# When is a Markov chain a Markov chain?

I’ve been taking tutorials on the third quarter of the second-year probability course, in which the student have met discrete-time Markov chains for the first time. The hardest aspect of this introduction (apart from the rapid pace – they cover only slightly less material than I did in Cambridge, but in half the time) is, in my opinion, choosing which definition of the Markov property is most appropriate to use in a given setting.

We have the wordy “conditional on the present, the future is independent of the past”, which is probably too vague for any precise application. Then you can ask more formally that the transition probabilities are the same under two types of conditioning, that is conditioning on the whole history, and conditioning on just the current value

$\mathbb{P}(X_{n+1}=i_{n+1} \,\big|\, X_n=i_n,\ldots,X_0=i_0) = \mathbb{P}(X_{n+1}=i_{n+1} \,\big |\, X_n=i_n),$ (*)

and furthermore this must hold for all sets of values $(i_{n+1},\ldots,i_0)$ and if we want time-homogeneity (as is usually assumed at least implicitly when we use the word ‘chain’), then these expressions should be functions of $(i_n,i_{n+1})$ but not n.

Alternatively, one can define everything in terms of the probability of seeing a given path:

$\mathbb{P}(X_0=i_0,\ldots,X_n=i_n)= \lambda_{i_0}p_{i_0,i_1}\ldots p_{i_{n-1}i_n},$

where $\lambda$ is the initial distribution, and the $p_{i,j}$s are the entries of the transition matrix P.

Fortunately, these latter two definitions are equivalent, but it can be hard to know how to proceed when you’re asked to show that a given process is a Markov chain. I think this is partly because this is one of the rare examples of a concept that students meet, then immediately find it hard to think of any examples of similar processes which are not Markov chains. The only similar concept I can think of are vector spaces, which share this property mainly because almost everything in first-year mathematics is linear in some regard.

Non-examples of Markov chains

Anyway, during the tutorials I was asking for some suggestions of discrete-time processes on a countable or finite state space which are not Markov chains. Here are some things we came up with:

• Consider a bag with a finite collection of marbles of various colours. Record the colours of marbles sampled repeatedly without replacement. Then the colour of the next marble depends on the set you’ve already seen, not on the current colour. And of course, the process terminates.
• Non-backtracking random walk. Suppose you are on a graph where every vertex has degree at least 2, and in a step you move to an adjacent vertex, chosen uniformly among the neighbours, apart from the one from which you arrived.
• In a more applied setting, it’s reasonable to assume that if we wanted to know the chance it will rain tomorrow, this will be informed by the weather over the past week (say) rather than just today.

Showing a process is a Markov chain

We often find Markov chains embedded in other processes, for example a sequence of IID random variables $X_1,X_2,\ldots$. Let’s consider the random walk $S_n=\sum_{i=1}^n X_i$, where each $X_i =\pm 1$ with probability p and (1-p). Define the running maximum $M_n=\max_{m\le n}S_m$, and then we are interested in $Y_n:=M_n-S_n$, which we claim is a Markov chain, and we will use this as an example for our recipe to show this in general.

We want to show (*) for the process $Y_n$. We start with the LHS of (*)

$\mathbb{P}(Y_{n+1}=i_{n+1} \,\big|\, Y_n=i_n,\ldots,Y_0=i_0),$

and then we rewrite $Y_{n+1}$ as much as possible in terms of previous and current values of Y, and quantities which might be independent of previous values of Y. At this point it’s helpful to split into the cases $i_n=0$ and $i_n\ne 0$. We’ll treat the latter for now. Then

$Y_{n+1}=Y_n+X_{n+1},$

so we rewrite as

$=\mathbb{P}(X_{n+1}=i_{n+1}-i_n \, \big |\, Y_n=i_n,\ldots, Y_0=i_0),$

noting that we substitute $i_n$ for $Y_n$ since that’s in the conditioning. But this is now ideal, since $X_{n+1}$ is actually independent of everything in the conditioning. So we could get rid of all the conditioning. But we don’t really want to do that, because we want to have conditioning on $Y_n$ left. So let’s get rid of everything except that:

$=\mathbb{P}(X_{n+1}=i_{n+1}-i_n\, \big |\, Y_n=i_n).$

Now we can exactly reverse all of the other steps to get back to

$= \mathbb{P}(Y_{n+1}=i_{n+1} \,\big|\, Y_n=i_n),$

which is exactly what we required.

