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!


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.

  • $\begingroup$ @Didier: Thanks! But I think it doesn't because of the following. First $P(X(t_3) | X(t_2), X(t_1)) = P(X(t_3)-X(t_2)|X(t_2), X(t_2)-X(t_1))$. Next $P(X(t_3)-X(t_2)|X(t_2), X(t_2)-X(t_1)) = P(X(t_3)-X(t_2)|X(t_2))$, if and only if $X(t_3)-X(t_2)$ and $X(t_2)-X(t_1))$ are conditionally independent given $X(t_2)$, which can not be implied by $X(t_3)-X(t_2)$ and $X(t_2)-X(t_1))$ are independent. Any mistake? $\endgroup$ – Tim Apr 29 '11 at 20:54
  • $\begingroup$ What is $P(W|U,V)$ for three random variables $U$, $V$, $W$? $\endgroup$ – Did Apr 29 '11 at 22:43
  • $\begingroup$ Why should "independent increments" require that $Y_j$ are independent of $X_0$? $X_0$ is not an increment. $\endgroup$ – Robert Israel Apr 29 '11 at 23:08
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    $\begingroup$ @Didier: Thanks! 1) I still have no clue how to explain and correct (2) in my last comment. Would you point me where in what texts/materials? 2) Generally when saying increments of a stochastic process, is $X_0$ an increment? Does the definition of an independent-increment process require $X_0=0$? $\endgroup$ – Tim May 3 '11 at 12:29
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    $\begingroup$ Invoking "smartness" here is a way to avoid following the explicit suggestions I made, which would lead you to understand the problem. It is also a cheap shot at my advice, considering the time and work I spent on your questions. // Since once again you are stopped by matters of definitions I suggest to come back to definitions: consider random variables $\xi$ and $\eta$ and a sigma-algebra $G$ such that $\xi$ is independent on $H=\sigma(\eta)\vee G$. Why is $E(u(\xi+\eta)\mid H)=E(u(\xi+\eta)\mid\eta)$ for every bounded $u$? Why is this related to your question? .../... $\endgroup$ – Did Nov 6 '11 at 8:48

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