The problem is a power-law asymptotics of the probability that a self-similar process does not exceed a fixed level during long time. The exponent in such asymptotics is estimated for some Gaussian processes, including the fractional Brownian motion (FBM) in , and the integrated FBM in , .

1. The Problem

Let be a real-valued stochastic process with the following asymptotics: where is the so-called survival exponent of . Below we focus on estimating for some self-similar Gaussian processes in extended intervals and , . Usually the estimation of the survival exponents is based on Slepian’s lemma. The estimation requires reference processes with explicit or almost explicit values of . Unfortunately, the list of such processes is very short. This includes the fractional Brownian motion (FBM), , of order both with one- and multidimensional time. According to Molchan ([1]) Another important example is the integrated Brownian motion with the exponent (Sinai [2]).

The nature of this result is best understood in terms of a series of generalizations where the integrand is a random walk with discrete or continuous time (see, e.g., Isozaki and Watanabe [3]; Isozaki and Kotani [4]; Simon [5]; Vysotsky [6, 7]; Aurzada and Dereich [8]; Dembo et al. [9]; Denisov and Wachtel [10]. The extension of (1.3) to include the case of the integrated fractional Brownian motion, , remains an important; but as yet unsolved problem.

Below we consider the survival exponents for the following Gaussian processes: ; FBM in , ; the Laplace transform of white noise with ; the fractional Slepian’s stationary process whose correlation function is , .

Our approach to the estimation of is more or less traditional. Namely, any self-similar process in generates a dual stationary process , , where is the self-similarity index of . For a large class of Gaussian processes, relation (1.1) induces the dual asymptotics with the same exponent , [1, 11]. More generally, the dual exponent is defined by the asymptotics To formulate the simplest condition for the exponents to be equal, we define one more exponent by means of the asymptotics where is the position of the maximum of in , that is, .

Lemma 1.1 (see [1, 11]). Let be a self-similar continuous Gaussian process in and be the reproducing kernel Hilbert space associated with . Suppose that there exists such an element of that and . Then , and can exist simultaneously only; moreover, the exponents are equal to each other.

The equality reduces the original problem to the estimation of . Nonnegativity of the correlation function of guarantees the existence of the exponent , [12]. In turn, the inequality of two correlation functions, , , implies, by Slepian’s lemma, the inverted inequality for the corresponding exponents: .

An essentially different approach is required to find the explicit value of for FBM in and to estimate in (1.4) for the fractional Slepian process with a small parameter .

2. Examples

2.1. Integrated Fractional Brownian Motion

Consider the process where is the fractional Brownian motion, that is, a Gaussian random process with the stationary increments: , . Molchan and Khokhlov [13, 14] analyzed the exponent theoretically and numerically and formulated the following Hypothesis: for and for .

The unexpected symmetry for caused some doubt as to the numerical results. To support the hypothesis, Molchan [11] derived the following estimates of for : where is a small constant and (+) and are indicators of the intervals and , respectively. Note that, in the case of and , it is unknown whether the exponent exists. In such cases we have to operate with upper and lower exponents. Therefore, in (2.2) for is any number from the interval . The relation (2.2) can be improved as follows.

Proposition 2.1. For , one has(a) ,(b) ,(c) .

Proof. The identity of the dual exponents for follows from [14]; the dual survival exponent exists because the dual correlation function, is positive. The inequality (a) is a consequence of the relation
To prove (b, c), we use the correlation function of the process , that is, and the respective exponent (see (1.3)). The relation implies for . Using (a) in addition, we come to the lower bound in (b) because for .
Similarly, the relation implies (c) for all . A test of the purely analytical facts (2.4), (2.6), and (2.7) is given in the appendix.

Remark 2.2. Proposition 2.1(a) follows from the more informative relation: This inequality is important for understanding the numerical result by Molchan and Khokhlov [13] represented in the form of empirical estimates of in Figure 1. We can see that the empirical estimates show small but one-sided deviations from the hypothetical curve before and after . The signs of these deviations are consistent with (2.8), while the amplitudes are compatible with the model and (more can be found in [13]).

2.2. The Laplace Transform of White Noise

Consider the process , where is Brownian motion. The dual stationary process has the correlation function . Using (2.5) as a majorant of , we improve the lower bound of as follows.

Proposition 2.3. .

Proof. That the exponents for the dual processes and are equal follows from Lemma 1.1 with , where . For indeed, , where . By definition of the Hilbert space , we have the desired estimate: By (1.3) and Slepian’s lemma, the relation has as a consequence the estimate . The opposite inequality implies . The test of ((2.11), ) and ((2.12), ) is very simple and yields the Li and Shao [12, 15] estimates: . The appendix contains a proof of (2.11), (2.12) for all interesting values of : , and .