The key idea is that we stuck to the definition in terms of Y, and held all the conditioning in terms of Y, since that what actually determines the Markov property for Y, rearranging the event until it’s in terms of one of the underlying Xs, at which point it’s easy to use independence.

Showing a process is not a Markov chain

Let’s show that $M_n$ is not a Markov chain. The classic mistake to make here is to talk about possible paths the random walk S could take, which is obviously relevant, but won’t give us a clear reason why M is not Markov. What we should instead do is suggest two paths taken by M, which have the same ‘current’ value, but induce transition probabilities, because they place different restrictions on the possible paths taken by S.

In both diagrams, the red line indicates a possible path taken by $(M_0,M_1,\ldots,M_4)$, and the blue lines show possible paths of S which could induce these.

In the left diagram, clearly there’s only one such path that S could take, and so we know immediately what happens next. Either $X_5=+1$ (with probability p) in which case $M_5=S_5=3$, otherwise it’s -1, in which case $M_5=2$.

In the right diagram, there are two possibilities. In the case that $S_4=0$, clearly there’s no chance of the maximum increasing. So in the absence of other information, for $M_5=3$, we must have $X_4=X_5=+1$, and so the chance of this is $p^2$.

So although the same transitions are possible, they have different probabilities with different information about the history, and so the Markov property does not hold here.

# Hitting Probabilities for Markov Chains

This continues my previous post on popular questions in second year exams. In the interest of keeping it under 2,500 words I’m starting a new article.

In a previous post I’ve spoken about the two types of Markov chain convergence, in particular, considering when they apply. Normally the ergodic theorem can be used to treat the case where the chain is periodic, so the transition probabilities do not converge to a stationary distribution, but do have limit points – one at zero corresponding to the off-period transitions, and one non-zero. With equal care, the case where the chain is not irreducible can also be treated.

A favourite question for examiners concerns hitting probabilities and expected hitting times of a set A. Note these are unlikely to come up simultaneously. Unless the hitting probability is 1, the expected hitting time is infinite! In both cases, we use the law of total probability to derive a family of equations satisfied by the probabilities/times. The only difference is that for hitting times, we add +1 on the right hand side, as we advance one time-step to use the law of total probability.

The case of hitting probabilities is perhaps more interesting. We have:

$h_i^A = 1,\; i\in A, \quad h_i^A=\sum_{j\in S}p_{ij}h_j^A,\; i\not\in A.$

There are two main cases of interest: where the chain is finite but has multiple closed communicating classes, and where the chain is infinite, so even though it is irreducible, a trajectory might diverge before hitting 0.

For the case of a finite non-irreducible Markov chain, this is fairly manageable, by solving backwards from states where we know the values. Although of course you could ask about the hitting probability of an open state, the most natural question is to consider the probability of ending up in a particular closed class. Then we know that the hitting probability starting from site in the closed class A is 1, and the probability starting from any site in a different closed class is 0. To find the remaining values, we can work backwards one step at a time if the set of possible transitions is sparse enough, or just solve the simultaneous equations for $\{h_i^A: i\text{ open}\}$.

We therefore care mainly about an infinite state-space that might be transient. Typically this might be some sort of birth-and-death chain on the positive integers. In many cases, the hitting probability equations can be reduced to a quadratic recurrence relation which can be solved, normally ending up with the form

$h_i=A+B\lambda^i$,

where $\lambda$ might well be q/p or similar if the chain is symmetric. If the chain is bounded, typically you might know $h_0=1, h_N=0$ or similar, and so you can solve two simultaneous equations to find A and B. For the unbounded case you might often only have one condition, so you have to rely instead on the result that the hitting probabilities are the minimal solution to the family of equations. Note that you will always have $h^i_i=1$, but with no conditions, $h^i_j\equiv 1$ is always a family of solutions.

It is not clear a priori what it means to be a minimal solution. Certainly it is not clear why one solution might be pointwise smaller than another, but in the case given above, it makes sense. Supposing that $\lambda<1$, and A+B=1 say, then as we vary the parameters, the resulting set of ‘probabilities’ does indeed vary monotically pointwise.

