Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 1 / 27J (Applied Probability)

For the natural filtration ${\mathcal{F}}_t=\sigma (X_s:0\le s\le t)$, a nonnegative random time $\tau$ is a stopping time when $\{\tau \le t\}\in {\mathcal{F}}_t$ for every $t\ge 0$: whether it has occurred can be decided from the observations already made.

For a time-homogeneous finite-state continuous-time Markov chain, the Strong Markov property says that, on $\{\tau <\infty \}$, conditional on ${\mathcal{F}}_{\tau }$, the process $(X_{\tau +s}{)}_{s\ge 0}$ has the law of a fresh chain started at $X_{\tau }$ and is independent of the earlier history given that state. In particular, with transition matrix $P_s=e^{sQ}$,The property applies also to bounded functions of several future times or of the entire future path.

Let $Z_t={\int }_0^{t}X(s)ds$ and set $a=q+\lambda$. Starting from $-1$, the first holding time has exponential distribution of rate $a$. Its joint density with a jump to $+1$ is $q{e}^{-au}du$. Before such a jump $Z$ decreases to $-u$, so the remaining rise needed to exceed $c$ is $c+u$. If instead the jump is to the absorbing state zero, the level stays nonpositive and never exceeds $c>0$.

Conditioning on this first jump and using the Strong Markov property gives

Starting from $+1$, the level reaches $c$ at time $c$ if the first holding time exceeds $c$. This has probability $e^{-ac}$; surviving beyond $c$ also makes the level strictly greater than $c$, as required by the strict event. A switch to $-1$ at time $u<c$ has density $q{e}^{-au}du$, leaves level $u$, and requires a later rise of $c-u$. Absorption before reaching $c$ cannot succeed. ThereforeHolding times have continuous distributions, so the single endpoint $u=c$ has no effect on the formula.

Assume ${\psi }_{+}(c)=e^{-Ac}$ with $A\ge 0$. Part (ii) givesInsert this into (iii). If $A\ne a$, the convolution becomesEquality for all $c>0$ forces $q^2=(a+A)(a-A)$, hence $A^2=a^2-q^2=\lambda (2q+\lambda )$. The case $A=a$ would instead give $e^{-ac}$ times a nonzero term linear in $c$, so is impossible. Thus

A path that reaches level $c$ from below does so while in state $+1$, because $Z$ is continuous and has slope $X$. At the first crossing time, the residual exponential holding time is memoryless and the Strong Markov property restarts the chain from $+1$. Conditional on ever reaching $c$, the probability of subsequently exceeding $c+d$ is therefore exactly the original probability of exceeding $d$. HenceThe zero-level limit is one because the initial $+1$ holding time is positive almost surely. Also $0<{\psi }_{+}(c)\le 1$, and it is nonincreasing. Consequently $h(c)=-\log {\psi }_{+}(c)$ is nondecreasing and additive. First $h$ is linear on positive rationals, then monotonicity gives the same result on all positive reals. ThusAbsorption can be viewed as killing at rate $\lambda$, independent of sign changes at rate $q$; this ensures the maximum is finite almost surely and makes the positive decay rate found in (iv) natural.