Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 4 / 26I (Applied Probability)

The embedded birth-death chain has upward probability $p_k=(k+2)/(2(k+1))$ and downward probability $q_k=k/(2(k+1))$. To calculate a return probability, solve the harmonic difference equation on $\{0,\dots ,m\}$ with boundary values one at zero and zero at $m$. Successive differences have ratio $q_k/p_k=k/(k+2)$, so their weights areTheir total infinite sum is $2$. Letting $m\to \infty$ gives the probability of ever hitting zero from $k$:In particular the chain leaves zero to one and returns with probability $1/2<1$, so it is transient. This finite-interval calculation selects the minimal nonnegative hitting-probability solution, rather than merely observing that $h_k$ is harmonic.

The continuous-time chain has the same embedded transitions and constant total exit rate $2/3$, so it is nonexplosive and has exactly the same return events. It too is transient, neither positive recurrent nor null recurrent.

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 givesAt 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.

For level $k$, let $H_k$ be the horizontal generator: neighboring horizontal sites exchange at rate $1/6$, with reflecting endpoints. It is symmetric and has zero row sums, so the uniform row vector $u_k=(1/(k+1),\dots ,1/(k+1))$ satisfies $u_k{H}_k=0$. Removing paths as soon as they jump vertically gives the killed horizontal generator $H_k-(2/3)I$. ThereforeConditional on any elapsed waiting time without a vertical jump, the horizontal position remains uniform. This proves the exit-position uniformity as well as independence of that position from the vertical holding time.

Now consider the upward rate block $U_k$ from level $k$ to $k+1$. For $k>0$ each of its $k+2$ destination columns has sum $1/3$: an endpoint receives its single doubled edge jump, and each other site receives two jumps of rate $1/6$. HenceLikewise every destination column of the downward block $D_k$ has sum $1/3$, so $(u_k{D}_k{)}_j={\mu }_k/k$. Thus, conditional on either next vertical level, the arrival position is uniform on that new level. At the corner the two upward rates are equal, giving uniform arrival on level one directly.

Starting with the single site at level zero, induction now proves the uniform conditional law for a triangular-wedge walk: conditional on every attainable past-level sequence, arrival is uniform; the killed semigroup makes departure uniform; and the rate-block column sums make arrival at the following level uniform again. This proves both equalities of the requested property and justifies the averaging in part (b). The statement does not condition departure on the future level, which would in general bias the exit positions.