 Solution

For $0<i<N$, the total rate out of $i$ is ${\lambda }_i$. The holding time of the continuous-time Markov chain and its next state satisfyThe holding time is independent of the direction of this jump: the conditional probability of destination $j$ is the rate $q_{ij}$ divided by the total rate ${\lambda }_i$, independently of the elapsed holding duration. This is the exponential-clock construction of a continuous-time Markov chain with the given infinitesimal generator matrix.

At $i=0$ or $i=N$, all rates out are zero. These are absorbing states; the path remains at its initial endpoint and $J_i=\infty$ almost surely. There is no next visited state, so $X_{J_i}$ is not a finite-time jump value at either endpoint. If one separately uses $X_{\infty }$ for the limiting state, that limit is the same endpoint.

The embedded jump chain in the interior is a simple symmetric random walk on $\{0,\dots ,N\}$, stopped at the endpoints. From any interior state it has probability at least $2^{-N}$ of hitting a boundary in the next $N$ jumps, by following successive leftward choices. Thus its number of jumps before absorption is finite almost surely. Each of this finite number of holding times is finite almost surely, so absorption also occurs in finite continuous time. The finite set of positive interior rates causes no explosion; in particular their maximum bounds the total jump rate. When $N=1$, the chain starts at an already absorbing endpoint.

The rates alter the clock but not the sequence of visited states. The gambler's ruin calculation from part (a), translated to an initial state $i$ between endpoints $0,N$, therefore givesThis includes probability one at zero and probability zero at $N$.