The maximum of Brownian motion with parabolic drift

We study the maximum of a Brownian motion with a parabolic drift; this is a random variable that often occurs as a limit of the maximum of discrete processes whose expectations have a maximum at an interior point. We give series expansions and integral formulas for the distribution and the first two moments, together with numerical values to high precision.


Introduction
Let W (t) be a two-sided Brownian motion with W (0) = 0; i.e., (W (t)) t≥0 and (W (−t)) t≥0 are two independent standard Brownian motions. We are interested in the process W γ (t) := W (t) − γt 2 (1.1) for a given γ > 0, and in particular in its maximum We also consider the corresponding one-sided maximum Since the restrictions of W to the positive and negative half-axes are independent, we have the relation where N γ is an independent copy of N γ . Note that (a.s.) W γ → −∞ as t → ±∞, so the maxima in (1.2) and (1.3) exist and are finite; moreover, they are attained at unique points and M γ , N γ > 0. It is easily seen (e.g., by Cameron-Martin) that M γ and N γ have absolutely continuous distributions. The purpose of this paper is to provide formulas for the distribution function of M and, in particular, its moments. The main results are given in Section 2, with proofs and further details in Sections 3-5. The numerical computations are discussed in Section 6. We use many more or less wellknown results for Airy functions; for convenience we have collected them in Appendices A-B. Finally, Appendix C discusses an interesting integral equation, while Appendix D contains an alternative proof of an important formula used in our proofs.

1.2.
Background. The random variable M is studied by Barbour [4], Daniels and Skyrme [8] and Groeneboom [9]. It arises as a natural limit distribution in many different problems, and in many related problems its expectation E M enters in a second order term for the asymptotics of means or in improved normal approximations. For various examples and general results, see for example Daniels [6;7], Daniels and Skyrme [8], Barbour [4;5], Smith [17], Louchard, Kenyon and Schott [14], Steinsaltz [18], Janson [12]. As discussed in several of these papers, the appearance of M in these limit results can be explained as follows, ignoring technical conditions: Consider the maximum over time t of a random process X n (t), defined on a compact interval I, for example [0, 1], such that as n → ∞, the mean E X n (t), after scaling, converges to deterministic function f (t), and that the fluctuations X n (t) − E X n (t) are of smaller order and, after a different scaling, converge to a gaussian process G(t). If we assume that f is continuous on I and has a unique maximum at a point t 0 ∈ I, then the maximum of the process X n (t) is attained close to t 0 . Assuming that t 0 is an interior point of I and that f is twice differentiable at t 0 with f (t 0 ) = 0, we can locally at t 0 approximate f by a parabola and G(t) − G(t 0 ) by a two-sided Brownian motion (with some scaling), and thus max t X n (t) − X n (t 0 ) is approximated by a scaling constant times the variable M , see Barbour [4]. In the typical case where the mean of X n (t) is of order n and the Gaussian fluctuations are of order n 1/2 , it is easily seen that the correct scaling is that n −1/3 (max t X n (t) − X n (t 0 )) d −→ cM , for some c > 0, which for the mean gives E max t X n (t) = nf (t 0 ) + n 1/3 c E M + o(n 1/3 ), see [4; 6; 7]. As examples of applications in algorithmic and data structures analysis, this type of asymptotics appears in the analysis of linear lists, priority queues and dictionaries [13; 14] and in a sorting algorithm [12].

Main results
The mean of M can be expressed as integrals involving the Airy functions Ai and Bi, for example as follows.
Theorem 2.1 (Daniels and Skyrme [8]). 2 (2.1) Ai(t) 2 + Bi(t) 2 2 t dt. (2.4) The expressions (2.1) and (2.2) (unfortunately with typos in the latter) are given by Daniels and Skyrme [8]. Since detailed proofs of the formulas are not given there, we for completeness give a complete proof in Section 5. (The proof includes a direct analytical verification of the equivalence of (2.1) and (2.2), which was left open in [8].) By (A.1) and (A.8), |Ai(iy)| increases superexponentially as y → ±∞, while Ai(t) decreases superexponentially and Bi(t) increases superexponentially as t → ∞; hence, the integrands in the integrals in Theorem 2.1 all decrease superexponentially and the integrals converge rapidly, so they are suited for numerical calculations. We obtain by numerical integration (using Maple), improving the numerical values in [4; 5; 8; 7], E M = 0.99619 30199 28363 11660 37766 . . . (2.5) We do not know any similar integral formulas for the second moment of M (or higher moments). Instead we give expressions using infinite series, summing over the zeros a k , k ≥ 1, of the Airy function, see Appendix A.
