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\begin{document}


{\bf Question}

A simple random walk has the infinite set $(a,\ a-1,\ a-2,\
\cdots)$ as possible states. State $a$ is an upper reflecting
barrier, for which reflection is certain, i.e., if the random walk
is in state $a$ at step $n$ then it will be in state $a-1$ at step
$n+1$. For all other states, transitions of $+1,\ -1,\ 0$ take
place with probabilities $p,\ q,\ 1-p-q$ respectively.

Let $p_{j,\ k}^{(n)}$ denote the probability that the random walk
is in state $k$ at step $n$, having starter in state $j$. Obtain
difference equations relating to these probabilities, for $k=a,\
k=a-1$ and $k<1-1$.

Assuming that there is a long-term equilibrium distribution
$\pi_k)$, where

$$\pi_k=\lim_{n\to \infty}p_{j,\ k}^{(n)}\ \rm{for}\ j=a,\ a-1,\
a-2,\ \cdots\ ,$$

use the difference equations for $p_{j,\ k}^{(n)}$ to obtain a set
of difference equations for $\pi_k$ for $k<a-1$, and also deduce
that

$$p\pi_{a-2}=q\pi_{a-1}\ \rm{and}\ p\pi_{a-1}=\pi_a.$$

Solve this set of equations for the case $p<q$ to obtain explicit
expressions for $\pi_k$ in terms of $p,\ q$ and $a$.




\vspace{.25in}

{\bf Answer}

${}$

$\begin{array}{llcl}k=a: & p_{j\ a}^{(n)} & = & p\,p_{j\
a-1}^{(n-1)}\\k=a-1: & p_{j\ a-1}^{(n)} & = & p\,p_{j\
a-2}^{(n-1)}+p_{j\ a}^{(n-1)}+(1-p-q)p_{j\ a-1}^{(n-1)}\\ k<a-1: &
p_{j\ k}^{(n)} & = & p\,p_{j\ k-1}^{(n-1)}+q\,p_{j\
k+1}^{(n-1)}+(1-p-q)p_{j\ k}^{(n-1)}\end{array}$

Assuming an equilibrium distribution $(\pi_k)$, taking limits in
the above equations gives

$\begin{array}{rcl} \pi_a & = & p\pi_{a-1}\\ \pi_{a-1} & = &
p\pi_{a-2}+\pi_a+(1-p-q)\pi_{a-1}\\ \pi_k & = &
p\pi_{k-1}+q\pi_{k+1}+(1-p-q)\pi_k\ \rm{for}\ k<a-1\end{array}$

substituting the first equation in the second gives

$$p\pi_{a-2}=q\pi_{a-1}.$$

The solution of the difference equation for $k<a-1$ is
$\pi_k=A\left(\df{p}{q}\right)^k +B$.

Now if $B \ne 0,\ \ds\sum_{\pi_k}$ diverges, so $B=0$

$\pi_{a-1}=\left(\df{p}{q}\right)\pi_{a-2}=A\left(\df{p}{q}\right)^{a-1}$

$\pi_a=p\pi_{a-1}=p\cdots A\left(\df{p}{q}\right)^{a-1}$

We require $\ds\sum \pi_k=1$

i.e.
$A\left[p\left(\df{p}{q}\right)^{a-1}+\ds\sum_{k=-\infty}^{a-1}
\left(\df{p}{q}\right)^k\right]=1$

${}$

$A\left[\df{p\cdot\left(\frac{p}{q}\right)^{a-1}
+\left(\frac{p}{q}\right)^{a-1}}{(1-\frac{q}{p})}\right]=1$

${}$

$A\left(\df{p}{q}\right)^{a-1}\left[p+\df{1}{1-\frac{q}{p}}\right]=1$

${}$

$A=\left(\df{p-q}{p^2-pq+p}\right)\left(\df{q}{p}\right)^{a-1}$

${}$

Hence $\pi_a=\df{p-q}{p-q+1}$

and
$\pi_k=\left(\df{p-q}{p^2-pq+p}\right)\left(\df{q}{p}\right)^{a-1-k}$
for $k<a$

Can also be done by recursion.


\end{document}
