 Solution

Assume identical dwell states and memoryless forward/backward exits with constant rates $k_{+}$ and $k_{-}$; put $K=k_{+}+k_{-}$. The probability density for the first exit to be forward at $t_1$ is $k_{+}{e}^{-K{t}_1}$. For the very next exit also to be forward at time $t$, multiply by $k_{+}{e}^{-K(t-t_1)}$ and integrate over the unobserved first time:This density integrates to $(k_{+}/K{)}^2$, the probability that both directions are forward. Conditional on that event, the normalized total-time density is the Erlang distributionIts mean is $2/K$, so the most probable time is not the mean. Since both backward-rate conventions in part (c) have $k_{-}\ll k_{+}$, neglecting backwards steps gives the usual approximation $p(t)=k_{+}^{2}t{e}^{-k_{+}t}$ and mode $1/k_{+}$. If “two forward steps” instead permits intervening backwards events, thinning independent Poisson exits gives that latter rate-$k_{+}$ distribution directly; it describes a different conditioning. The low barrier noted in part (c) limits the physical accuracy of all these memoryless dwell assumptions.