Probability and Time Symmetry in Classical Markov Processes

Abstract

Definitions of time symmetry and examples of time-directed behaviour are discussed in the framework of discrete Markov processes. It is argued that typical examples of time-directed behaviour can be described using time-symmetric transition probabilities. Some current arguments in favour of a distinction between past and future on the basis of probabilistic considerations are thereby seen to be invalid.

Appendix

We now prove that symmetry of the transition probabilities (2.19), together with the further assumptions that the state space is finite and that the transition probabilities are continuous, implies equilibrium of the process.

We proceed by induction on the size n of the state space. The case $n=1$ is trivial. Assume that the result has been proved for all sizes $1\leq m<n$. We now prove it for n by reductio.

Assume that the single-time distribution is not invariant, i.e.

$$\exists s \exists t\geq s,\quad \textbf{p}(t)=P(t|s)\textbf{p}(s)\neq \textbf{p}(s)\ .$$

((2.51))

For the rest of the proof we now fix such an s.

Since we assume (2.19), i.e. $P(t|s)=P(s|t)$, we also have

$$\textbf{p}(s)=P(t|s)\textbf{p}(t)\ ,$$

((2.52))

and therefore

$$P(t|s)^2\textbf{p}(s)=\textbf{p}(s)\qquad\textrm{and}\qquad P(t|s)^2\textbf{p}(t)=\textbf{p}(t)\ .$$

((2.53))

Now fix a time $t\geq s$ and consider the matrix $P:=P(t|s)^2$. This is an $n\times n$ stochastic matrix that we can consider as the transition matrix of a homogeneous Markov process with discrete time. By (2.53), $\textbf{p}(t)$ and $\textbf{p}(s)$ are both invariant distributions for this Markov process, and by (2.51) they are different.

By the ergodic theorem for discrete-time Markov processes, existence of at least two different invariant distributions implies that there are at least two ergodic classes. Therefore (whether or not there are any transient states), P must have a block diagonal form

$$P=\left(\begin{array}{ll} P' &\ \ \textbf{0} \\ \textbf{0} &\ \ P'' \end{array}\right)\ ,$$

((2.54))

where P ^′ is an $m\times m$ matrix and P ^″ an $(n-m)\times (n-m)$ matrix, for some $0<m <n$.

For fixed s, $P=P(t|s)^2$ depends on t, and so a priori could m; but in fact $m(t)$ is independent of t. Indeed, assume there is an $m\neq m(t)$ such that for all $\varepsilon>0$ there is a $t'$ with $|t-t'|<\varepsilon$ and $m(t')=m$. The matrix elements of $P=P(t|s)^2$, in particular the ones off the diagonal blocks, are continuous functions of the transition probabilities. Therefore, by the continuity of the transition probabilities, $P(t|s)^2$ must also have zeros off the same diagonal blocks, i.e. $m=m(t)$, contrary to assumption. Therefore, for each $m\neq m(t)$ there is an $\varepsilon(m)>0$ such that for all t ^′ with $|t-t'|<\varepsilon(m)$ we have $m(t')\neq m$. Taking the smallest of these finitely many $\varepsilon(m)>0$, call it ɛ ₀, it follows that $m(t')=m(t)$ for all $t'$ in the open ɛ ₀-neighbourhood around t. However, again by the continuity of the matrix elements, this ɛ ₀-neighbourhood is also closed, and therefore it is the entire real line. Since t was arbitrary, $P(t|s)^2$ has the form (2.54) with the same m for all $t\geq s$.

We now focus on the matrix $P(t|s)$ itself rather than on $P(t|s)^2$. Assume that for some $t\geq s$ it has some element $p_{k|l}(t|s)$ outside of the $m\times m$ and $(n-m)\times (n-m)$ diagonal blocks. In order for $P(t|s)^2$ to have the given block diagonal form, several other elements of $P(t|s)$ have to be zero, in particular all elements in the k-th column of $P(t|s)$ that lie inside the corresponding diagonal block.

Since $P(t|s)$ is a stochastic matrix and every column sums to 1, it follows that already those elements of the k-th column that lie outside the diagonal blocks sum to 1, and therefore the sum of all elements in the diagonal blocks of $P(t|s)$, call it $d(t)$, is at most $n-1$, i.e.

$$d(t)=\sum_{i,j\leq m}p_{i|j}(t|s)+ \sum_{i,j\geq m+1}p_{i|j}(t|s)\leq n-1\ ,$$

((2.55))

for any $t\geq s$ such that $P(t|s)$ has some element outside of the diagonal blocks. Let t ₀ be the infimum of such t. By continuity, we have also

$$d(t_0)\leq n-1\ .$$

((2.56))

Now, if $t_0\neq s$, then for all $t<t_0$ we have that $d(t)=n$, but then by continuity $d(t_0)=n$, contradicting (2.56). If instead $t_0=s$, since $P(s|s)=\textbf{1}$, we again have $d(t_0)=n$, contradicting (2.56). For all $t\geq s$, thus, $P(t|s)$ has the same block diagonal form as $P(t|s)^2$ with fixed m.

But then, our original Markov process decomposes into two sub-processes, with state spaces of size m and $n-m$, respectively. If $\textbf{p}(t)\neq\textbf{p}(s)$ (assumption (2.51)), then the same must be true for at least one of the two sub-processes, but, by the inductive assumption, this is impossible. Therefore, (2.51) is false and

$$\forall s \forall t\geq s,\ \textbf{p}(t)=\textbf{p}(s)\ ,$$

((2.57))

QED.