A Non-markovian Process with Unbounded P -variation

A recent theorem in [3] provided a link between a certain function of transition probabilities of a strong Markov process and the boundedness of the p-variation of its trajectories. Here one assumption of that theorem is relaxed and an example is constructed to show that the Markov property cannot be easily dispensed with.


Introduction
Let ξ t , t ∈ [0, T ], be a strong Markov process defined on some complete probability space (Ω, F, P ) and with values in a complete separable metric space (X, ρ).Denote the transition probability function of ξ t by P s,t (x, dy), 0 ≤ s ≤ t ≤ T , x ∈ X.For any h ∈ [0, T ] and a > 0 consider the function α(h, a) = sup{P s,t (x, {y : ρ(x, y) ≥ a}) : x ∈ X, 0 ≤ s ≤ t ≤ (s + h) ∧ T }. ( The behavior of α(h, a) as a function of h gives sufficient conditions for regularity properties of the trajectories of the process ξ t .As Kinney showed in [1], ξ t has an almost surely càdlàg version if α(h, a) → 0 as h → 0 for any fixed a > 0, and an almost surely continuous version if α(h, a) = o(h) as h → 0 for any fixed a > 0. (See also an earlier paper by Dynkin [2]).
Recently we have established an interesting connection between the function α(h, a) and the p-variation of the trajectories of the process ξ t (see Theorem 2 below).The objective of this paper is twofold: first to relax an assumption on the parameter β in Theorem 2 and then to show that if ξ t is no longer Markov, the claim of Theorem 2 is false.To be more precise, we will construct an example of a càdlàg non-Markov process η t on [0, 1] with unbounded p-variation for any p > 0 and such that the inequality (2) below holds up to a logarithmic factor and hence satisfies the relaxed condition on β stated in Theorem 3. Some of the properties of this η t are given in Theorem 4.
Recall that for a p ∈ (0, ∞) and a function f defined on the interval [0, T ] and taking values Electronic Communications in Probability in (X, ρ) its p-variation is The range of the parameter β in the following definition is extended from "β ≥ 1" used in [3] to "β > 0" needed here.

Preliminaries and Results
As mentioned above, we first relax condition "β ≥ 1" in Theorem 2 as follows: Remark 1 Most of the interesting processes (e.g.Brownian motion, Levy processes on R n ) satisfy (2) with β ≥ 1 and the condition "p > γ/β" in Theorem 2 is sharp.Yet this need not be true in general.To see this consider a symmetric real-valued α-stable Lévy motion X t , t ∈ [0, 1], with α ∈ (0, 2).Furthermore, let f (x) = x δ for some δ ∈ (0, 1) and define Since the function f is a strictly increasing continuous bijection of [0, 1] onto itself, both the strong Markov and the p-variation properties are preserved (for the latter see e.g.Lemma 4.4 of [4]).This implies that Y t , like X t , is also a strong Markov process with bounded p-variation for any p > α (see e.g.[3]).On the other hand, X t has independent increments, so we also get for h ∈ (0, 1], a > 0 and some constant K > 0 So Theorem 3 (but not Theorem 2) applies if δ ∈ ((3−e)/(e−1), 1) and yields the boundedness of p-variation of paths of Y t only for any p > α/δ > α.
The following is needed for the construction of the example described previously.Let {δ m } ∞ m=1 be a monotone decreasing sequence of positive numbers such that ∞ m=1 δ m = 1.Set δ 0 := 0. In the future we will conveniently choose δ m := 2 • 3 −m , for m ≥ 1, but most of the arguments can be adapted for any sequence {δ m } which sums to one.Denote the partial sums by ∆ m , i.e. for any m = 1, 2, . . .let ∆ m := m k=0 δ k , and set J m = [∆ m−1 , ∆ m ).This way we get a partition of [0, 1) into a union of disjoint subintervals J m of length δ m .Next consider an increasing sequence of integers {n m } ∞ m=1 such that for any m ≥ 1, and let Furthermore, let {c m } ∞ m=0 be an increasing sequence of real numbers, converging to 1/2 as m → ∞, but rather slowly.To be more specific we take This choice is influenced by the desire to have a process with unbounded p-variation for any p > 0 and is, of course, not unique.Any sequence which converges to 1/2 even slower than {c m } will suit as well.Let X 0 , X 1 , . . .be a sequence of independent (but not identically distributed) Bernoulli random variables such that Consider a sequence {Z m } ∞ m=1 of independent random variables, independent from {X k } ∞ k=0 .Assume that for each m = 1, 2, . . ., a random variable Z m is distributed uniformly on the interval [a m , (1 + n m )a m ].Without loss of generality we will assume that all random variables X i , i ≥ 0, and Z m , m ≥ 1, are defined on the same probability space (Ω, F, P ).Furthermore, for m = 1, 2 . . ., set V m := d m + Z m and V m := V m + a m .Now on the interval [0, 1] define a random process η t ≡ η(t) as follows: set η 0 = X 0 , η 1 = 1/2 and on a subinterval J m for any m = 1, 2, . . .let where we set η(0−) ≡ η 0 and η(s−)(ω) = lim u↑s η u (ω), i.e. we use the left hand side limit ω-pointwise, for s ∈ (0, 1).The idea behind this definition is simple: we consider a process which with probability 1/2 starts either at 0 or at 1 and shortly before the end of the first subinterval J 1 returns back to η(0), if X 1 = 0, or, depending on X 0 , jumps to c 1 or 1 − c 1 , in case X 1 = 1.Inside J 2 , η t starts where it left J 1 , and at time t = ∆ 2 remains at η(∆ 1 ), if X 2 = 0, or, depending on X 0 , jumps to c 2 or 1 − c 2 , in case X 2 = 1, i.e. closer to 1/2.The same happens in any other interval J m .Another way to look at what is happening to η t at t = ∆ m , m = 1, 2, . . . is via Markov chains.Indeed, set W 0 := X 0 , W m := η(∆ m ), for m = 1, 2, . . ., and notice that {W k } ∞ k=0 defines a Markov chain on a state space For each m = 1, 2, . . ., W m is either equal to W m−1 , c m or 1 − c m .Furthermore, Later we will also use the sets The following theorem lists some of the interesting properties of the process η t .
Theorem 4 Let η t be defined by (7).Then (i) η t is almost surely left-continuous at t = 1, and hence the paths of η t are almost surely càdlàg; (ii) the p-variation v p (η) of η t is unbounded for any p > 0; (iii) there exists a constant K > 0 such that for any h ∈ (0, 1) and a > 0 Remark 2 Slightly abusing notation we define α(h, a) in (10) as For Markov processes this definition agrees with (1).

