MAXIMAL WEAK-TYPE INEQUALITY FOR STOCHASTIC INTEGRALS

Assume that X is a real-valued martingale starting from 0, H is a predictable process with values in [−1, 1] and Y is the stochastic integral of H with respect to X. The paper contains the proofs of the following sharp weak-type estimates. (i) If X has continuous paths, then


Introduction
Suppose that (Ω, F, P) is a complete probability space, filtered by a nondecreasing right-continuous family (F t ) t≥0 of sub-σ-fields of F. Assume in addition that F 0 contains all the events of probability 0. Assume further that X = (X t ) t≥0 is an adapted realvalued martingale with right-continuous paths that have limits from the left.Let Y be the Itô integral of H with respect to X, where H is a predictable process with values in [−1 , 1].Let X * = sup t≥0 X t , |X| * = sup t≥0 |X t | be the one-sided and two-sided maximal functions of X, respectively.We will also use the notation X * t = sup 0≤s≤t X s and |X t | * = sup 0≤s≤t |X s |, t ≥ 0.
One of the objectives of this paper is to study maximal weak-type inequalities between X and Y .Let us take a look at some related results from the literature, which may serve as a motivation.In the seminal paper [4], Burkholder invented a method of in which the constant 2 is optimal.These results, as well as the methodology, have been extended in many directions and applied in many areas of mathematics: see e.g.[1], [2], [5], [6], [7], [10], [11], [15], [16], [17] and references therein.
One can also ask about maximal versions of the above statements.In [8], Burkholder enhanced his method from [4], and applied it to this new class of estimates.As an illustration, he showed the bound where γ = 2.536 . . . is the unique solution of the equation This constant is the best possible in (1.2).For further results in this direction and much more, see [12], [13] and Chapter 7 in [14].The purpose of this paper is to study the maximal version of (1.1), i.e., the modification which involves the maximal function of X on the right.In view of (1.1) or (1.2), the inequality holds with some constant if |X| * is used.However, what may be a little unexpected, the estimate is also valid if only the one-sided maximal function is involved.Let us formulate the precise statement; for clarity, we have decided to consider the setting of arbitrary and continuous-path martingales separately.
Theorem 1.1.Suppose that X is a real-valued continuous-path martingale starting from 0 and Y is the integral, with respect to X, of a certain process H with values in [−1, 1].Then we have the inequality and the constant 2 is the best possible.
This result is perhaps not that surprising, especially in the light of the following reasoning: any continuous-path martingale can be represented as a time-changed Brownian motion B; furthermore, by Levy's theorem, the distributions of |B| and B * coincide.
Consequently, this suggests that the constants in the above maximal weak-type bound and in (1.1) should coincide.Actually, we will not try to formalize the above reasoning and present an alternative proof.
For martingales with possibly discontinuous trajectories, the situation gets much more interesting and the description of the corresponding optimal constant is far more complicated (see Section 3 below).Here is the precise statement.Our approach will depend heavily on the technique of Burkholder from [8], or rather its appropriate modification.We establish Theorem 1.1 in the next section, and devote the final part of the paper to the proof of Theorem 1.2.

Weak-type inequality for continuous-path martingales
We have decided to divide this section into two subsections.(1.3)As announced in the introduction, the key ingredient of the reasoning is a certain special function.Let U : R 3 → R be given by the formula

Proof of
It is easy to check that U is continuous.Furthermore, we have the following fact.
Lemma 2.1.Suppose that x ≤ z.Then To deal with the bound on the left, assume first that |y| ≥ This completes the proof.
We are ready to establish the estimate of Theorem 1.1.
Proof.It is convenient to split the reasoning into several parts.
Step 2. We proceed to the proof of (2.2).With no loss of generality we may assume that EX * < ∞; otherwise, there is nothing to prove.Introduce the process Z = (Z t ) t≥0 = ((X t , Y t , X * t )) t≥0 , fix T ≥ 0 and consider the stopping time Weak-type inequality We will show that (the integrability of the variables on the left and on the right follows from (2.1) and the assumption EX * < ∞).To do this, write the trivial equality Next, observe that where in the last passage we have used (2.1).Now take expectation throughout and add the obtained bound to the preceding equality; we get (2.3).
Step 3. Now we will handle the term EU (X τ ∧T , Y τ ∧T , X * τ ∧T ).By the very definition of τ , if t ∈ [0, τ ∧ T ), then Z t takes values in the set {(x, y, z) : |y| < x − z + 1}, on which U is of class C ∞ .Therefore, we are allowed to apply Itô's formula to get where where [X, X], [X, Y ], [Y, Y ] denote the square brackets of the processes indicated (see Dellacherie and Meyer [9] for details).Let us take a look at the terms I 1 − I 3 .The first of them has expectation zero, by the properties of stochastic integrals.To handle the second one, note that the process X * increases on the set {t : X t = X * t }.However, we easily derive that for such t, U z (X t , Y t , X * t ) = 0; this proves that I 2 = 0. Finally, note that U xx (x, y, z) = −2, U xy (x, y, z) = 0 and U yy (x, y, z) = 2 on {(x, y, z) : |y| < x − z + 1}. Consequently, Putting all the above facts together and integrating both sides of (2.4), we obtain Step 4. Now we combine the latter estimate with (2.3) and get By the left majorization in (2.1), this implies Since T was arbitrary and EX * T ≤ EX * , the inequality (2.2) follows.This completes the proof of (1.3).