Why is this true? Why should the minimum solution give the true hitting probability values? To see this, take the equations, and every time an $h_i^A$ appears on the right-hand side, substitute in using the equations. So we obtain, for $i\not\in A$,

$h_i^A=\sum_{j\in A}p_{ij}+\sum_{j\not\in A} p_{ij}h_j^A,$

and after a further iteration

$h_i^A=\sum_{j_1\in A}p_{ij_1}+\sum_{j_1\not\in A, j_2\in A}p_{ij_1}p_{j_1j_2}+\sum_{j_1,j_2\not\in A}p_{ij_1}p_{j_1j_2}h_{j_2}^A.$

So we see on the RHS the probability of getting from i to A in one step, and in two steps, and if keep iterating, we will get a large sum corresponding to the probability of getting from i to A in 1 or 2 or … or N steps, plus an extra term. Note that the extra term does not have to correspond to the probability of not hitting A by time N. After all, we do not yet know that $(h_{i}^A)$ as defined by the equations gives the hitting probabilities. However, we know that the probability of hitting A within N steps converges to the probability of hitting A at all, since the sequence is increasing and bounded, so if we take a limit of both sides, we get $h_i^A$ on the left, and something at least as large as the hitting probability starting from i on the right, because of the extra positive term. The result therefore follows.

It is worth looking out for related problems that look like a hitting probability calculation. There was a nice example on one of the past papers. Consider a simple symmetric random walk on the integers modulo n, arranged clockwise in a circle. Given that you start at state 0, what is the probability that your first return to state 0 involves a clockwise journey round the circle?

Because the system is finite and irreducible, it is not particularly interesting to consider the actual hitting probabilities. Also, note that if it is convenient to do so, we can immediately reduce the problem when n is even. In two steps, the chain moves from j to j+2 and j-2 with probability ¼ each, and stays at j with probability ½. So the two step chain is exactly equivalent to the lazy version of the same dynamics on n/2.

Anyway, even though the structure is different, our approach should be the same as for the hitting probability question, which is to look one step into the future. For example, to stand a chance of working, our first two moves must both be clockwise. Thereafter, we are allowed to move anticlockwise. There is nothing special about starting at 0 in defining the original probability. We could equally well ask for the probability that starting from j, the first time we hit 0 we have moved clockwise round the circle.

The only thing that is now not obvious is how to define moving clockwise round the circle, since it is not the case that all the moves have to be clockwise to have experienced a generally clockwise journey round the circle, but we definitely don’t want to get into anything complicated like winding numbers! In fact, the easiest way to make the definition is that given the hitting time of 0 is T, we demand that the chain was at state n at time T-1.

For convenience (ie to make the equations consistent) we take $h_0=0, h_n=1$ in an obvious abuse of notation, and then

$h_j=\frac12h_{j-1}+\frac12 h_{j+1},$

from which we get

$h_j=a+bj \Rightarrow h_j=\frac{j}{n}.$

Of course, once we have this in mind, we realise that we could have cut the circle at 0 (also known as n) and unfolded it to reduce the problem precisely to symmetric gambler’s ruin. In particular, the answer to the original problem is 1/2n, which is perhaps just a little surprising – maybe by thinking about the BM approximation to simple random walk, and that BM started from zero almost certainly crosses zero infinitely many times near we might have expected this probability to decay faster. But once it is unfolded into gambler’s ruin, we have the optimal stopping martingale motivation to reassure us that this indeed looks correct.

# Convergence of Transition Probabilities

As you can see, I haven’t got round to writing a post for a while. Some of my reasons for this have been good, and some have not. One reason has been that I’ve had to give a large number of tutorials for the fourth quarter of the second year probability course here in Oxford. The second half of this course concerns discrete-time Markov chains, and the fourth problem sheet discusses various modes of convergence for such models, as well as a brief tangent onto Poisson Processes. I’ve written more about Poisson Processes than perhaps was justifiable in the past, so I thought I’d say some words about convergence of transition probabilities in discrete-time Markov chains.

Just to be concrete, let’s assume the state space K is finite, and labelled {1,2,…,k}, so that it becomes meaningful to discuss

$p_{12}^{(n)}:=\mathbb{P}(X_n=2|X_0=1).$

That is, the probability that if we start at state 1, then after n ‘moves’ we are at state 2. We are interested in the circumstances under which this converges to the stationary distribution. The heuristic is that we can view a time-step of a Markov chain as an operation on the space of distributions on K. Note that this operation is deterministic. If this sounds complicated, what we mean is that we specify an initial distribution, that is the distribution of $X_0$. If we consider the distribution of $X_1$, this is given by $\lambda P$, where $\lambda$ is the initial distribution, and P the transition matrix.

