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I am reading 'Green, Brown & Probability' by Kai Lai Chung, and on chapter 5 (Markov Property) page 30, it says that the Markov property DOES NOT says that for $T=2$,

$P\big(3.14< \lvert X_3 \rvert <3.15 \big| 3<\lvert X_1 \rvert<4 , 3.1 <\lvert X_2 \rvert< 3.2 \big)=P\big(3.14< \lvert X_3 \rvert <3.15 \big| 3.1 <\lvert X_2 \rvert< 3.2\big)$

'namely that the past $3 < \lvert X_1 \rvert < 4$ may well have an after- effect on the future when the present \X2\ is given as shown'.

But I don't see it, is this related to independence of increments?

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    $\begingroup$ No, "this is related to" the fact that Markov property does not say that and that simple examples show this stronger property needs not hold. What is your question? $\endgroup$ – Did Oct 6 '16 at 6:01
  • $\begingroup$ I don't see why this does not hold (perhaps you could point me to another example or reference?). I'm using this book as a companion for self-study in probability, and before it was using the brownian motion (hence my wrong question about the increments). Should I ask then, What does it means to say that $P(\Lambda | \mathcal{F}_t)=P(\Lambda| X_T)$?, where $\Lambda\in \sigma {X_s, T\leq s < \infty}$? In the discrete case, the definition asks for equality on the right hand side, so I suppose I can't ask for an inequality as in the question right? $\endgroup$ – gcardona Oct 6 '16 at 6:19
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This has nothing to do with independent increments. The Markov property works when we have an equality assumption for the "present", e.g. $$P(X_3\in A\mid X_2 = x, X_1\in B) = P(X_3\in A\mid X_2 = x). $$ However, when there is a range of values, as in your question, it is possible that the condition on the "past" value $X_1$ makes some particular values of the "present" $X_2$ more probable.

For example, let $X$ be a Poisson process. Consider e.g. the probability $$ P(X_3 \in\{100,50\}\mid X_2\in\{100,1\}, X_1>50). $$ Clearly, it is equal to $$ P(X_3 \in\{100,50\}\mid X_2=100, X_1>50) = P(X_3 =100\mid X_2=100) = e^{-1}. $$ On the other hand, the conditional probability $$ P(X_3 \in\{100,50\}\mid X_2\in\{100,1\}) $$ is close to $P(X_3 \in\{100,50\}\mid X_2=1)$ (since $X_2=100$ is highly unlikely) and is extremely small. Consequently, $$ P(X_3 \in\{100,50\}\mid X_2\in\{100,1\}) \neq P(X_3 \in\{100,50\}\mid X_2\in\{100,1\}, X_1>50). $$

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  • $\begingroup$ Thank you, I have this clear now, I need to keep working on this. $\endgroup$ – gcardona Oct 6 '16 at 18:27

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