Sharpness of (1.3)
We will exhibit an appropriate example.Suppose that B = (B t ) t≥0 is a standard, onedimensional Brownian motion starting from 0. Fix a small δ ∈ (0, 1/2) and introduce the following stopping times: τ 0 ≡ 0 and, inductively, almost surely) and the predictable process Finally, for any t ≥ 0, put Clearly, Y is a stochastic integral of H with respect to X. Let us describe briefly the behavior of the pair (X, Y ).It evolves according to the following, two-stage procedure.
It starts from (0, 0) and moves along the line of slope 1 until Y gets to −1/2 or to δ.If the first possibility occurs, the first stage τ is over; if the latter case happens, the pair starts moving along the line of slope −1, and does so until Y reaches 1/2 or 0. If Y gets to 1/2, the first stage τ is over; otherwise, (X, Y ) begins evolving along the line of slope 1, until Y visits −1/2 or δ.The pattern of movement is then repeated.Thus, at the end of the first stage τ , we see that |Y | = 1/2; furthermore, we also easily check that To describe the second stage, assume that Y τ = 1/2 (when Y τ = −1/2, the evolution is symmetric).Then (X, Y ) evolves along the line of slope 1 until Y reaches 1 or −1; by the previous considerations, we see that X cannot exceed X * + δ.Now the remainder of the proof is straightforward.By the above analysis, (in the last passage we used the integrability of τ ) and P(Y * ≥ 1) ≥ P(|Y σ | = 1) = 1.So, letting δ → 0, we see that the constant 2 is indeed the best possible in (1.3).

Proof of Theorem 1.2
In the non-continuous setting, the reasoning is much more involved.As previously, we have divided the section into a few parts.
Proof.Let us expand the function h into power series: Directly from the initial condition, we get that a 0 = −1 and a 1 = 1.Furthermore, the differential equation (3.1) gives the recurrence To do this, note that the above recurrence for (a k ) k≥0 gives Thus, (3.6) is proved.This property shows that the sequence (b k ) k≥0 is bounded, and hence a k is a sequence of negative numbers with a k = O(k −2 ).Consequently, the series ∞ k=0 a k converges and by Abel's theorem, we have as y ↑ 1.To deal with h (1−), note that this limit exists since h is concave on [0, 1) (see the proof of the previous lemma).The finiteness of this limit follows from (3.3) and the finiteness of h(1) we have just proved above.
We are ready to introduce the constant of inequality (1.4).Namely, put where the approximation comes from computer simulations.