Anyway, the heuristic is that the stationary distribution is the unique fixed point of this operation on the space of distributions. It is therefore not unreasonable to assume that unless there are some periodic effects, we expect repeated use of this operation to move us closer to this fixed point.

We can further clarify this by considering the matrix form. Note that a transition P always has an eigenvalue equal to 1. This is equivalent to say that there is a solution to $\pi P=\pi$. Note it is not immediately equivalent to saying that P has a stationary distribution, as the latter must be non-negative and have elements summing to one. Only the first property is difficult, and relies on some theory or cleverness to prove. It can also be shown that all eigenvalues satisfy $|\lambda|\le 1$, and in general, there will be a single eigenvalue (ie dimension 1 eigenspace) with $|\lambda|=1$, and the rest satisfies $|\lambda|<1$. Then, if we diagonalise P, it is clear why $\pi P^n$ converges entry-wise, as $\pi UP^n U^{-1}$ converges. In the latter, only the entries in the row corresponding to $\lambda=1$ converge to something non-zero.

In summary, there is a strong heuristic for why in general, the transition probabilities should converge, and if they converge, that they should converge to the stationary distribution. In fact, we can prove that for any finite Markov chain, $p_{ij}^{(n)}\rightarrow \pi_j$, provided we two conditions hold. The conditions are that the chain is irreducible and aperiodic.

In the rest of this post, I want to discuss what might go wrong when these conditions are not satisfied. We begin with irreducibility. A chain is irreducible if it has precisely one communicating class. That means that we can get from any state to any other state, not necessarily in one step, with positive probability. One obvious reason why the statement of the theorem cannot hold in this setting is that $\pi$ is not uniquely defined when the chain is not irreducible. Suppose, for example, that we have two closed communicating classes A and B. Then, supported on each of them is an invariant distribution $\pi^A$ and $\pi^B$, so any affine combination of the two $\lambda \pi^A+(1-\lambda) \pi^B$ will give a stationary distribution for the whole chain.

In fact, the solution to this problem is not too demanding. If we are considering $p_{ij}^{(n)}$ for $i\in A$ a closed communicating class, then we know that $p_{ij}^{(n)}=0$ whenever $j\not\in A$. For the remaining j, we can use the theorem in its original form on the Markov chain, with state space reduced to A. Here, it is now irreducible.

The only case left to address is if i is in an open communicating class. In that case, it suffices to work out the hitting probabilities starting from i of each of the closed communicating classes. Provided these classes themselves satisfy the requirements of the theorem, we can write

$p_{ij}^{(n)}\rightarrow h_i^A \pi^A_j,\quad i\not\in A, j\in A.$

To prove this, we need to show that as the number of steps grows to infinity, the probability that we are in closed class A converges to $h_i^A$. Then, we decompose this large number of steps so to say that not only have we entered A with roughly the given probability, but in fact with roughly the given probability we entered A a long time in the past, and so there has been enough time for the original convergence result to hold in A.

Now we turn to periodicity. If a chain has period k, this says that we can split the state space into k classes $A_1,\ldots,A_k$, such that $p_{ij}^{(n)}=0$ whenever $n\not\equiv j-i \mod k$. Equivalently, the directed graph describing the possible transitions of the chain is k-partite. This definition makes it immediately clear that $p_{ij}^{(n)}$ cannot converge in this case. However, it is possible that $p_{ij}^{(kn)}$ will converge. Indeed, to verify this, we would need to consider the Markov chain with transition matrix $P^k$. Note that this is no longer irreducible, as it there are no transitions allowed between classes $A_1,\ldots,A_k$. Indeed, a more formal definition of the period, in terms of the lcd of possible return times allows us to conclude that there is no finer reducibility structure. That is, $A_1,\ldots,A_k$ genuinely are the closed classes when we consider the chain with matrix $P^k$. And so the Markov chain with transition matrix $P^k$ restricted to any of the $A_i$s satisfies the conditions of the theorem.

