A <Lévy process> $(X_t)_{t\geq0}$ starts at zero, has <independent random variables>[independent] and stationary increments, is <stochastic continuity>[stochastically continuous], and is taken with <càdlàg function>[càdlàg] sample paths. Thus for $0\leq t_0<\cdots<t_n$, 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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Solution

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

{wiki=Càdlàg}

A càdlàg function is right-continuous and has a finite left limit at every positive time. The name abbreviates the French phrase /continue à droite, limites à gauche/.


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 Incoming links: Càdlàg function