Conditional persistence of Gaussian random walks

Let {Xn}n≥1 be a sequence of i.i.d. standard Gaussian random variables, let Sn = ∑n i=1 Xi be the Gaussian random walk, and let Tn = ∑n i=1 Si be the integrated (or iterated) Gaussian random walk. In this paper we derive the following upper and lower bounds for the conditional persistence: P { max 1≤k≤n Tk ≤ 0 ∣∣∣ Tn = 0, Sn = 0} . n, P { max 1≤k≤2n Tk ≤ 0 ∣∣∣ T2n = 0, S2n = 0} & n−1/2 logn , for n→∞, which partially proves a conjecture by Caravenna and Deuschel [3].


Introduction
Suppose that X n , n ≥ 1, are i.i.d.random variables with mean zero and finite positive variance.Denote S n = X 1 + X 2 + • • • + X n and T n = S 1 + S 2 + • • • + S n , n ≥ 1.In this paper, we study the following conjecture of Caravenna and Deuschel [3] which is motivated from their study of sticky particles in a random polymer: Here and throughout this paper, the following symbols are used for positive sequences α(n) and β(n where c 1 and c 2 are two positive constants.Furthermore, we denote α(n) β(n) if α(n) β(n) and α(n) β(n).We refer to [3] for the significance of the conjecture and its application in wetting and pinning models.
Here we remark that the question is indeed quite natural, by presenting a practical example.Suppose that a person holds n units of shares of a certain stock, of which options to sell the stock: either he sells all the n units of shares to get cash now, or he sells one unit of share per period for n periods.If the average rate of increase of the stock price during the n periods is the same as the constant simple interest rate r, and these two options make no difference at the end, then what is the probability that the person never regrets during the n periods after choosing the first option?By the assumptions, the stock price in the period k is P k = P 0 + S k + kr, where P 0 is the current stock price and S k = X 1 + X 2 + . . .+ X k is the random price after k periods with {X n } n≥1 being i.i.d.symmetric random variables.The person would not regret in the period k if P 1 + P 2 + . . .+ P k ≤ (P 0 + kr) Since there is no difference between the two options after n periods, we have S 1 + S 2 + . . .+ S n = 0. Furthermore, the average rate of increase of the stock price during the n periods is the same as the constant simple interest rate r, therefore S n = 0. Thus, the conditional probability that the person never regrets during the n periods can be expressed exactly as The conjecture is quite challenging.In their original paper [3], Caravenna and Deuschel showed that n −11/2 P {max 1≤k≤n T k ≤ 0 | T n = 0, S n = 0} (log n) −α for some positive α under a mild assumption on {X n }.Recently Aurzuda, Dereich and Lifshits [1] proved that the conjecture holds for the case when {X n } are i.i.d.Bernoulli random variables.Then, Denisov and Wachtel [6] announced an extension of the main result in [1], whose formal proof was not given but claimed to follow from the arguments in [5].While we believe that the methods proposed in [1] and in [6] for discrete random variables {X n } may be adapted with some appropriate modifications to handle continuous random variables, in this paper we use a more elementary method to study this conjecture for the case when {X n } are i.i.d.standard Gaussian random variables.
More precisely, we will prove the following: The main idea of our approach is to write the conditional probability as a ratio of two expectations.For the proof of the upper bound, we write the conditional probability as a ratio of expectations by singling out the middle two random variables X n/2 and X n/2 +1 , and then reduce the problem to the product of two unconditional persistence probabilities P max 1≤k≤ n/4 T k ≤ 0 and P max 3n/4 ≤k≤n T k ≤ 0 (where T is defined similarly as T using random variables {X k } k≥ 3n/4 instead of {X k } 1≤k≤ n/4 ).Since both unconditional persistence probabilities are of order n −1/4 (cf.[4]; see also [8], [2] and reference therein for other related persistence), the original conditional persistence is of order n −1/2 .This method works for any continuous random variables {X n } satisfying the corresponding inequality (3.4).For the proof of the lower bound, we rewrite the conditional probability as a ratio of expectations using the last two random variables X 2n−1 and X 2n .Then by the symmetry between the first n − 1 random variables X 1 , . . ., X n−1 and the last n − 1 random variables X n , . . ., X 2n−2 , we arrive at n −1/2 / log n.This proof can be also extended to some other random variables (such as exponential random variables) by using central limit theorem.However, a new method Conditional persistence of Gaussian random walks seems to be needed to remove the log n factor.