Recall that these zeros all are real and negative, so we have 0 > a 1 > a 2 > . . . , see [1, (10.4.94)] and Appendix A; note that |a k | k 2/3 , see (A.30). (We use x n y n , for two sequences of positive numbers x n and y n , to denote that 0 < lim inf n→∞ x n /y n ≤ lim sup n→∞ x n /y n < ∞; this is also denoted x n = Θ(y n ).) We first introduce more notation. Let F N (x) be the distribution function of N , i.e., F N (x) := P(N ≤ x), and let F M (x) be the distribution function of M ; further, let be the corresponding tail probabilities. Then, by (1.4), and, equivalently, If we know G(x), we thus know the distribution of both N and M , and we can compute moments by Two formulas for the distribution function are given in the following theorem. Others are given in (3.3) and Lemma 3.5. The proof is given in Section 3.
Hi(a k ) Ai (a k ) Ai(a k + 2 1/3 x), x > 0. (2.10) The sum converges conditionally but not absolutely for every x > 0. Alternatively, with an absolutely convergent sum, for x ≥ 0, The function G(x) is plotted in Figure 1.
Remark 2.3. By (A.4) and (A.30), for any fixed x > 0, |Ai(a k + 2 1/3 x)| is usually of the order |a k | −1/4 k −1/6 , and using also (A.31) and (A.33), the summands in (2.10) are (typically) of the order k −1/6−1/6−2/3 = k −1 , so this sum is not absolutely convergent. (For some values of k, the term may be smaller than k −1 because a k + 2 1/3 x may be close to another zero, but such cases are infrequent and do not prevent the series from being absolutely divergent.) On the other hand, by (A.22), πHi(z) + z −1 = O(|z| −4 ) on the negative real axis, and it follows that the terms in (2.11) are O(k −3 ), so the series is absolutely convergent. Moreover, since Ai is bounded on the real axis (see (A.1) and (A.4)), the sum in (2.11) converges uniformly for x ≥ 0, and is thus a continuous function of x; this is no surprise since we already  have remarked that N has an absolutely continuous distribution, so G is continuous. Note also that (2.11) for x = 0 is the trivial G(0) = 1, since each Ai(a k ) = 0, while (2.10) does not hold for x = 0.
The sum in (2.11) can be differentiated termwise and we have the following result, proved in Section 3.
Theorem 2.4. N and M have absolutely continuous distributions with infinitely differentiable density functions, for x > 0, Integral formulas for f N (x) will be given in (3.10) and (5.10). The density functions f N (x) and f M (x) are plotted in Figures 2 and 3.
Remark 2.5. In contrast, the sum Hi(a k ) Ai (a k ) Ai (a k + 2 1/3 x) (2.14) obtained by termwise differentiation of (2.10) is not convergent for any x ≥ 0, as will be seen in Section 3.
Moments of M and N now can be obtained from (2.8) and (2.9) by integrating (2.11) termwise. This yields the following result; see Section 4 for proofs as well as related integral formulas. For higher moments, see Remark 4.4.  We define for convenience Theorem 2.6. The means and second moments of M and N are given by the absolutely convergent sums

Distributions
Salminen [16, Example 3.2] studied the hitting time and gave the formula [16, (3.10)], for x, β > 0, (with α = −β in his notation), for the density function of τ . Note that τ is a defect random variable, and that τ = ∞ if and only if min t≥0 (x + W (t) + βt 2 ) > 0. By symmetry, Hence, choosing β = 1/2, If we formally integrate termwise we obtain (2.10), from (A.16). However, as seen in Remark 2.3, the sum is not absolutely convergent, so we cannot use e.g. Fubini's theorem, and we have to justify the termwise integration by a more complicated argument. 3) converge rapidly for each fixed t, because of the negative term a k t in the exponent. But the convergence rate is small for small t, and when integrating we have the problem just described.