Remark 2.4. The dual survival exponent of is of interest as a parameter of the following asymptotic relation: for random polynomials with the standard Gaussian independent coefficients [16]. A continuous analogue of the polynomial on any of four intervals is the Laplace transform of white noise, which partially explains the appearance of in the asymptotic relation (2.13). Simulations suggest , [16] and , [17].

2.3. Fractional Slepian’s Process

We reserve this term for a Gaussian stationary process with correlation function because is known as the Slepian process and , is equal in distribution to the fractional Brownian motion on the interval (0,1). By the Polya criterion, the fractional Slepian process exists because is a nonincreasing and a convex function on the semiaxis. The fact of the correlation function being nonnegative guarantees the existence of in (1.4). can be useful as a reference process in estimation of the survival exponents. Therefore it is important to have accurate estimates of the exponent for . The case of small is the most interesting because it describes a transition of to white noise. Our estimates of are based on two lemmas, where we use the following notation:

Lemma 2.5 (see [12]). Let be a centered Gaussian stationary process with a finite nonnegative correlation function, that is, and for . Then the limit exists for every . Moreover,

Remark 2.6. Lemma 1.1 was derived by Li and Shao [12] for the Slepian process, , but the proof remains valid for the general case. There is an explicit but very complicated formula for with [18]. In case of , this result reduces to and gives .

Lemma 2.7 (see [8]). Let be a centered Gaussian process in an interval with a correlation function and be the Hilbert space with the reproducing kernel on . If , then

Remark 2.8. Lemma 2.7 is a version of Proposition  1.6 from the paper by Aurzada and Dereich [8]; relation (2.19) successfully supplements the original Lemma 1.1.

Proposition 2.9. The persistence exponent of process has the following estimates: where the left inequality holds for .

Corollary 2.10. If is the odd component of the fractional Brownian motion, then

Proof. The dual stationary process has the correlation function which is positive. Hence the exponent exists. The inequality and Proposition 2.9 immediately imply the corollary.

Remark 2.11. The following estimates of are due to Krug et al. [19]: For small these estimates are one-sided only.

Remark 2.12. A considerable difference in the behavior of and for small is expected. Heuristically this can be explained as follows. As , the discrete processes and have different weak limits: and , respectively, where and are independent standard Gaussian variables. The probability (1.4) for the limiting processes is quite different: Unfortunately, this argument fails to predict the behavior of for small , because the step cannot be arbitrary and is a function of .

2.4. Khanin’s Problem

The survival exponent for fractional Brownian motion in the intervals is independent of the parameter : . This interesting fact follows from both self-similarity of and the stationarity of its increments [1].

In the case , the variables and are positive correlated. Therefore, a possible power-law asymptotics where we change sign before for negative only, may have a radically different exponent compared with . The question of finding bounds on the exponent for the process was asked by K. Khanin. The next proposition contains a partial answer to this question.

Proposition 2.13. In the case , the exponent for exists and admits of the following estimates: in addition, .(2) Let be the lower exponent in (2.26), then

Remark 2.14. To clarify why is unbounded for small in the case , we consider again the limiting sequence for as . This is , where the are independent standard Gaussian variables. The probability (1.1) for the limit sequence is where is a slowly varying function, whereas for the limit sequence of we have where is the Gaussian distribution function. As in Remark 2.12, we have nontrivial exponential asymptotics where the threshold for is constant or bounded. Indeed, if the event in (2.31) is true, then

2.5. An Explicit Value of

We have two explicit but isolated results for the fractional Brownian motion: for and for . These results can be combined as follows.

Proposition 2.15. If , , then .

Remark 2.16. The result is based on the following properties of the position of the maximum of in : has a continuous probability density in (0, 1) and as . In the case of multidimensional time, the behavior of near is a key to the survival exponent of for and . By (1.2), in the case , and in the degenerate case: .

