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.

  • $\begingroup$ 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

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,

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


\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.

  • 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

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