Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 33 / 5

The PDF defines multiplication by an indicator: $X_t=U{1}_{\{V\ge e^{-t}\}}$. PutThen $T>0$ almost surely, $\{T\le t\}=\{X_t>0\}\in {\mathcal{F}}_t$, and $T$ is a stopping time. It has exponential distribution of rate one and is independent of $U$. The path stays at zero until $T$, jumps to $U$ at $T$, and remains there.

Let $p=\mathbb{P}(U\in B)$ and $N_t=1_{\{T\le t,U\in B\}}$. We verify the martingale property directly. On $\{T\le s\}$ both increments $N_t-N_s$ and $(T\wedge t)-(T\wedge s)$ vanish. On $\{T>s\}$, every observation up to $s$ is zero, so the past reveals only that survival event. Conditional on it, $R=T-s$ is rate-one exponential by memorylessness and remains independent of the mark $U$. With $h=t-s$,whileSubtracting $p$ times the second identity from the first gives zero. The process is adapted and integrable, since its absolute value is bounded by $1+pt$ on a finite horizon. ThereforeOn $\{T\le t\}$ we have $X_t=U$, so this is exactly the requested process. This first-principles argument uses the natural filtration, which does not disclose the future mark or lifetime. It describes a single marked exponential jump process.

At a positive state $x$, the future is constant. At zero, the residual holding time is independent rate-one exponential and the next state is an independent uniform mark. Thus the conditional future given the natural filtration depends only on the present state: $X$ is a Markov jump process. For any bounded Borel $g$, its transition semigroup isThe Lévy kernel of a Markov jump process, written in destination-state convention, isIn displacement convention it has the same expression $K(x,dz)=1_{\{x=0\}}{1}_{(0,1)}(z)dz$, since its only nonzero jump rate is from zero. The Markov jump-process generator is $Lg(x)=1_{\{x=0\}}({\int }_0^{1}g(y)dy-g(0))$.

We state the generator martingale theorem used here: for a nonexplosive Markov jump process and a bounded $g$ with bounded $Lg$, the process $g(X_t)-g(X_0)-{\int }_0^{t}Lg(X_{s-})ds$ is a true martingale. In this one-jump model it also follows directly from part (a), first for mark indicators and then by bounded-function approximation.

Take $g_{\theta }(x)=e^{\theta x}$, and putHere $L{g}_{\theta }(x)/g_{\theta }(x)=r_{\theta }{1}_{\{x=0\}}$. Therefore choose the previsible processIt is continuous adapted and thus predictable. To verify the exponential, write $d{g}_{\theta }(X_t)=d{N}_t+L{g}_{\theta }(X_{t-})dt$ with the generator martingale $N$. The semimartingale integration by parts formula, using the continuous finite-variation factor $e^{-Y_t}$, givesThus $Z_t=e^{\theta X_t-Y_t}$ is a local martingale. On each fixed horizon it is bounded by $e^{|\theta |+|r_{\theta }|t}$, so it is a true martingale with $Z_0=1$. This is the exponential martingale from a jump generator; for $\theta =0$ it reduces to the constant process one.