There remains one case which I’ve casually brushed over. When we were discussing the irreducible case, I said that if we had at least one communicating classes, then we could work out the limiting transition probabilities from a state in an open class to a state in a closed class by calculating the hitting probability of that closed class, then applying the standard version of the theorem to that closed class. This relies on the closed class being aperiodic.

Suppose otherwise that the destination closed class A has period k as before. If it were to be the case that the number of steps required to arrive at A had some fixed value mod k, or modulo a non-trivial divisor of k, then we certainly wouldn’t have convergence, for the same reasons as in the globally periodic case. However, we should ask whether we can ever have convergence?

In fact, the answer is yes. For concreteness, and because it’s easier to write ‘odd’ and ‘even’ than $m \mod k$, let’s assume A has size 2 and period 2. That is, once we arrive in A, thereafter we alternate deterministically between the two states. Anyway, for some large time n, we can write $p_{ca}^{(n)}$ for $a\in A, c\not\in A$ as:

$p_{ca}^{(n)}=h_i^A(n),$

where the latter term is the probability that we arrive in A at a time-step which has the same parity as n. It’s not terribly hard to come up with an example where this holds, and this idea holds in greater generality, where A has period k (and not necessarily just k states), we have to demand that the probability of arriving at a time which is a mod k is equal for all a in [0,k-1].

Of course, for applications, we don’t normally care much about irreducible chains, and we can easily remove periodicity by introducing so-called laziness, whereby on each time-step we flip a coin (biased if necessary) and stay put if it comes up heads, and apply the transition matrix if it comes up tails. Then it’s possible to get from any state to itself in one step, and so we are by construction aperiodic.

# Mixing Times 1 – Reversing Markov Chains

A small group of us have started meeting to discuss Levin, Peres and Wilmer’s book on Markov Chains and Mixing Times. (The plan is to cover a couple of chapters every week, then discuss points of interest and some of the exercises – if anyone is reading this and fancies joining, let me know!) Anyway, this post is motivated by something we discussed in our first session.

Here are two interesting facts about Markov Chains. 1) The Markov property can be defined in terms of products of transition probabilities giving the probability of a particular initial sequence. However, a more elegant and general formulation is to say that, conditional on the present, the past and the future are independent. 2) All transition matrices have at least one equilibrium distribution. In fact, irreducible Markov Chains have precisely one equilibrium distribution. Then, if we start with any distribution, the distribution of the chain at time t converges to the equilibrium distribution.

But hang on. This might be a fairly serious problem. On the one hand we have given a definition of the Markov property that is symmetric in time, in the sense that it remains true whether we are working forwards or backwards. While, on the other hand, the convergence to equilibrium is very much not time-symmetric: we move from disorder to order as time advances. What has gone wrong here?

We examine each of the properties in turn, then consider how to make them fit together in a non-contradictory way.

Markov Property

As many of the students in the Applied Probability course learned the hard way, there are many ways to define the Markov property depending on context, and some are much easier to work with than others. For a Markov chain, you can find a way to say that the transition probability $\mathbb{P}(X_{n+1}=x_{n+1}\,|\,X_n=x_n,\ldots,X_0=x_0)$ is independent of $x_0,\ldots,x_{n-1}$. Alternatively, you can use this to give an inductive specification for the probability of the first n values of X being some sequence.

It requires a moment’s checking to see that the earlier definition of past/future independence is consistent with this. Let’s first check that we haven’t messed up a definition somewhere, and that the time-reversal of a general Markov chain does have the Markov property, even as defined in the context of a Markov chain.

For clarity, consider $X_0,X_1,\ldots, X_N$ a Markov chain on some finite state space, with N some fixed finite end time. We aren’t losing anything by reversing over a finite time interval – after all, we need to know how to do it over a finite time interval before it could possibly make sense to do it over $(-\infty,\infty)$. We examine $(Y_n)_{n=0}^N$ defined by $Y_n:= X_{N-n}$.