For (large) integers N ∈ N, let R N := ( 3 2 πN ) 2/3 , and let Γ N := Γ N (θ 0 , x 0 ) be the closed contour obtained from Γ by cutting the infinite rays at r = R N and connecting them by the arc Γ N : Note that by (A.30), |a N | < R N < |a N +1 | (at least for large N ; in fact for all N ≥ 1). Thus, Γ N goes around the N first zeros of Ai; moreover, Γ N does not come too close to any of the zeros; this is made more precise by the estimates in Lemma A.2 and Lemma A.3.
Lemma 3.2. Let τ = τ x be the hitting time (3.1) for β = 1/2 and some x > 0. Then the defect random variable τ has the density function, for t > 0, Proof. For x > 0 and z ∈ Γ or z ∈ Γ N , Lemmas A.1 and A.2 yield Ai(z + 2 1/3 x)/Ai(z) = O(1), and thus the integrand in (3.4), Φ(z) say, is bounded by This shows both that Γ Φ(z) dz is absolutely convergent, and that Φ(z) has simple poles at the zeros a k of Ai, and evaluating Γ N Φ(z) dz by residues, we see that it equals the partial sum of the N first terms of (3.2). Consequently, (3.2) yields Γ N Φ(z) dz → f τ (t) as N → ∞, and the result follows by (3.5).
Remark 3.3. The contour Γ in (3.4) can be deformed to the imaginary axis, using the estimate in Lemma A.1. Hence, setting z = 2 1/3 si, Moreover, this holds also for t ≤ 0, with f τ (t) = 0, since the right-hand side of (3.6) then easily is shown to vanish: writing it again as a line integral along the imaginary axis we can move the line of integration to Re z = σ, for any σ > a 1 , and for t ≥ 0 we may let σ → +∞, again using Lemma A.1. This exhibits e t 3 /6 f τ (t) as the inverse Fourier transform of s → Ai(2 1/3 (si+ x))/Ai(2 1/3 si). By Lemma A.1, this function is integrable and in L 2 , and using (A.1) and (A.2), it is seen that so is its derivative, which implies that the Fourier transform e t 3 /6 f τ (t) is integrable. The Fourier inversion formula yields and, more generally, by analytic continuation, (and, in fact, for Re z > a 1 ). This formula for the Laplace transform is (in a more general version) given by Groeneboom [9, Theorem 2.1], where e t 3 /6 f τ (t) is denoted h 1/2,x (t); see also (5.5) below. Conversely, this formula from [9] yields by Fourier inversion (3.6) and (3.4), so we could have used it instead of (3.2) from [16] as our starting point. We give an alternative proof of (3.8) in Appendix D, which thus gives us a self-contained proof of Lemma 3.2. (Groeneboom [9], Salminen [16] and our Appendix D use similar methods. See also Appendix C for another approach.) Remark 3.4. For our purposes we consider only x > 0 in (3.1). For x < 0, the hitting time is a.s. finite; its distribution is found in Martin-Löf [15].
) with respect to t and interchange the order of integration, which is allowed by Fubini's theorem and the estimate Lemma A.1, which implies that for z ∈ Γ, the integrand in (3.4) is bounded by O exp(−cx|z| 1/2 − t 3 /6) . The result follows by (A.16) (and a change of variables t = 2 1/3 t 1 ).
The integral in (3.9) is absolutely convergent and converges rapidly for any fixed x > 0 by Lemma A.1 and (A.21). We denote the integrand in (3.9) by Ψ(z) = Ψ(z; x).
Proof of Theorem 2.2. Fix x > 0. By (A.21) and Lemmas A.1 and A.2, To obtain (2.11), we take out the first term in the expansion (A.20) of Hi(z), and write (3.9) as The first integral can be converted to a sum of residues by the argument just given for (3.9), which yields the sum in (2.11). Indeed, we have better estimates now, and the resulting sum is absolutely convergent as seen in Remark 2.3.
For the second integral we instead close the contour on the right, by a large circular arc {Re it } for t from π − θ 0 to −(π − θ 0 ); it follows by Lemma A.1 that the error tends to 0 as R → ∞. Inside this closed contour, Ai has no zeros, so the only pole is at z = 0 where the residue is Ai(2 1/3 x)/Ai(0). The result follows, noting that we go around this contour in the negative direction.
