Consider the stochastic process $(X_t)_{t \in \mathbb{R}}$ and show the equivalence of the following two Markov properties:

(a) $P(X_t \in A \mid X_u, u \leq s) = P(X_t \in A\mid X_s) \qquad \forall A \in \mathcal{S}, \forall t>s \in \mathbb{R}$;

(b) $P(A\cap B\mid X_t) = P(A\mid X_t)P(B\mid X_t) \text{ almost surely} \qquad \forall t \in \mathbb{R}, \forall A \in \mathcal{F}_t, \forall B \in \mathcal{T}_t$.

"The future and the past are independent given the present."

As suggested I try to show the discrete case first, where the random variables take discrete values and time is discrete. So far with the given comments and answers:

(b) => (a)

Set $A := \cap_k [X_{k < n} = i_k], B := [X_n+1 = i], C := [X_n = i_n]$, then (b) implies

$P(A\cap B|C) = P(A|C)P(B|C) \iff \frac{P(A\cap B\cap C)}{P(C)} = \frac{P(A\cap C) P(B\cap C)}{P(C)^2} $ $\iff P(B|C) = P(B|A\cap C)$

For these special $A,B,C$ the latter is equivalent to (a) (as we have shown for the discrete case in a previous evercise).

(a) => (b)

Analog/reverse to above reasoning, (a) implies (b) for events $A := \cap_k [X_{k < n} = i_k], B := [X_n+1 = i], C := [X_n = i_n]$ since for such events in above reasoning all steps where biimplications.

It remains to conclude that (a) implies (b) for events of the form $A=\bigcap_i[X_{u_i}=x_i]$, $B=\bigcap_j[X_{s_j}=z_j]$ where $u_i\leqslant n\leqslant s_j$ as well:

Again $P(A\cap B | C) = P(A|C)P(B|C) \iff P(B|C) = P(B|A\cap C)$

so we need to show that the latter holds also for the general events $A$ and $B$ only assuming (a) and the first part we have already shown; and it indeed follows from the first bit: $P(B|A\cap C) = P(\bigcap_j[X_{s_j}=z_j] | A\cap C) = P(\bigcap_j[X_{s_j}=z_j] | C)$ since $s_j > n$.

Is this the discrete case completed or am I missing something? I think somehow I need to handle the special cases where we have one $s_j = n$ or $u_i=n$.

How can I derive the continuous case from the discrete case then? And where does the "almost surely" some in in the continuous case?

  • 3
    $\begingroup$ "I do not have any clue how to get started." This is VERY odd. Which similar exercises did you solve? $\endgroup$ – Did Jan 12 '14 at 9:10
  • $\begingroup$ Well, sorry. Similar a similar exercise was to reformulate this general formulation to the discrete case. But otherwise we did not further work with the general continuous case.. $\endgroup$ – user20183 Jan 12 '14 at 11:50
  • $\begingroup$ And what is different in the continuous case, you think? $\endgroup$ – Did Jan 12 '14 at 11:53
  • $\begingroup$ I wasn't precise. We reformulated (a) in the discrete case to something else than (b). For the discrete case we reformulated it, so that it says $P(X_n=k|X_i=k_i,i < n) = P(X_n=k|X_{n-1}=i_{n-1})$. That's not really analog to that kind of exercise I guess $\endgroup$ – user20183 Jan 12 '14 at 11:57
  • $\begingroup$ Except that $P(X_t\in A\mid X_u,u\leqslant s)$ actually involves conditionings very similar to the ones you manipulated in the discrete time setting. Agreed? $\endgroup$ – Did Jan 12 '14 at 12:38

About your try in the discrete-time discrete-space time, let $u\leqslant t\leqslant s$, $x$, $y$ and $z$ in the state space, $A=[X_u=x]$ and $B=[X_s=z]$ then $$ P[A\cap B\mid X_t=y]=\frac{P[X_u=x,X_t=y,X_s=z]}{P[X_t=y]}, $$ while $$ P[A\mid X_t=y]P[B\mid X_t=y]=\frac{P[X_u=x,X_t=y]\,P[X_t=y,X_s=z]}{P[X_t=y]^2}, $$ hence these coincide if and only if $$ \frac{P[X_t=y,X_s=z]}{P[X_t=y]}=\frac{P[X_u=x,X_t=y,X_s=z]}{P[X_u=x,X_t=y]}, $$ that is, $$ P[X_s=z\mid X_t=y]=P[X_s=z\mid X_u=x,X_t=y]. $$ Well, does (a) imply this? Because if it does, you showed that (a) implies a special form of (b), namely, (b) when $A$ and $B$ depend only on the value of the process at one time.

Then, extend this reasoning to show that (b) implies (a) in full. You shall consider events $$ A=\bigcap_i[X_{u_i}=x_i],\qquad B=\bigcap_j[X_{s_j}=z_j], $$ for finitely many $u_i\leqslant t\leqslant s_j$, prove (a) in this case and show this is enough to deduce (a) in full.

  • $\begingroup$ Thanks! With help of your answer I tried again to fix the discrete case (added to question). I am not quite sure if it is complete already. If it is, I wonder how to extend this to the continuous case.. $\endgroup$ – user20183 Jan 12 '14 at 14:14
  • $\begingroup$ I did not manage, since I can not simply evaluate $X_t$ in the continuous case and check the equality for every $X_t=a$...I somehow need an approach for those $P(X|Y)$ style random variables... $\endgroup$ – user20183 Jan 13 '14 at 21:42

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