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\begin{document}


{\bf Question}

A random walk has the infinite set \{0, 1, 2, ...\} as possible
states.  State 0 is a partially reflecting barrier.  If state 0 is
occupied at step n then states 0 and 1 are equally likely to be
occupied at step n+1 of the random walk.  For all other states,
transitions of +1, -1, 0 take place with the probabilities $p,$
$q,$ $1-p-q$ respectively.  Let $\ds p_{j,k}^{(n)}$ denote the
probability that the random walk is in state $k$ at step $n$,
having started in state $j$.  Derive the difference equation
$$p_{j,k}^{(n)} = p \cdot p_{j,k-1}^{(n-1)} + q \cdot
p_{j,k+1)}^{(n-1)} + (1-p-q) \cdot p_{j,k}^{(n-1)} \, \, \, (k
\geq 2)$$ giving clear explanation of the reasoning leading to the
equation.  Write down analogous equations for $k=0$ and $k=1$. The
long-term equilibrium distribution is given by $$ \pi_k = \lim_{n
\to \infty}p_{j,k}^{(n)} \hspace{.5in} (j = 0, 1, 2,...)$$ when
these limits exist.  Obtain a set of difference equations for
$(\pi_k).$  Solve these equations, recursively or otherwise,
showing that if $p \geq q$ there is no solution, and finding
explicit expressions for $\pi_k$ in the case $p<q$.  You may
assume that $q\not= 0.$



\vspace{.25in}

{\bf Answer}

Arguing conditionally on the last step gives \begin{eqnarray*}
p_{jk}^{(n)} & = & p p_{j,k-1}^{(n-1)} + qp_{j,k+1}^{(n-1)} +
(1-p-q)p_{j,k}^{(n-1)} \\ p_{j0}^{(n)} & = &
\frac{1}{2}p_{j0}^{(n-1)} + qp_{j1}^{(n-1)} \\ p_{j1}^{(n)} & = &
\frac{1}{2}p_{j0}^{(n-1)} + qp_{j,2}^{(n-1)} +
(1-p-q)p_{j,1}^{(n-1)} \end{eqnarray*} Taking limits as n $\to
\infty$gives \begin{eqnarray*} \pi_k & = & p\pi_{k-1} +
q\pi_{k+1}+ (1-p-q)\pi_k  \hspace{.2in} k \geq 2 \\ \pi_0 & = &
\frac{1}{2}\pi_0 + q\pi_1 \\ \pi_1 & = & \frac{1}{2} \pi_0 +
q\pi_2 + (1-p-q)\pi_1 \end{eqnarray*} Rewriting these equations
gives, \begin{eqnarray} q\pi_1 & = & \frac{1}{2} \pi_0 \\ q\pi_2 +
(-p-q)\pi_1 + \frac{1}{2}\pi_0 & = & 0 \\ q\pi_{k+1} + (-p-q)\pi_k
+ p\pi_{k-1} & = & 0 \end{eqnarray}  Using (1) in (2) gives $\ds
q\pi_2 = p\pi_1$  Assuming that $\ds q\pi_k = p\pi_{k-1}$ gives,
using (3) $\ds q\pi_{k+1} = p\pi_k$


Hence by induction this is true for $k \geq 1$

 ${}$

Thus $\ds \pi_k = \left(\frac{p}{q}\right)^{k-1} = \frac{1}{2q}
\left(\frac{p}{q} \right)^{k-1}\pi_0 \hspace{.2in} k \geq 1$

Now for $(\pi_k)$ to be a probability distribution, $\ds\sum \pi_k
= 1$

i.e. $\ds \left(1 + \frac{1}{2q} \sum_{k=1}^\infty
\left(\frac{p}{q} \right)^{k-1}\right) = 1$

If $p \geq q$ the series diverges, so there is no solution.

If $p<q,$  $$\pi_0 \left( 1 + \frac{1}{2q} \frac{1}{\left(1 -
\frac{p}{q}\right) }\right) = 1; \hspace{.2in} \pi_0 \left(1 +
\frac{1}{2(q-p)}\right) = 1$$ So $$\pi_0 = \frac{2(q-p)}{1+
2(q-p)} \hspace{.2in}{\rm and} \hspace{.2in}\pi_k = \frac{1}{2q}
\left(\frac{p}{q}\right)^{k-1} \cdot\frac{2(q-p)}{1+2(q-p)}
\hspace{.1in}(k \geq 1)$$



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