Remark 3.6. We may also use the expansion (A.20) of Hi with more terms. In general, subtracting the sum with L terms in (A.20) from Hi in (3.9) yields an integral that can be converted to a sum of residues as above; this sum is similar to the ones in (2.10) and (2.11), and the terms are now of order k −1−2L . We also have the integral with the subtracted terms; this is a linear combination of terms of the type Γ z −n−1 Ai(z + 2 1/3 x)/Ai(z), which as above equals −2πi times the residue at 0, so this integral can be written as a combination of derivatives of Ai at 2 1/3 x and 0; by the equation Ai (z) = zAi(z), and successive derivations of this equation, the result can be written as p 1 (x)Ai(2 1/3 x) + p 2 (x)Ai (2 1/3 x) for some polynomials p 1 and p 2 (depending on L), whose coefficients are rational functions in Ai(0) and Ai (0). We leave the details to the reader.
Proof of Theorem 2.4. If | arg z| < π−δ and |z| ≥ 1, say, then by Lemma A.3, |Ai (z)/Ai(z)| |z| 1/2 . More generally, using also Ai (z) = zAi(z), and differentiating this equation further, by induction, It follows by Lemma A.1 and (A.21) that for every fixed Consequently, we can differentiate (3.9) under the integral sign an arbitrary number of times; this shows that G is infinitely differentiable on (0, ∞) and that N has an infinitely differentiable density f N = −G given by This integral can be evaluated as a sum of residues as for (2.11) in the proof of Theorem 2.2 above, by first adding π −1 z −1 to Hi(z), which yields (2.12). Alternatively, and perhaps simpler, (A.5) and (A.31) imply that, , and thus the terms in the sum in (2.12) Hence we can integrate the sum in (2.12) termwise; equivalently, we can differentiate (2.11) termwise, which yields (2.12).
The result for M and (2. To see that the sum (2.14) does not converge for any x > 0, let y := 2 1/3 x. Take x = |a k | − y in (A.5). Since then, by Taylor's formula and (A.30), and thus Hence the term in (2.14), t k say, satifies, for some constants c 1 , c 2 > 0 and k ∈ I n , Since there are Θ(n 2 ) integers in I n , the sum over them is Θ(1), and thus the sum in (2.14) diverges. (The case x = 0 is simple.)

Moments
Proof of Theorem 2.6. Write (2.11) as where we for convenience define a 0 = 0 and , k ≥ 1.
We have AI(0) = Thus, by (2.9) and (2.17),  2 x dx, and (2.21) follows. The numerical evaluation is done by Maple, using the method discussed in Section 6.
, and thus the formula can be written This can be seen as an instance of Parseval's formula, see Remark B.1. However, although simpler than our expression above, this sum converges more slowly and is less suitable for our purposes.
. We prefer to keep Ai(0) and Ai (0) in our formulas.
Remark 4.4. Higher moments can be computed by the same method, with Airy integrals evaluated as shown in Appendix B, but in order to get convergence, one may have to use a version of (2.11) with more terms taken out of the expansion (A.20) of Hi, as discussed in Remark 3.6. We do not pursue the details.
We can also give integral formulas based on Lemma 3.5.
Theorem 4.5. The moments of M and N are given by, for any real p > 0, Proof. Immediate from (2.8)-(2.9) and Lemma 3.5 (with a change of variables x → 2 −1/3 x). The double and triple integrals converge absolutely by Lemma A.1 and (A.21).
For integer p, the integrals over x in Theorem 4.5 can be evaluated by the formulas in Appendix B. In particular, by (B .1) and (B.22), Although there is no singularity when z = w in the double integral in (4.6), it may be advantageous to use different, disjoint, contours for z and w.
Hi(a k )(Hi(a k ) − 2Bi(a k )). (4.9) These formulas are closely related to (2.17)-(2.18). They are simpler, but less suitable for numerical calculations since they do not even converge absolutely; the terms in the sums decrease as k −5/6 by (A.32) and (A.33). (However, they alternate in sign, and the sums converge.) The formulas (4.8) and (4.9) are what we obtain if we substitute (2.10) in (2.8) and (2.9) (with p = 1) and integrate termwise; however, since the resulting sums are not absolutely convergent, termwise integration has to be justified carefully, and we use a detour via complex integration.
