Markov property misunderstanding

Question

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?

Answer 1

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, $$ P(X_3 \in\{100,50\}\mid X_2\in\{100,1\}) \neq P(X_3 \in\{100,50\}\mid X_2\in\{100,1\}, X_1>50). $$

Answer 2

As a more extreme example, if $S$ denotes the state space, the Markov property does not assert that $$P(X_3\in A\mid X_2\in S, X_1\in B)=P(X_3\in A\mid X_2 \in S)\ ;$$ the LHS is the conditional probability $P(X_3\in A\mid X_1\in B)$, while the RHS is the unconditional probability $P(X_3\in A)$; these need not be equal. OTOH the Markov property does assert $$P(X_3\in A\mid X_2=x, X_1\in B)=P(X_3\in A\mid X_2=x)$$ for each $x$, as remarked by @zhoraster.