= Solution
Let $u(i)$ be the expected number of visits to state $3$ before absorption, counting the present state when $i=3$. Its reward equations are
$$
u(0)=0,
\qquad
u(1)=\frac12u(2),
$$
$$
u(2)=\frac12u(1)+\frac12u(3),
\qquad
u(3)=1+\frac12u(2)+\frac12u(3).
$$
The last equation gives $u(3)=2+u(2)$, so the second nontrivial equation gives $u(2)=2+u(1)$. Hence $u(1)=1+u(1)/2$, and
$$
\boxed{u(1)=2}.
$$
\b[Thus the expected occupation count is two; together with the previous parts, this is the <four-state reflecting absorbing random walk> calculation.]
Back to article page