A special function
In the proof of the inequality (1.4) and its sharpness, we will exploit the following auxiliary function u : (here h (1) is understood as the limit h (1−), and C is as in (3.8)).We easily see that u is continuous.Some further properties are studied in the lemmas below.Proof.The function u is differentiable at any point of the form (x, 0), x < 1, and u(x, y) = u(x, −y) for all x, y.Therefore, it is enough to prove the concavity along line segments contained in (−∞, 1] × (0, ∞).Fix any point (x, y) from this set, a number a ∈ [−1, 1] and consider the function G(t) = G x,y,a (t) = u(x + t, y + at) (defined for those t, for ECP 19 (2014), paper 25. which x + t ≤ 1 and y + at ≥ 0).The claim will follow if we manage to show that G is concave.To do this, we will prove that G (t) ≤ 0 for those t for which the second derivative exists, and G (t−) ≥ G (t+) for the remaining t.Actually, it is enough to show this for t = 0, due to the translation property G x,y,a (r + s) = G x+r,y+ar,a (s).If x < y, then G (0) = 0; if x > y, then G (0) = u xx (x, y) + 2au xy (x, y) + a 2 u yy (x, y).(3.4).Consequently, if we fix (x, y), then G (0) in (3.10) is the largest when a = 1.However, then G (0) = 0, since u is linear over line segments of slope 1 contained in (−∞, 1] × [0, ∞).The final case we need to consider is that of x = y.Then the second derivative of G at t = 0 does not exist (unless a = 1: but then G (0) = 0, as we have just pointed out), so we need to compare the left-and the right-hand derivative.We compute that G (0−) = 0 and This completes the proof.Some comments concerning the partial derivatives above are in order.Namely, the symbol u x (1, y) means the left-sided derivative lim h→0 (u(1, y) − u(1 − h, y))/h; moreover, the derivative u y (1, ±1) does not exist, and we take this symbol to be 0.
Proof.Let us first gather some information on the function y → u(1, y).This function is even and continuous on R.Moreover, it is equal to 1 − C when |y| ≥ 1, and is convex on [−1, 1] (see the above formula for u yy ).Consequently, if |y| ≥ 1, the estimate (3.11) is clear: the left-hand side is at most 1 − C, while the right is equal to 1 − C. Suppose then that |y| < 1; by symmetry, it is enough to study (3.11) for y ∈ (0, 1).Rewrite the bound as By continuity, we may assume that a = 0.As we have already written above, the function y → u(1, y) is continuous on R, convex on [−1, 1] and equal to 1 − C outside (−1, 1).On the other hand, the right-hand side of (3.12) is linear in d.Since both sides are equal for d = 0, we will be done if we show the estimate for d such that y + ad ∈ {−1, 1}.Assume that a and d satisfy this condition and look back at (3.12).Note that C + u x (1, y) = C yh (1 − y) + h (1 − y) ≥ 0. Indeed, by (3.1) and some manipulations, we rewrite this bound as (see the end of the proof of Lemma 3.1).Consequently, if ad and y are kept fixed, the right-hand side of (3.12) is minimal when d is minimal.Thus it is enough to prove this estimate when a ∈ {−1, 1}.We consider two cases separately.
or, after some manipulations, By (3.1), we actually have equality here: we will need this fact later on.

Proof of (1.4)
Using approximation arguments of Bichteler [3], it suffices to focus on the following discrete-time version of the maximal bound.Suppose that (Ω, F, P) is equipped with a filtration (F n ) n≥0 and let f = (f n ) n≥0 be an adapted real-valued martingale satisfying f 0 = 0 almost surely.Define the corresponding difference sequence (df n ) n≥0 by df 0 = f 0 and df be a predictable sequence, i.e., the sequence of random variables such that for each n, the term v n is F (n−1)∨0 measurable.A sequence g = (g n ) n≥0 is a transform of f by v, if for any n = 0, 1, 2, . . .we have the equality dg n = v n df n .Note that by the predictability of v, the sequence g is again a martingale.By the aforementioned results of Bichteler, (1.4) will follow if we establish the bound P(|g| * ≥ 1) ≤ CEf * , for all f , g as above, where v is assumed to take values in [−1, 1].Here, in analogy with the continuous-time setting, f * = sup n≥0 f n and |g| * = sup n≥0 |g n |.Actually, using some straightforward approximation and a stopping-time argument as in the continuous-time case, it is enough to show for simple f and g with f 0 = 0.Here by simplicity of f we mean that for any n, the random variable f n takes only a finite number of values and there exists a deterministic number N such that f N = f N +1 = . ... Clearly, if f and g are simple, then the almost sure limits f ∞ and g ∞ exist and are finite.Now we will describe the appropriate modification of Burkholder's method from [8], which will be useful in the proof of (3.13).
Suppose we want to prove the inequality for all pairs (f, g), where f is a simple martingale starting from 0 and g is its transform by a predictable sequence with values in [−1, 1].The key idea is to study a class U(V ) which consists of all functions U : D → R satisfying the four properties below: ECP 19 (2014), paper 25.
V (x, y, z) ≤ U (x, y, z) if (x, y, z) ∈ D (3.17) and the following further condition: for all (x, y, z) The interplay between the maximal inequality (3.14) and the class U(V ) is described in the theorem below.It is a simple modification of Theorems 2.2 and 2.3 in [8] (see also Section 11 in [4] and Chapter 7 in [14]).We omit the proof, as it requires only some straightforward minor changes.
Theorem 3.5.The inequality (3.14) holds for all n and all pairs (f, g) as above if and only if the class U(V ) is nonempty.
To handle (3.13), we will apply the above technique to the function We will prove the following fact.
Lemma 3.6.The function U is an element of U(V ).
(vii) All the other states are absorbing.
We did not specify the transition probabilities, they are uniquely determined by the requirement that the pair (F n , G n ) n≥0 is a martingale.To gain some intuition about the process, it is best to look at a few first steps.The pair starts from (1, y 0 ).In the first step, it moves, along the line of slope −1, either to (2 + y 0 , −1) (and then stops), or to (1 − δ, y 0 + δ).If the second possibility occurs, the next step begins, in which the pair moves along the line of slope 1.It moves either to (1, y 0 + 2δ) (and then it evolves as previously -along the line of slope −1, to one of the points (2 + y 0 + 2δ, −1), (1 − δ, y + 3δ), and so on), or to (1 − y 0 − 2δ, 0).In the latter case, it moves, again along the line of slope 1, to (1, y 0 + 2δ) (and then evolves according to the above scheme), or to (1 − y 0 − 3δ, −δ).
If the latter occurs, the pair moves along the line of slope −1, to (1, −y 0 − 4δ) (and then its movement is symmetric to that corresponding to the point (1, y 0 + 4δ)), or to (1 − y 0 − 4δ, 0) (in which the evolution is analogous to that corresponding to the starting point (1 − y 0 − 2δ, 0)).The pattern of the movement is then repeated.It is clear from the above description that for any n ≥ 1 we have dG n = v n dF n , for some predictable sequence v with values in {−1, 1}.Next, look at the process H = (|F n − 1| + |G n |) n≥0 .A careful inspection of the dynamics described in (i)-(viii) shows that this process is nondecreasing; this in particular gives that the pair (F n , G n ) n≥0 is bounded away from (1, 0).Now, note that the martingale (F n , G n ) n≥0 is simple: it is clear that for each n the random variable (F n , G n ) takes a finite number of values, and, as we will prove now, the number of nontrivial steps is bounded by 3N .Suppose on contrary that there is a trajectory of length exceeding 3N ; then the first 3N moves of (F n , G n ) n≥0 must be of type (ii)-(vii).However, if we look back at the process H = (|F n − 1| + |G n |) n≥0 and the conditions (ii)-(vii), we see that during each three consecutive moves of this type, the process H increases in at least 2δ.Since H 0 = y 0 , we get H 3N ≥ y 0 + 2N δ = 1, and this means that the evolution of (F n , G n ) n≥0 is over after 3N steps, a contradiction.Our final observation is that G ∞ ∈ {−1, 1} with probability 1.
The key fact is that the sequence (EU Here is the precise statement.

