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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!

Tim
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1 Answers1

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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.

Did
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  • @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? – Tim Apr 29 '11 at 20:54
  • What is $P(W|U,V)$ for three random variables $U$, $V$, $W$? – Did Apr 29 '11 at 22:43
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    Why should "independent increments" require that $Y_j$ are independent of $X_0$? $X_0$ is not an increment. – Robert Israel Apr 29 '11 at 23:08
  • @Didier: Thanks! (1) In my last comment, $P(W|U,V)$ means conditional distribution of $W$ given $U$ and $V$, i.e., $P(W|\sigma({U,V})$. Any mistake in my last comment? (2) I was wondering why "$X_n$ is $\mathcal{F}^X_n$ measurable and $Y_{n+1}$ is independent on $\mathcal{F}^X_n$", implies $P(X_{n}+Y_{n+1}∈dy|\mathcal{F}^X_n)=P(X_n+Y_{n+1}∈dy)$? (3) In "$(Y_n)n≥1$ is a sequence of independent random variables independent on $X_0$", do you mean $(Y_n)n≥1,X_0$ are independent? Why is $X_0$ there, since it is not an increment as Robert pointed out? – Tim Apr 30 '11 at 13:25
  • Your (1) is not compatible with your first comment, where you wrote something like P(W|U,V)=P(W-U|U,U-V) (an assertion I would be in serious trouble to give a meaning to). Do you know how the conditional distribution of W conditionally on U is defined? // In the context of your questions here, I suggest that you try to answer completely and rigorously your (2) since this would allow you to gain a deeper understanding of the texts you read. To begin with, you might want to explain as precisely as possible why your (2) has nothing to do with what I wrote... – Did Apr 30 '11 at 22:01
  • I modified my answer to get rid of $X_0$ since this anecdotic ingredient seemed to divert you from the heart of the question. – Did Apr 30 '11 at 22:07
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    @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 $X_0=0$? – Tim May 03 '11 at 12:29
  • Re your (2), you equate a random variable (the LHS) with a number (the RHS). As I said, considering this question and your other questions here on MSE, I think that here you at last have an entrance to notions around which, up to now, you have been making circles. But this supposes that you try to fully explain this infamous point (2), otherwise you will continue to stumble on questions of definitions and to equate numbers with random variables with probability distributions... For example, what is the conditional distribution of a random variable conditionally on a sigma-algebra? – Did May 03 '11 at 13:07
  • Okay, I now understand why my understanding in (2) is wrong. I also know how conditional probability is defined from conditional expectation. But I still have no idea how you got $$\mathbb{P}(X_{n+1}\in\mathrm{d}y|\mathcal{F}^X_n)=Q_n(X_n,\mathrm{d}y)$$ from "$X_n$ is $F^X_n$ measurable and $Y_{n+1}$ is independent on $F^X_n$." Obviously I am not smart, and you can ignore me if you like. – Tim Nov 06 '11 at 07:15
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    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? .../... – Did Nov 06 '11 at 08:48
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    .../... All this is explained in my posts but you might have more success with Probability with martingales by David Williams (a book I think I already mentioned) which explains it very clearly. – Did Nov 06 '11 at 08:48
  • 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? – Tim Nov 08 '11 at 06:57
  • See https://math.stackexchange.com/questions/73353/conditional-expectation-of-function-of-two-rvs-one-measurable-one-independent for the proof of $\mathbb{P}(X_{n+1}\in dy|\mathscr{F}_n^X)=Q_n(X_n,dy)$. – No-one May 31 '22 at 22:21