1
$\begingroup$

In Shreve's Stochastic Calculus in Finance, the Markov property is defined as

Definition 2.3.6. Let $(\Omega,\mathcal F,P)$ be a probability space, let $T$ be a fixed positive number, and let $\mathcal F(t), 0 \leq t \leq T$, be a filtration of sub-$\sigma$-algebras of $\mathcal F$. Consider an adapted stochastic process $X(t), 0 \leq t \leq T$. Assume that for all $0 \leq s \leq t \leq T$ and for every nonnegative, Borel-measurable function $f$, there is another Borel-measurable function $g$ such that $$E(f(X(t))| \mathcal F(s))=g(X(s)). (2.3.29)$$ Then we say that the $X$ is a Markov process.

That is different from the Markov property I learned before, which was defined directly for the conditional distributions of a stochastic process: the conditional distribution of future state given the past and present states is equal to the conditional distribution of the future state given the past.

Are Shreve's definition and the one I learned before equivalent, because of Riesz representation theorem which states the equivalence between measures and integrals relative to the measures?

Is it always possible to apply Riesz representation theorem to rephrase statements about measures equivalently into statements about their integrals?

In Shreve's book, $g$ is said to be dependent on $s$, but I think it also depends on $t$ because the RHS of the equation depends on both $s$ and $t$. Am I correct?

Thanks!

$\endgroup$
  • $\begingroup$ math.stackexchange.com/questions/183833/… $\endgroup$ – user940 Feb 16 '13 at 15:00
  • $\begingroup$ @ByronSchmuland: Thanks! Your reply there gives an equivalent definition which is also in terms of integral/expectation. Are they equivalent to the other definition in terms of conditional distribution (not conditional expectation), because of Riesz representation theorem? $\endgroup$ – Ethan Feb 16 '13 at 15:07
  • $\begingroup$ Yes, using the theory of regular conditional distributions as described for example in Chapter 6 of Foundations of Modern Probability by Olav Kallenberg. $\endgroup$ – user940 Feb 16 '13 at 15:24
  • $\begingroup$ @ByronSchmuland: I forgot to ask in my post. In Shreve's book, $g$ is said to be dependent on $s$, but I think it also depends on $t$ because the RHS of the equation depends on both $s$ and $t$. Am I correct? $\endgroup$ – Ethan Feb 16 '13 at 15:32
  • $\begingroup$ Yes, as in the other question we would normally express $g$ as $P_{s,t}f$ where $P_{s,t}$ is the transition kernel of the Markov process. $\endgroup$ – user940 Feb 16 '13 at 15:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.