Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 201 / 6

A Lévy process $(X_t{)}_{t\ge 0}$ starts at zero, has independent and stationary increments, is stochastically continuous, and is taken with càdlàg sample paths. Thus for $0\le t_0<\cdots <t_n$, the increments $X_{t_j}-X_{t_{j-1}}$ are independent, and the law of $X_{s+t}-X_s$ depends only on $t$.

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Write $\mathbb{E}{X}_1=\mu$ and $\mathrm{Var}(X_1)={\sigma }^2<\infty$. For rational $t=m/n$, stationarity and independence of the $n$ increments over intervals of length $1/n$ giveusing variance additivity for independent random variables. Stochastic continuity extends both identities from rational to real $t$. In the centered case $\mu =0$, this becomes $\mathbb{E}{X}_t=0$ and $\mathbb{E}{X}_t^2=t{\sigma }^2$.

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For $s<t$, write $X_t=X_s+Y$, where the increment $Y=X_t-X_s$ is independent of the natural filtration at time $s$, has mean zero, and has variance $(t-s){\sigma }^2$. HenceIt follows that $M_t=X_t^2-t{\sigma }^2$ is a martingale, as asserted by the centered square-integrable Lévy martingale identity.

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Let $T=\inf \{t\ge 0:X_t\notin (-a,b)\}$, and suppose the jumps of $X$ have absolute value at most $c$. A nonconstant centered finite-variance Lévy process oscillates, so $T<\infty$ almost surely. Before $T$ the process lies in $(-a,b)$, and at $T$ its bounded overshoot gives $X_T\in [-a-c,b+c]$. Thus the variables $X_{T\wedge n}^2$ are uniformly bounded.

Apply the optional sampling theorem for a supermartingale to the martingale from part (c):Bounded convergence theorem on the left and monotone convergence theorem on the right yield

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Let $X_t=N_t^{+}-N_t^{-}$, where $N^{+}$ and $N^{-}$ are independent Poisson processes of rate $\lambda$. This symmetric Poisson difference process is centered, has jumps $\pm 1$, and has variance rate ${\sigma }^2=2\lambda$. If $a,b$ are positive integers, then $X_T\in \{-a,b\}$ exactly. Optional sampling of the martingale $X_t$ givesConsequentlyPart (d) now gives the mean exit time

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