# Markov chains and conditioning on impossible events

Consider a Markov chain $(X_0,X_1,\ldots)$ with a state space $S\equiv\{s_1,s_2\}$ and the following matrix of “transition probabilities” (I will explain the use of quotation marks below): \begin{align*} \begin{array}{c|cc} &s_1&s_2\\ \hline s_1&1&0\\ s_2&1&0 \end{array} \end{align*} That is, no matter what initial state the system starts in, it will always end up in state $s_1$ in one period and stay there forever.

Rigorously speaking, these “transition probabilities” are to be interpreted as follows: \begin{align*} \mathbb P\,(X_{n}=s_1\,|\,X_{n-1}=s_1)=&\,1,\\ \mathbb P\,(X_{n}=s_2\,|\,X_{n-1}=s_1)=&\,0,\\ \mathbb P\,(X_{n}=s_1\,|\,X_{n-1}=s_2)=&\,1,\\ \mathbb P\,(X_{n}=s_2\,|\,X_{n-1}=s_2)=&\,0 \end{align*} for each $n\in\mathbb N$.

My concern is that the last two probabilities are ill-defined (except possibly for $n=1$), because for any given initial probabilities, the condition events $\{X_{n-1}=s_2\}_{n=2}^{\infty}$ have zero probability! Strictly speaking, therefore, the above matrix cannot be interpreted as conditional probabilities because of the problem of conditioning on events that never occur.

What is the standard resolution of this technical problem? Does one make the hand-waving assumption of defining conditional probabilities that depend on impossible events anyway, or is there a more sophisticated and rigorous way around this issue?

Any input is appreciated.

In fact one defines the transition kernels: $$K(\omega, j) = \Bbb{E}(1_{X_n = j} \mid \mathcal{F}_{n-1})(\omega) \quad \Bbb{P } \,a.s.$$

This means that we have a regular conditional probability that allows us to talk about the jumps of our process.

The markov property consists in saying that $K(\cdot, j)$ is $\sigma(X_{n-1})$ measurable

that is

$$K(\omega, j) = \phi_j(X_{n-1})(\omega) \quad \Bbb{P}\, a.s.$$

Hence there is no hand waving.

• I see, thank you. So basically a rigorous way of operationalizing conditioning on null events is to regard the conditional probabilities $$\mathbb P\,(X_n=j\,|\,\sigma(X_0,\ldots X_{n-1}))$$ as being defined up to almost-everywhere equivalence (with respect to the unconditional probability measure $\mathbb P$) and impose the Markov property as the equivalence classes of $$\mathbb P\,(X_n=j\,|\,\sigma(X_0,\ldots X_{n-1}))$$ and $$\mathbb P\,(X_n=j\,|\,\sigma(X_{n-1}))$$ being identical. Does that sound right? – triple_sec Aug 13 '15 at 0:47

As you point out, there is no guarantee that each of the states in a Markov Chain can be visited at any particular step $n$. Another good example with lots of impossible visits is a periodic chain where any one state can only be visited on every $d$th step, where $d$ is the period.

More subtle are examples on countably infinite state spaces. For example, consider a random walk on the non-negative integers with a 'drift' towards 0 and a reflecting barrier at 0. The higher-numbered states may be visited extremely rarely.

Some texts do not address the interpretation of conditional probability for conditions with probability 0 at all. Others agree to ignore the conditional probability in cases where it is moot. Yet others, instead of your "for all $n$" seem to say "for all applicable $n$." In any case, this issue is not a difficulty in applications.