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Independent increments and Markov property.do not imply each other. I was wondering

  • if being one makes a process closer to being the other?
  • if there are cases where one implies the other?

Thanks and regards!

1 Answers 1

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Independent increments do imply Markov property.

To see this, assume that $(X_n)_{n\ge0}$ has independent increments, that is, $X_0=0$ and $X_n=Y_1+\cdots+Y_n$ for every $n\ge1$, where $(Y_n)_{n\ge1}$ is a sequence of independent random variables. The filtration of $(X_n)_{n\ge0}$ is $(\mathcal{F}^X_n)_{n\ge0}$ with $\mathcal{F}^X_n=\sigma(X_k;0\le k\le n)$. Note that $ \mathcal{F}^X_n=\sigma(Y_k;1\le k\le n), $ hence $X_{n+1}=X_n+Y_{n+1}$ where $X_n$ is $\mathcal{F}^X_n$ measurable and $Y_{n+1}$ is independent on $\mathcal{F}^X_n$. This shows that the conditional distribution of $X_{n+1}$ conditionally on $\mathcal{F}^X_n$ is $ \mathbb{P}(X_{n+1}\in\mathrm{d}y|\mathcal{F}^X_n)=Q_n(X_n,\mathrm{d}y), \quad \mbox{where}\quad Q_n(x,\mathrm{d}y)=\mathbb{P}(x+Y_{n+1}\in\mathrm{d}y). $ Hence $(X_n)_{n\ge0}$ is a Markov chain with transition kernels $(Q_n)_{n\ge0}$.

On the other hand, Markov property does not imply independent increments.

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    Thanks! (1) I now understand why $\mathbb{P}(X_{n+1}\in\mathrm{d}y|\mathcal{F}^X_n)=Q_n(X_n,\mathrm{d}y)$ in your original reply. Section 9.10 "Conditioning under independence assun1ptions" in Williams' book has a conclusion that can explain this. (2) But I don't know how to understand the conclusion "consider random variables ξ and η and a sigma-algebra$G$such that ξ is independent on H=σ(η)∨G. Why is E(u(ξ+η)∣H)=E(u(ξ+η)∣η) for every bounded u?" Some references? How is it related to my question?2011-11-08