2
$\begingroup$

Given a Markov process $(X_t)_{t \geq 0}$ I would like to verify that $P(X_{t_3} \in A | X_{t_2} \in B \; X_{t_1} \in C)= P(X_{t_3} \in A | X_{t_2} \in B)$ with $t_3 > t_2 > t_1$. I know it should be obvious, but I would like to prove thaat using (if possible) only the definition of a Markov process, that is $E[\phi(X_{t+h})| \mathcal{F}_t]= E[\phi(X_{t+h})|X_t]$ or equivalently $ P(X_{t+h}| \mathcal{F}_t)= P(X_{t+h}|X_t)$. It should be really easy, but I just don't get how to use conditional probability to make it work.

$\endgroup$
2
  • $\begingroup$ What does ${F}_t$ mean in your question? $\endgroup$ – soobster Jul 11 '19 at 20:49
  • $\begingroup$ It is the filtration at a time $t$ $\endgroup$ – tommy1996q Jul 11 '19 at 20:58
2
$\begingroup$

Firstly, this is not actually true. For example, define an i.i.d. sequence of random variables $\{X_i\}$ defined by,

$$X_i = \begin{cases} 1 &\text{ with probability }1/2\\ -1 &\text{ with probability }1/2 \end{cases}.$$

Let $S_n = \sum_{i=1}^n X_i$ (here $S_0 = 0$). Then $\{S_n:n\in\mathbb{N}\}$ is a Markov chain. Let $t_1 = 1$,$t_2 = 2$ and $t_3 = 3$. Let $C = \{S_1 = 1\}$, $B = \{|S_2|\leq 2\}$ and $A = \{S_3 = -3\}$. If at time 1, $S_1 = 1$, then it's impossible for $S_3 = -3$, so

$$P(S_{t_3} \in A|S_{t_2}\in B,S_{t_1}\in C) = 0.$$

On the other hand, $|S_2| \leq 2$ with probability 1, so

$$P(S_{t_3} \in A|S_{t_2}\in B) = P(S_3= -3) = 1/8 \neq 0 = P(S_{t_3} \in A|S_{t_2}\in B,S_{t_1}\in C).$$

I think what you actually wanted to prove was,

$$P(X_{t_3} \in A|X_{t_2},X_{t_1}) = P(X_{t_3} \in A|X_{t_2}).$$

Notice, in the discrete setting, we can write this as,

$$P(X_{t_3} \in A|X_{t_2}=x_2,X_{t_1}=x_1) = P(X_{t_3} \in A|X_{t_2}=x_2),$$

whenever $P(X_{t_2}=x_2,X_{t_1}=x_1) > 0$. But we cannot replace $(x_2,x_3)$ with non-atomic sets. That's why the definition of the conditional expectation is a general statement about integrals over different filtrations rather than a straight-forward conditional probability as defined in an undergraduate probability course.

We can do this through repeated iteration of the tower property. Notice that,

$$\sigma(X_{t_2}) \subset \sigma(X_{t_2},X_{t_1}) \subset \mathcal{F}_{t_2}.$$

Then,

\begin{align*} P(X_{t_3}|X_{t_2},X_{t_1}) &= E[P(X_{t_3}|X_{t_2},X_{t_1})|\mathcal{F}_{t_2}] \\ &= E[P(X_{t_3}|\mathcal{F}_{t_2})|X_{t_2},X_{t_1}] \\ &= E[P(X_{t_3}|X_{t_2})|X_{t_2},X_{t_1}] \\ &= P(X_{t_3}|X_{t_2}). \end{align*}

Hope that helps!

*edits: Fixed some typos.

$\endgroup$
2
  • 1
    $\begingroup$ Great answer! Thanks! $\endgroup$ – tommy1996q Jul 20 '19 at 8:57
  • $\begingroup$ No problem. I remember specifically working through this exact question while I was preparing for prelims a couple years ago, so it was a pleasure to go over that again. $\endgroup$ – forgottenarrow Jul 20 '19 at 11:11

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.