= Solution
For $k>0$, an interior horizontal position has total upward and downward rates $1/3$ each. At either edge the upward rate is $1/3+1/6=1/2$ and the downward rate is $1/6$. Thus every position has total vertical-jump rate $2/3$. Averaging over the uniform conditional horizontal law gives
$$
\lambda_k=\frac{(k-1)/3+2(1/2)}{k+1}=\frac{k+2}{3(k+1)},
\qquad
\mu_k=\frac{(k-1)/3+2(1/6)}{k+1}=\frac{k}{3(k+1)}.
$$
At zero there are two upward jumps of rate $1/3$, so $\lambda_0=2/3$ and $\mu_0=0$, agreeing with these formulas. Dividing by $2/3$ yields the two embedded transition <probabilities> in part (a).
The constant vertical hazard makes every vertical holding time exponential with rate $2/3$. Conditional uniformity at a new level, together with preservation of uniformity during horizontal motion, makes the next direction depend only on that level. Consequently the vertical process has the <Markov property> in its own history and the generator in part (a). Part (c) proves the preservation assertion and the required conditioning on past levels, rather than assuming strong lumpability for every possible horizontal starting point.
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