$\mathbb{P}(X_n=x_n|X_{n+1}=x_{n+1},\ldots,X_N=x_N)=\mathbb{P}(X_n=x_n|X_{n+1}=x_{n+1})$

is the statement of the Markov property for $(Y_n)$. We rearrange the left hand side to obtain:

$=\frac{\mathbb{P}(X_n=x_n,X_{n+1}=x_{n+1},\ldots,X_N=x_N)}{\mathbb{P}(X_{n+1}=x_{n+1},\ldots,X_N=x_N)}$

$=\frac{\mathbb{P}(X_N=x_N|X_n=x_n,\ldots,X_{N-1}=x_{N-1})\mathbb{P}(X_n=x_n,\ldots,X_{N-1}=x_{N-1})}{\mathbb{P}(X_N=x_N|X_{n+1}=x_{n+1},\ldots,X_{N-1}=x_{N-1})\mathbb{P}(X_{n+1}=x_{n+1},\ldots,X_{N-1}=x_{N-1})}.$

Now, by the standard Markov property on the original chain $(X_n)$, the first probability in each of the numerator and denominator are equal. This leaves us with exactly the same form of expression as before, but with one fewer term in the probability. So we can iterate until we end up with

$\frac{\mathbb{P}(X_n=x_n,X_{n+1}=x_{n+1})}{\mathbb{P}(X_{n+1}=x_{n+1})}=\mathbb{P}(X_n=x_n|X_{n+1}=x_{n+1}),$

as required.

So there’s nothing wrong with the definition. The reversed chain Y genuinely does have this property, regardless of the initial distribution of X.

In particular, if our original Markov chain starts at a particular state with probability 1, and we run it up to time N, then saying that the time-reversal is a Markov chain too is making a claim that we have a non-trivial chain that converges from some general distribution at time 0 to a distribution concentrated at a single point by time N. This seems to contradict everything we know about these chains.

Convergence to Equilibrium – Markov Property vs Markov Chains

It took us a while to come up with a reasonable explanation for this apparent discrepancy. In the end, we come to the conclusion that Markov chains are a strict subset of stochastic processes with the Markov property.

The key thing to notice is that a Markov chain has even more regularity than the definition above implies. The usual description via a transition matrix says that the probability of moving to state y at time t+1 given that you are at state x at time t is some function of x and y. The Markov property says that this probability is independent of the behaviour up until time t. But we also have that the probability is independent of t. The transition matrix P has no dependence on time t – for example in a random walk we do not have to specify the time to know what happens next. This is the property that fails for the non-stationary time-reversal.

In the most extreme example, we say $X_0=x_0$ with probability 1. So in the time reversal, $\mathbb{P}(Y_N=x_0|Y_{N-1}=y_{N-1})=1$ for all $y_{N-1}$. But it will obviously not be the case in general that $\mathbb{P}(Y_n=x_0|Y_{n-1}=y_{n-1})=1$ for all $y_{n-1}$, as this would mean the chain Y would be absorbed after one step at state $x_0$, which is obviously not how the reversal of X should behave.

Perhaps the best way to reconcile this difference is to consider this example where you definitely start from $x_0$. Then, a Markov chain in general can be thought of as a measure on paths, that is $\Omega^N$, with non-trivial but regular correlations between adjacent components. (In the case of stationarity, all the marginals are equal to the stationary distribution – a good example of i.d. but not independent RVs.) This is indexed by the transition matrix and the initial distribution. If the initial distribution is a single point mass, then this can be viewed as a restriction to a smaller set of possible paths, with measures rescaled appropriately.

What have we learned?

Well, mainly to be careful about assuming extra structure with the Markov property. Markov Chains are nice because there is a transition matrix which is constant in time. Other processes, such as Brownian motion are space-homogeneous, where the transitions, or increments in this context, are independent of time and space. However, neither of these properties are true for a general process with the Markov property. Indeed, we have seen in a post from a long time ago that there are Markov processes which do not have the Strong Markov Property, which seems unthinkable if we limit our attention to chain-like processes.

Most importantly, we have clarified the essential point that reversing a Markov Chain only makes sense in equilibrium. It is perfectly possibly to define the reversal of a chain not started at a stationary distribution, but lots of unwelcome information from the forward chain ends up in the reversed chain. In particular, the theory of Markov Chains is not broken, which is good.

# Markovian Excursions

In the previous post, I talked about the excursions of a Brownian motion. Today I’m thinking about how to extend these ideas to more general Markov chains. First we want to rule out some situations. In particular, we aren’t hugely interested in discrete time Markov chains. The machinery is fairly well established for excursions, whether or not the chain is transient. Furthermore, if the state space is discrete, as for a Poisson process for example, the discussion is not hugely interesting either. Remember that the technical challenges in the constructions of local time arise because of the Blumenthal 0-1 law property that Brownian motion visits 0 infinitely often in the small window after the start time. We therefore want the process to be regular at the point of the state space under discussion. This is precisely the condition described above for BM about 0.