The integral converges absolutely by (A.1) for all complex z and w, uniformly in compact sets, and thus Q is an entire function of two variables; moreover, (B.5) and (B.9) yield the explicit formulas By Theorem 4.5 and (4.6), using Γ as discussed above, First consider the simple integral, Γ Φ(z) dz say. It follows from Lemma A.3 and (A.21) that Γ Φ(z) dz − Γn Φ(z) → 0 as n → ∞, and we find by the residue theorem applied to Γ n , letting n → ∞, together with (A.26) and (A.17), Gi(a k )Hi(a k ) Hi(a k ) 2 .
The sums converge, e.g. by the argument just given, but only the final sum Hi(a k ) 2 converges absolutely, by (A.32)-(A.34). This yields the result (4.8) for N .
Next consider the double integral. If z ∈ Γ and w ∈ Γ , or z ∈ Γ n and w ∈ Γ m with m ≥ 2n, then by (4.7), (4.11) and Lemma A.3, It follows that Γ Γ − Γn Γ m → 0 as m ≥ 2n → ∞. Using the residue theorem for first Γn and then Γ m , with m = 2n, we find that the double integral above equals lim n→∞ (2πi) 2 n j=1 2n k=1 Q(a j , a k )Hi(a j )Hi(a k ) Ai (a j )Ai (a k ) .
Remark 5.1. Setting s = 0 and γ = 1/2 in (5.8), we obtain another formula for the density of N : For y > 0, By residue calculus, as with similar integrals in e.g. the proof of Theorem 2.2, this may be written as a sum of residues of 2 1/3 Ai(2 1/3 y + z)/Ai(z) 2 at the poles a k ; however, now the poles are double and we omit the details.

The integral formulas (2.1)-(2.4) can be transformed into each other by properties of the Airy functions as follows.
Proof of (2.4). The integrand in (2.1) is analytic except at the zeros of Ai, which lie on the negative real axis. Furthermore, by (A.1), |Ai(z)| is exponentially large, so the integrand in (2.1) is exponentially small, as |z| → ∞ with π/3 + δ ≤ | arg(z)| ≤ π − δ; in particular when π/2 ≤ | arg(z)| ≤ 2π/3. Hence, we can deform the integration path from the imaginary axis to any reasonable path in this domain. We choose to integrate along the two rays from the origin with arg z = ±2π/3, and obtain thus which proves (2.4).
By (A.1) and (A.8), this converges rapidly to i as z → ∞ along the positive real axis. Hence an integration by parts yields The integral along the line from e −2πi/3 ∞ to 0 equals −1 times the complex conjugate of the integral in (5.15), so we obtain from (5.11) and (5.15), which is (2.2).
For the second moments, we compute the sums in (2.20) and (2.21) in the same way. For the double sum in (2.21), we compute the sum with k ≤ 199 exactly, and find using asymptotics the tail sum (We may, for example, use the first term asymptotics for ϕ(j) and a j for j ≥ 200, since the sum is small and only a low relative precision is needed.) The (complementary) distribution function G(x) and the density function f N (x) may, for any given x > 0, be computed to high precision from (2.11) and (2.12) in the same way; for the tails of the sums ∞ k=200 ϕ(k)Ai(a k + 2 1/3 x)/Ai (a k ) and ∞ k=200 ϕ(k)Ai (a k + 2 1/3 x)/Ai (a k ) we use the asymptotic expansion of ϕ(k) given above together with, see [1, (10.4 Ai(a k + 2 1/3 x) ∼ (−1) k+1 2 1/6 3 1/6 π 2/3 sin (3π) 1/3 xk 1/3 k −1/6 + . . .
Appendix B. Some Airy integrals Integrals of the Airy functions (and their derivatives) times powers of x are easily reduced using the relations Ai (x) = xAi(x) and Bi (x) = xBi(x) and integration by parts. We have, for example, using also the definition (A.23), and in general the recursion Integrals of products of two Airy functions and powers of x can be treated similarly, see [2]; we quote the following (that are easily verified by differentiation): and in general the recursion By the same method, we can also treat products involving two different translates of Airy functions; this gives for example the following (again, these are easily verified by differentiation), if a = b and c = (a + b)/2: (B.10) and the recursion In particular, if a k and a are zeros of Ai, with k = , the formulas above yield, recalling the rapid decay (A.1) and (A.2) at ∞: Ai (B.23) In particular, where the coefficients are given by (B.24), and thus, using (B.5), We will also use the Laplace transform of the Airy function, which is easily found. (Taking z imaginary, we obtain the Fourier transform e iξ 3 /3 of Ai; this is sometimes taken as the definition of Ai, see e.g. [10, Definition 7.6.8].) Proof. By (A.1) and (A.4), the integral converges absolutely for every z with Re z > 0, and thus the integral is an analytic function of z in the right halfplane, say F (z). We have, for Re z > 0, by Ai (t) = tAi(t) and two integrations by parts, Hence, F (z) = Ce z 3 /3 for some C. By analytic extension, this holds for any complex t.