3. Proofs

Proof of Proposition 2.9
Lower Bound. Let be a dual fractional Brownian motion with the parameter , that is, a Gaussian stationary process with correlation function . We prove in the appendix that for , Applying Slepian’s lemma, one has because .
Upper Bound. The random variable corresponds to an element of the Hilbert space, , with the reproducing kernel . By definition of , we have
It is easy to see that . Therefore, Let be the median of the random variable , where . Then because . Setting in Lemma 2.5 and using notation (2.15), one has
Using Lemma 1.1 and the inequality , we have It is well known (see, e.g., [20]) that , where and is the Dudley entropy integral related to the semimetrics on : .
In our case and therefore where and is the standard Gaussian distribution. Hence,

Proof of Proposition 2.13
Part . In the case of , the process has nonnegative correlations on . In the standard manner, this implies the existence of for . More precisely, starting from a self-similar 2D process on , we consider the dual 2D stationary process whose correlation matrix has positive elements. By [12], we conclude that the exponent for exists.
The equality for . We will use Lemma 1.1. By the relation , the map is an isometry between the Hilbert spaces and associated with and on , respectively. To prove the equality of the dual exponents, it is enough to find such that for . We can use because (see [14]).
Estimation of . Since for any , , we have, by Slepian’s lemma, Using (1.2), one has .
Obviously, . Therefore, for any .
Part . Let , then for . Hence, Finally, But then, for all . If , then we get a lower bound of for .
The equality .Let and be the reproducing Kernel Hilbert spaces associated with and , respectively. By the definition of , the map is an isometric embedding of in . To prove that the exponents are equal, it is enough to find such that , for , and . As we showed above, this can be .

Proof of Proposition 2.15
Consider the fractional Brownian motion in . By Lemma 1.1, we can focus on the exponent related to the position of the maximum of in .
Distribution of . We remind the main properties of the distribution function, , of related to the normalized interval (see [1, 14]):(i) has a continuous density such that decreases and increases on ;(ii) have the following estimates: where .
Due to monotonicity of and , one has By (3.15), (3.16), Using (3.15), (3.17), one has If we set , then where , .
Distribution of . Let , where , then the processes and on are equal in distribution. Hence, and have the same distribution as well. Therefore, where . We have used here the existence and continuity of .
Exponent . Set . Then (3.21) implies .
Let , then as , and (3.20), (3.21) give a lower bound on : Here and below .
Using (3.18), (3.21), we get an upper bound on : By substituting , we have Hence, that is, .
The equality . Consider the Hilbert space related to FBM and a function The standard spectral representation of the kernel and the representation (3.26) yield where . Setting , the desired statement follows from Lemma 1.1 because and .


Relation (2.4). .
By (2.3), one has for small and large where . Therefore, we have the following asymptotics for : These relations support (2.4) both for small and large enough . To verify (2.4) in the general case, we consider the following test function: . Using new variables: , , the test function is transformed to a function on the square . Namely, , where We have to show that . It is easy to see that at the boundary of . By (A.1), in vicinities of two sides of : and . The same is true for the other sides: and because Here To verify , note that , where . Obviously, has a single zero in (0,1), that is, has a single extreme point. But and for small . Therefore, .
Numerical testing supports the desired inequality for interior points of .

Comment 1. Our preliminary numerical test was concerned with points on a grid with a step of 0.005. The first derivatives of are uniformly bounded from above on . This fact helps us to find the final grid step to prove for all interior points of . The relevant analysis is cumbersome and so has been omitted.

Relation (2.6). .
To verify the inequality, we consider the following test function: . Using (2.3), (2.5), and new variables , we will have the following representation for the test function:
One has in vicinities of two sides of : and , because
The same is true for the other sides: and .

One has and because To prove (A.9), note that and as . Hence, (A.9) holds if has a single extremum in (0,1). By we conclude that is a concave function with two zeroes in (0,1), because and as or 1.
This means that has two extremums in (0,1) only. But , and for small because as . Hence has a single zero in (0,1) and has a single extremum.
We have proved that for small .

Here and because .
Hence, .
As a result near the boundary of . Numerical testing supports the desired inequality for the interior of (see more in the Comment 1 from the appendix section “Relation (2.4)”).

Relation (2.7). .
Let . By change of variables: and , we get a test function on and the relation between and is One has In addition, Finally, and where . By (A.9), .
Therefore near the boundary of . The numerical testing supports this conclusion for the interior of (see more in the Comment 1 from the appendix section “Relation (2.4)”).

Relations (2.11), (2.12)
Consider , where and is given in (2.5). By the change of variables , we transform the test function to a function on such that Taking into account the asymptotics of near 0, we come to a necessary condition for to be negative, namely, . Let , then , that is, .
The Case . Here, as . An additional condition on we can get from the relation as . One has , where By , we have and . Thus if .
The Case . Here, is a polynomial, , and , that is, is a concave function with . Therefore, and as a result, .
Consider . One has and . Therefore, , if is convex, that is, . To verify this property, note that where .
Obviously, if . This holds for 0<.
For , The right part here is positive for , that is, for .
Let . Then where . We have and