Why is it harder in general?

The informal notion of a local time should transfer to a more general Markov chain, but there are some problems. Firstly, to define something in terms of an integral is not general enough. The state space E needs some topological structure, but any meaningful definition must be in terms of functions from E into the reals. There were also all sorts of special properties of Brownian motion, including the canonical time-space rescaling that came in handy in that particular case. It turns out to be easiest to consider the excursion measure on a general Markov chain through its Laplace transform.

Definition and Probabilistic interpretations

The resolvent is the Laplace transform of the transition probability $P_t(x,A)$, viewed as an operator on functions $f:E\rightarrow \mathbb{R}$.

$R_\lambda f(x):=\mathbb{E}_x\left[\int_0^\infty e^{-\lambda t}f(X_t)dt\right]=\int_0^\infty e^{-\lambda t}P_tf(x)dt$.

We can interpret this in terms of the original process in a couple of ways which may be useful later. The motivation is that we would like to specify a Poisson process of excursions, for which we need to know the rate. We hope that the rate will in fact be constant, so it will in fact to suffice to work out the properties of the expected number of excursions (or whatever) up to various random times, in particular those given by exponential RVs.

So, we take $\zeta\sim\text{Exp}(\lambda)$ independent of everything else, and assume that we ‘kill the chain’ at time $\zeta$. Then, by shuffling expectations in and out of the integral and separating independent bits, we get:

$R_\lambda f(x)=\mathbb{E}_x\int_0^\zeta f(X_s)ds = \frac{1}{\lambda}\mathbb{E}_xf(X_\zeta)$.

As in the Brownian local time description

$R_\lambda 1_A(x)=\mathbb{E}(\text{time spent in }A\text{ before death at time }\zeta_\lambda)$.

Markovian property

We want to show that excursions are Markov, once we’ve sorted out what an ‘excursion’ actually means in this context. We do know how to deal with the Markovian property once we are already on an excursion. It is relatively straightforward to define an extension of the standard transition probability operator, to include a condition that the chain should not hit point a during the transition. That is

$_aP_t(x,A):= \mathbb{P}_x(X_t\in A\cap H_a>t)$.

This will suffice to define the behaviour once an excursion has started. The more complicated bit will be the entrance law $n_t(A)$, being the probability of arriving at A after time t of an excursion. To summarise, as with BM, all the technical difficulties with an excursion happen at the beginning, ie bouncing around the start point, but once it is ‘up-and-running’, its path is Markovian and controlled by $_aP_t$.

Marking

The link between the resolvent and the excursions, is provided as in the Brownian case, by supplying a PPP of marks at uniform rate $\lambda$ to real time. This induces a mark process on excursions, weighted by an (exponential) function of excursion length. We make no distinction between an excursion including one mark or many marks. Then the measure on marked excursion is, in a mild abuse of notation:

$n_\lambda=(1-e^{-\lambda\zeta(f)})\cdot n.$

We compare with the Laplace transform $n_\lambda(dx)=\int_0^\infty e^{-\lambda t}dtn_t(dx)$ using a probabilistic argument.

We can apply the measure to a function in the usual way: $\lambda n_\lambda(1_A)$ is the measure of those excursions for which the first mark occurs in $A$. So by taking $A=E$, we get

$\lambda n_\lambda(1)=\text{ Excursion measure }=\int_U n(df)(1-e^{-\lambda\zeta(f)}).$

We have therefore linked the exponential mark process on excursion measure with the Laplace transform of the entrance law. So in particular:

$\frac{\lambda n_\lambda(A)}{\lambda n_\lambda(1)}=\mathbb{P}(\text{first mark when in }A)=\int_0^\infty \lambda e^{-\lambda t}P_t(0,A)dt=\lambda R_\lambda 1_A(0)$.

The resolvent is relatively easy to calculate explicitly, and so we can find the Laplace transform $n_\lambda(A)$. From this it is generally possible to invert the transform to find the entrance law $n_t$.

References

A Guided Tour Through Excursions – L. C. G. Rogers.

This pair of posts is very much a paraphrase of chapters 3 and 4 of this excellent text. The original can be found here (possibly not open access?)