Preparation
For convenience, we introduce some notations.We set Similarly, we denote Thus, S 1,m = S m and T 1,m = T m .With these notations, we now can write for n ≥ 4 and k + 3 < n, Therefore, under the conditions T 1,n = 0 and S 1,n = 0, we have Together with the fact that T n,k+2 = T n,k+3 + S n,k+3 + X k+2 , we obtain Furthermore, under the conditions T 1,n = 0 and S 1,n = 0, If the density function of X 1 is denoted as f (x) = (2π) −1/2 e −x 2 /2 , then we claim that ) . (2.1) ECP 19 (2014), paper 70.
Proof of (2.1).Before the formal proof of (2.1), let us first show an equality which gives a good motivation of (2.1).Suppose that two random variables X and Y are standard Gaussian random variables, and h is a differentiable function, then we will show where f is the density function of a standard Gaussian random variable.We can regard (2.2) as the simplest case of (2.1), and these two proofs are essentially the same.The second equality in (2.2) is trivial, so we now prove the first equality in (2.2).A version of the conditional probability can be written as (cf.Section 2.13 in [7]) •) denotes the joint density function of the two-dimensional random variable (X, Y − h(X)).For notational simplicity, if we let Z = Y − h(X), then the joint density f X,Z (x, z) can be obtained by change of variables from (X, Y ) to (X, Z).More precisely, the Jacobian determinant is equal to Now we come to the proof of (2.1).If we denote W = (X 1 , . . ., X k , X k+3 , . . ., X n ), then, we can write where u, v are functions on R n−2 .Let g be the density function of W .Because W and X k+1 and X k+2 are independent, the joint density of W, X k+1 and X k+2 is g(w)f (x k+1 )f (x k+2 ).Thus, as in (2.2), the conditional density of (W | X k+1 = u(W ), X k+2 = v(W )) could be given as , the denominator can be written as Therefore, where a m = {max 1≤i≤m t 1,i ≤ 0} , b m = {max n−m+1≤i≤n t n,i ≤ 0} , s k,m and t k,m are defined similarly as S k,m and T k,m with {X i } replaced by {x i }.

Upper Bound
To prove the upper bound, we choose k = n/2 −1 and m = k/2 .Because A k ⊆ A m and B n−k−2 ⊆ B m , it follows from (2.1) that ) . and With these notations, (3.1) can be rewritten as Note that a, b, 1 Am and 1 Bm only depend on X 1 , ..., X k−m , X k+m+3 , ..., X n , while U and V only depend on X k−m+1 , ..., X k , X k+3 , ..., X k+m+2 .Therefore, a, b, 1 Am and 1 Bm are independent of (U, V ).If we can show that there exists a constant C > 0 such that for all real numbers α and β, then by conditioning on the variables X 1 , ..., X k−m , X k+m+3 , ..., X n , we can bound the numerator on the right-hand side of (3.
Thus, we immediately obtain q n ≤ C • P{A m ∩ B m }.By the unconditional persistence estimate obtained in [4], we have Note that (U, V ) has the same distribution as (Y m,m , Z m,m ).Thus (3.3) is equivalent to the following claim: there exists a constant C such that for all real number α and β, It remains to show the claim.To this end, we prove the following lemma.Lemma 3.1.If U and V are two centered Gaussian random variables, then for any α, β ∈ R, Proof.Without loss of generality, we can assume U = σ U X, and where X and Y are independent N (0, 1) random variables, and ρ = corr(U, V ).Conditioning on X and using the identity Conditional persistence of Gaussian random walks which holds for all c, t ∈ R, we obtain Taking expectation and using (3.5) again, we obtain which proves the lemma after simplification.
In either case, since m = k/2 , we immediately obtain (3.4) for C ≈ √ 8.This finishes the proof of the upper bound.

Lower Bound
The idea of the proof of the lower bound is similar to that of the upper bound.We first introduce a few more notations.For a fixed large n, we define two functions F 1 and F 2 as F 2 (y n+3 , . . ., y 2n+2 ) = f (y n+3 − 2y n+4 + y n+5 ) . . .f (y 2n − 2y 2n+1 + y 2n+2 ) • f (y 2n+1 − 2y 2n+2 )f (y 2n+2 ), and four sets For notational simplicity, we will derive a lower bound for q 2n+4 instead of q 2n .This of course makes no essential difference.Note that The denominator can be directly computed using Lemma 3.1: We thus focus on the numerator which can be expressed as a multiple integral with respect to the joint distribution of {X 1 , . . ., X 2n+2 }.But here we choose a multiple integral with respect to the joint distribution of {T 1 , . . ., T 2n+2 }.We do the following change of variables It is then straightforward to check that the Jacobian determinant is 1.Thus, the numerator becomes where the last equality comes from the symmetry of {F i } i=1,2 and f.
In order to estimate the last integral, we consider a subset D of Ω + 3 defined as D = (y n+1 , y n+2 ) ∈ R 2 : y n+1 ≥ 0, y n+2 ≥ 0, and y n+1 < n 3/2 (log n) By definition and using the unconditional persistence probability of [4], the first integral can be estimated as  The second integral over Ω + 3 \ D can be estimated as follows.From definition, G 1 (y n+1 , y n+2 )dy n+1 dy n+2 Since T n+1 is a Gaussian random variable with mean zero and variance n 3 /3+n

Theorem 1 . 1 .
If {X n } n≥1 are i.i.d.standard Gaussian random variables, S n = n i=1 X i and T n = n i=1 S i , then the following estimates hold ECP 19 (2014), paper 70.Page 2/9 ecp.ejpecp.org