find the stationary distribution.
\end{description}




\vspace{.25in}

{\bf Answer}

$$P = \left(   \begin{array}{ccccccccc} 0 & 1 & 0 & 0 & 0 & \cdots
& & 0 & 0 \\ \frac{1}{d} & 0 & \frac{d-1}{d} & 0 & 0 & \cdots & &
0 & 0 \\  0 & \frac{2}{d} & 0 & \frac{d-2}{d} & 0 & \cdots & & 0 &
0 \\ 0 & 0 & \frac{3}{d} & 0 & \frac{d-3}{d} & \cdots & & 0 & 0
\\ \vdots \\  0 & 0 &\cdots & & & & \frac{d-1}{d} & 0 &
\frac{1}{d}\\ 0 &0 & \cdots & & & & 0 & 1 & 0
\end{array} \right)$$

\begin{description}
\item[(i)] All states are periodic with period 2.

\item[(ii)] Suppose $\ds {\bf p}^{(0)} = \left(\frac{1}{2}, 0,
\frac{1}{2}, 0, ..., 0\right)$

Then $\ds {\bf  p}^{(1)} = {\bf p}^{(0)}P = \left( 0,
\frac{d+2}{2d}, 0, \frac{d-2}{2d}, 0, ..., 0\right)$

$\ds {\bf p}^{(2)} = {\bf p}^{(1)}P  $

$\ds =\left(\frac{d+2}{2d^2}, 0, \frac{(d-1)(d+2)}{2d^2} +
\frac{3(d-2)}{2d^2}, 0, \frac{(d-2)(d-3)}{2d^2},0, ..., 0\right)$

\item[(iii)] The stationary distribution satisfies {\boldmath $\pi
= \pi$}$P$

Let {\boldmath$\pi$} = $(\pi_0, \pi_1, ..., \pi_d)$

So $\ds \pi_0 = \frac{1}{d}\pi_1$

$\ds \pi_j = \frac{d-(j-1)}{d} \pi_{j-1} + \frac{j+1}{d}\pi_{j+1}
\hspace{.2in} j = 1, 2, .., d-1$

$\ds \pi_d = \frac{1}{2}\pi_{d-1} \hspace{.2in} \left(i.e. \,
\pi_{j+1} = \frac{d}{j+1}\pi_j - \frac{d-(j-1)}{j+1}\pi_{j-1}
\right)$

$\pi_1 = d\pi_0,$

$\pi_2 = \frac{d}{s}\pi_1 - \frac{d}{2}\pi_0 =
\frac{d(d-1)}{2}\pi_0,$

$\ds\ \pi_3 = \frac{d}{3}\pi_2 - \frac{d-1}{3}\pi_1 = \ldots =
\left(
\begin{array}{c} d \\ 3 \\ \end{array} \right)\pi_0,\ldots$

$\ds \pi_j = \left( \begin{array}{c} d \\ j \\ \end{array}
\right)\pi_0.$

$\ds \sum \pi_j = 1,$ so $\ds \sum_{j=0}^d \left( \begin{array}{c}
d
\\ j \end{array} \right) \pi_0 = 1 $ hence $\ds \pi_0 =
\frac{1}{2^d}$

Therefore $$\pi_j = \frac{\left( \begin{array}{c} d \\ j
\end{array} \right)}{2^d}$$

\end{description}



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