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\begin{document}


{\bf Question}

A simple random walk can occupy states $(0,\ 1,\ 2,\ \cdots)$. For
states $j \geq 2$ there is a probability $p$ of making a step of
$+1$, and $q$ of making a step of -1, where $p+q=1$ and $pq \ne
0$. State 0 is an absorbing barrier. State 1 is partially
reflecting in the sense of the following conditional
probabilities:

$$\begin{array} {rcl} P(X_{n+1}=2\ |\ X_n=1) & = & p\\
P(X_{n+1}=1\ |\ X_n=1) & = & qk\\ P(X_{n+1}=0\ |\ X_n=1) & = &
p(1-k) \end{array}$$

where $0<k<1$.

${}$

Let $f(n,\ j)$ denote the probability of absorption at step $n$,
starting in state $j$. Let $F_j(s)$ denote the generating function
for the sequence of probabilities $f(n,j)\ (n=0,\ 1,\ 2,\
\cdots)$.

Explain why $F_j(1)$ gives the probability of eventual absorption,
starting in state $j$.

${}$

Write down a difference equation for $f(n,j)$, by arguing
conditionally on the result of the first step, for $j=1$ and for
$j>1$.

${}$

Hence obtain a difference equation for $F_j(1)$, the probability
of eventual absorption, for $j=1$ and for $j>1$. Find the general
solution of the difference equation for $j>1$. Show by considering
large $j$, that for $q>p$ and for $q=p,\ F_j(1)$ is constant,a nd
further that $F_j(1)=1$.

${}$

For $q>p$, using the assumption that $F_j(1) \to 0$ as $j \to
\infty$, show that

$$F_j(1) = \df{p(1-k)}{p-kq}\left(\df{q}{p}\right)^j\ \rm{or}
j>0.$$



\vspace{.25in}

{\bf Answer}

$F_j(s)=\ds\sum_{n=0}^\infty f(n,\ j)s^n$

$F_j(1)=\ds\sum_{n=0}^\infty f(n,\ j)$

which is the probability of absorption, either at step 0, or step
1, or step 2, $\cdots$, i.e. the probability of eventual
absorption.

Now arguing conditionally on the result of the first step, for
$j=1$, gives

$f(n,\ 1)=pf(n-1,\ 2)+q(1-k)f(n-1,\ 0)+qkf(n-1,\ 1)\ \ (A)$

For $j>1$ we obtain

$f(n,\ j)=pf(n-1,\ j+1)+qf(n-1,\ j-1)\ \ (B)$

$\begin{array}{lrcl} \rm{Now} & f(0,1) & = & 0\\ & f(n,\ 0) & = &
\left\{\begin{array}{cl}1 & n=0\\ 0 &
\rm{otherwise}\end{array}\right.\\ \rm{so} & f(n-1,\ 0) & = &
\left\{\begin{array}{cl}1 & n=1\\ 0 &
\rm{otherwise}\end{array}\right.\end{array}$

Summing $(A)$ for $n=1,\ 2,\ \cdots$ gives

$F(1)=p\ds\sum_{n=1}^\infty f(n-1,\
2)+q(1-k)+qk\ds\sum_{n=1}^\infty f(n-1,\ 1)$

$F_1(1)=pF_2(1)+q(1-k)+qkF_1(1)\ \ (C)$

For $j>1,\ f(0,\ j)=0$, so summing $(B)$ for $n=1,\ 2,\ \cdots$
gives

$$F_j(1)=pF_{j+1}(1)+qF_{j-1}(1)$$

Now let $F_j(1)=\lambda^j\ \ j \geq 1$

then $p\lambda^2-\lambda+q=0$,\ so $(p\lambda-q)(\lambda-1)=0$.

For $\begin{array}{ll} p \ne q &
F_j(1)=A\left(\df{q}{p}\right)^j+B\\ p=q & F_j(1)=Aj+B
\end{array}$

Now if $q>p$ or $q=p,\ A \ne C \to F_j(1)$ is outside $[0,1]$ for
large $j$, and so couldn't represent a probability. Hence $A=C$
and so for $q \geq p,\ F_j(1)=B$.

Using $(C)$ now gives

$\begin{array}{l} B=pB+q(1-k)+qkB\\ B(1-p-qk)=q-qk \end{array}$

$B=1$ (as $p+q=1$ and $0<k<1$)

Hence $F_j(1)=1$.

Now if $q<p,\ \left(\df{q}{p}\right)^j \to 0$ as $j \to \infty$ so
assuming $F_j(1) \to 0$ this gives $B=C$. Hence
$F_j(1)=A\left(\df{q}{p}\right)^j$ and using $(C)$ again gives

$\begin{array}{l}
A\left(\df{q}{p}\right)=pA\left(\df{q}{p}\right)^2+q(1-k)+qkA\left(\df{q}{p}\right)\\
A\left(\df{q}{p}-\df{q}{p}^2-\df{q \cdot 2 \cdot
k}{p}\right)=q(1-k) \end{array}$

so $A=\df{p(1-k)}{p-qk}$ using $P+q=1$.

${}$

Thus $F_j(1)=\df{p(1-k)}{p-kq}\left(\df{q}{p}\right)^j$ in this
case.



\end{document}