Appendix C. An integral equation for f τ (t) We give here another approach, based on Daniels [personal communication, 1993], to find the density function f τ of the defect stopping time τ = τ x , which was the basis of our development in Section 3. Unfortunately, we have not succeeded to make this approach rigorous, but we find it intriguing that it nevertheless yields the right result, so we present it here as an inspiration for further research.
Let as in (3.1) be the first passage time of W (t) to the barrier b(t) := −x − t 2 /2, where x > 0 is fixed. Let f τ (t) be the (defect) density of τ , and φ(y; t) = e −y 2 /2t / √ 2πt the density of W (t). The first entrance decomposition of W (t) to the region w < b(t) gives the integral equation (using the strong Markov property) for t > 0. (Similar arguments using the last exit decomposition, which leads to another functional equation involving also another unknown function, are used by Daniels [6] and Daniels and Skyrme [8].) Since we have by (C.1) The exponents can be written as −(t − u)(t + u) 2 /8 = −t 3 /6 + u 3 /6 + (t − u) 3 /24 and −(x + t 2 /2) 2 /2t = −x 2 /2t − xt/2 + t 3 /24 − t 3 /6, so the integral equation (C.2) can be transformed into This is a convolution equation of the form If the Laplace transforms g(s) := ∞ 0 e −st g(t) dt etc. were finite for Re s large enough, we could get the solution from g(s) = h(s)/ k(s). However, the factors e t 3 /24 in h(t) and k(t) grow too fast, so h(s) and k(s) are not finite for any s > 0 and this method does not work. Nevertheless, if we instead define h(s) and k(s) by h(s) := σ+∞i σ−∞i e −st h(t) dt and k(s) := σ+∞i σ−∞i e −st k(t) dt, integrating along vertical lines in the complex plane with real part σ > 0, then the formula g(s) = h(s)/ k(s) yields the correct formula for g(s) and thus for g(t). (Note that h and k can be seen as Fourier transform of h and k restricted to vertical lines. The value of σ > 0 is arbitrary and does not affect h and k.) Let us show this remarkable fact by calculating h(s) and k(s).
Consider first h(t) and express the Gaussian factor by Fourier inversion: The exponent b(t)u + tu 2 /2 can then be written as −ux + u 3 /6 + (t − u) 3 /6 − t 3 /6, so that This is indeed the Laplace transform of g(t) = f τ (t)e t 3 /6 given in (3.8), which by inversion yields the formulas (3.7) and (3.6) for f τ (t).
It seems likely that it should be possible to verify the crucial formula g(s) k(s) = h(s) by suitable manipulations of integrals, which would give another proof of the formulas (3.6)-(3.8) for f τ (t) and g(t). For example, if we define, for Re t > 0, F (t) := h(t) − Re t 0 g(u)k(t − u) du (note that F is not analytic), then F (t) = 0 for real t > 0, and it is easily verified that the equation g(s) k(s) = h(s) is equivalent to σ+∞i σ−∞i e −st F (t) dt → 0 as σ → ∞. However, we do not know how to verify this directly. We therefore leave finding a direct proof of g(s) k(s) = h(s) as an open problem.
Appendix D. An alternative derivation of (3.8) A proof of the formula (3.8) for the Laplace transform of the density of the passage time τ is given by Groeneboom [9] (in a more general form, allowing an arbitrary starting point and not just t = 0, or, equivalently, linear term bt in (1.1) or (3.1); for simplicity we do not consider this extension). His proof uses partly quite technical methods. For the service of the reader we here present an alternative proof based on the same ideas but from a different point of view; we believe that this yields a more straightforward proof for our purposes. As discussed in Remark 3.3, this implies (3.6) and (3.4), so it gives us a self-contained proof of the central Lemma 3.2.