As an easy corollary we have
Corollary 5 The conclusion of Theorem 3 no longer holds if ξ t , t ∈ [0, T ] is not necessarily Markov.

Proofs
The proof of Theorem 3 requires only minor adjustments to the proof of Theorem 2. To be specific, we first slightly improve Lemma 2.5(iii) of [3] as follows: Lemma 6 Let γ(a, x) = x 0 u a−1 e −u du be the incomplete gamma function defined for a > 0 and x ≥ 0. Then for any q ∈ (0, 1 − 1/e) and a ≥ 2/q − 3 0 < aγ(a, 1) − Proof.The proof is identical to that of Lemma 2.5(iii) except for the right-hand inequality.
Since the series alternates plus, minus, etc., and the terms decrease in absolute value, three terms provide an upper bound: .
Before modifying Corollary 2.6 of [3], recall a few definitions.Let F ξ be the natural filtration generated by the process ξ, i.e.
Proof.The proof of this lemma is identical to that of Lemma 2.7 of [3].Just use Corollary 7 instead of Corollary 2.6 of [3].
To present the necessary changes to Lemma 3.1 of [3] we once again need a few definitions from [3].Let P P := {κ : κ = t i : 0 = t 0 < t 1 < • • • < t mκ = T } be the set of all point partitions of [0, T ].For any integer r, recall M r = 2 −r−1 and let κ ∈ P P, κ = {t i } mκ i=0 be an arbitrary point partition.Define the random sets Let r 1 be the largest integer less or equal to −(log 2 a 0 + 3), so that M r+2 ≥ a 0 for all r ≤ r 1 .
And finally replace 1.95 in the bound of P 1 on page 2063 of [3] by q+1/e 1−q−1/e to complete the proof of Theorem 3. Now switch attention to the proof of Theorem 4. The first two properties of η t are easy to establish while the third is somewhat more involved.
Proof of property (ii): it is easy to see that for any fixed ω ∈ Ω where for m = 1 and for m ≥ 2 we have Plugging in these bounds into (11) we obtain for every p > 0 and ω ∈ Ω. Hence v p (η • ) = +∞ almost surely for every p > 0.
Proof of property (iii): inequality (10) will follow once we obtain a bound on transition probabilities P (|η t − η s | > a|η s = y) for 0 ≤ s < t ≤ (s + h) ∧ 1 and y ∈ A ∞ .This will be done using two lemmas: the first will handle the case when both s and t belong to the same subinterval J m for some m ≥ 1, and the second will tackle the case s ∈ J m and t ∈ J k for k > m.
Lemma 10 Suppose η t is defined by (7).Let s, t ∈ J m for some m ≥ 1.Then for any y ∈ A ∞ (see (8)) almost surely where ψ m (x) = P (x < V m ) is given by Proof.For given s and t from J m , we have η t ∈ {y, 1 − y, c m , 1 − c m } whenever η s = y.Moreover, For s, t ∈ J m we have q 1 = 0 if y ∈ A m−1 (see ( 9)), and if y ∈ A m−1 where we used the definition of η t to get Using the function ψ m we can rewrite a bound for q 1 with y ∈ A m−1 as follows Let us now evaluate q 2 .For m = 1, 2, . . ., set then by definition of η t we get q 2 = 0 if y ∈ A 0 m−1 , and if y ∈ A 0 m−1 then The argument for q 3 is essentially the same: if |1 − c m − y| > a, then by definition of η t we get q 3 = 0 if y ∈ A 1 m−1 , and if y ∈ A 1 m−1 then Using the fact that the sets A 0 m−1 and A 1 m−1 are disjoint and combining the bounds ( 14),( 15), (16) together with (13) we obtain the claim of this lemma.