Lemma 3.7.
There is a constant K depending only on y 0 such that for each n, Proof.The idea is that ((F n , G n , F * n )) n≥0 evolves along the directions along which U is linear, or almost linear.We will present a detailed analysis for the moves described in (ii) and (iii), the remaining ones can be handled similarly.As we have proved in Lemma (see the proof of (3.11) with a = −1 and d = 1 + y).Now, suppose that P( For the moves in (iii), the argumentation is much simpler, as then (F, G) moves along line segment of slope 1 contained in (−∞, 1] × R, along which u is linear.Consequently, if P((F n , G n ) = (1 − δ, y + δ)) > 0 for some y ∈ (0, 1), then i.e., the desired bound holds with K = 0. We leave the analogous analysis of the moves from (iv)-(vi) to the reader.
Equipped with the above statements, we can now prove the sharpness of (1.4).Consider a "shifted" modification of the process (F, G): assume that (f, g) is a Markov martingale starting from (0, 0), which, at its first move jumps to (−1, −1) or to (y 0 , y 0 ); next, if it went to (−1, 1), it stops; if it went to (y 0 , y 0 ), then it evolves as (F − 1 + y 0 , G).
We easily see that G is a transform of F by a predictable sequence with values in {−1, 1} and P(|G| * ≥ 1) = 1.To handle EF * , we use Lemma 3.7.Namely, .
However, y 0 was an arbitrary number; letting y 0 → 0, we see that the above expression converges to h(1) −1 = C.This establishes the desired sharpness.

(2. 1 )
Proof.Let us start with the estimate on the right.If |y| ≥ x − z + 1, then both sides are equal.On the other hand, if |y| <

(3. 7 )
This is true for k = 3, since b 4 = −0.078125.Assuming that this property holds for some k, we use the above recurrence to get

, and ( 3 . 7 )
follows.To show that the sequence (b k ) k≥2 is nondecreasing, note that we have b 2 < b 3 (recall the above values of b 2 and b 3 ).Now, assuming that b k+1 − b k ≥ 0 for some k ≥ 2, we apply the above recurrence and the negativity of b k to get that