Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 3 / 8H (Markov Chains)

The independent identically distributed sequence $(X_n)$ is itself a Markov chain, since the conditional law of $X_{n+1}$ is its common marginal law and does not depend on the past.

Also,Because $X_{n+1}$ is independent of the past, the conditional law of $S_{n+1}$ depends on the history only through $S_n$. Thus the partial sums $(S_n)$ are a Markov chain.

The running minima satisfySince $X_{n+1}$ is independent of the past, the conditional law of $L_{n+1}$ depends only on $L_n$. Hence $(L_n)$ is a Markov chain. This is the running minimum of an independent sequence is Markov property.

The moving sums $(K_n)$ are not necessarily Markov. For a counterexample, let the $X_n$ be independent Bernoulli variables with parameter $1/2$. On the eventwe must have $X_{n-1}=0$ and $X_n=1$, soOn the other hand, $K_n=1$ and $K_{n-1}=2$ force $X_{n-1}=1$ and $X_n=0$, whenceBoth conditioning events have positive probability. Knowledge of $K_n$ alone therefore does not determine the next-step law. The overlapping moving sum need not be Markov.