Lemma 11 Suppose η t is defined by (7).Let s ∈ J m and t ∈ J k for some 1 ≤ m < k.Then for any y ∈ A ∞ almost surely In particular, if δ m = 2 • 3 −m and n m is given by (4), then Proof.First notice that if η s = y, then we always have η t ∈ (A k \ A m−1 ) ∪ {y, 1 − y}, and so where For j < k we simply use the definition of η t and properties of X j to get v 0,j ≤ P (X j = 1|η s = y)1 {|cj −y|>a} ≤ δ j , v 1,j ≤ P (X j = 1|η s = y)1 {|1−cj −y|>a} ≤ δ j , and for j = k (7) yields where, using similar argument as for q 1 in the proof of Lemma 10, z∈Am−1 P (η s = y|η(∆ m−1 ) = z)P (η(∆ m−1 ) = z) Hence, for any y ∈ A m , by the definitions of δ m and ψ j , for j = m, k, we get Combining the above bounds we get the first inequality of this lemma.To get the second simply sum the tail of the geometric series with δ m+1 /δ m = 3 −1 , m ≥ 1.Furthermore, since 3 m > 2m for any m ≥ 0, the definitions of δ m and n m imply δ m−1 ≥ 4/n m , and the second inequality of this lemma follows.
Back to the proof of (iii) Lemmas 10 and 11 show that we need only bounds on various differences of the function ψ j for j = m, k to get a bound on transition probabilities.And the bound (10) will follow from the careful treatment of the supremum with respect to the index m.Therefore, we first provide the following inequalities which follow easily from the definition of ψ m (x): Thus for any 0 ≤ s < t ≤ 1 almost surely sup m: s,t∈Jm where K 2 can be taken to be 2K 1 (1 − δ 1 ) −1 and the last inequality follows from the choice of m(t − s) and δ m .Furthermore, t − s < a m(t−s)−1 = 3 3 m(t−s)−1 + 1 2 3 m(t−s)+1 −1 ≤ 3 −3 m(t−s)−1 , thus, for any 0 < t − s < h, since the function g(x) = x log 3 (1/x), x ∈ (0, 1], has the absolute maximum at x = 1/e equal to (e ln 3) −1 and is increasing for x ∈ (0, 1/e).Now let's look at the cases when s ∈ J m and t ∈ J k for m < k.Then it is clear that if 0 ≤ s < t ≤ 1 and η t = η s , then

Concluding remarks
One can choose larger n m 's to reduce the factor 1 ∨ ln 1 h in (10) to 1 ∨ ln ln 1 h , e.g. by taking log 3 (n m ) = 3 3 m − m − 1 + log 3 2, or something even smaller, but getting rid of this logarithmic factor completely poses a problem.The reason is simple: even though the functions ψ m are Lipschitz continuous, they are not uniformly Lipschitz for m ≥ 1 (the Lipschitz constants are of the order O(δ −1 m )).So the bound of q 1 (see ( 14)) cannot be essentially improved, unless one is willing to increase the speed of convergence to 1/2 of the sequence c m which would restrict the number of terms one needs to consider in (17) (see also (13)) and would also allow for the p-variation of η t to be almost surely finite for some p > 0.