Correction to: On maximum of Gaussian random fields having unique maximum point of its variance
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Correction to: On maximum of Gaussian random fields having unique maximum point of its variance Sergey G. Kobelkov1 · Vladimir I. Piterbarg1,2 · Igor V. Rodionov3
© Springer Science+Business Media, LLC, part of Springer Nature 2020
Correction to: Extremes (2019) 22: 413–432 https://doi.org/10.1007/s10687-019-00346-2 In the proof of Proposition 3 below formula (31), it was stated that (u) and (u) are integral sums for the integral 2 e−u f (t) dt, (1) I (u) = f (t)≤u−1 γ1 (u)
where f (t) = (1 − σ 2 (t))/2. In spite of the fact that (u) ≤ I (u) ≤ (u), the relation (u)/ (u) → 1, u → ∞, was not justified. To show this, we should significantly change the proof of Proposition 3. First, we modify Condition 5.
The online version of the original article can be found at https://doi.org/10.1007/s10687-019-00346-2. Igor V. Rodionov
[email protected] Sergey G. Kobelkov [email protected] Vladimir I. Piterbarg [email protected] 1
Lomonosov Moscow State University, Moscow, Russia
2
International Laboratory of Stochastic Analysis and its Applications National Research University Higher School of Economics, Moscow, Russia
3
Trapeznikov Institute of Control Sciences, Russian Academy of Sciences, Profsoyuznaya 65, 117997 Moscow, Russia
S.G. Kobelkov et al.
Condition 5∗ For any t there exists the function h1 (t) ∈ [0, ∞] such that lim u2 (1 − σ 2 (q(u)t)) = h1 (t).
u→∞
(2)
Moreover, if h1 (e) = 0 for all e ∈ Sd−1 , the unit sphere in Rd , then lim u2 σ 2 (s) − σ 2 (s + q(u)t) = 0 u→∞
uniformly in t from any closed set and s ∈ Bε . Let the latter assumption hold. Then, using the argument similar to the one on the regular variation property of 1 − r(t) (see page 416 in the original paper), we can prove that h1 (t) is continuous and that if h1 (e) ∈ (0, ∞) for some vector e ∈ Rd then h1 (te) is regular varying at zero. Thus the condition “h1 (e) = 0 for all e ∈ Sd−1 ” implies that h1 (t) = 0 for all t ∈ Rd and then the last condition may be replaced by the condition “h1 (e) = 0 for all e ∈ Sd−1 ” in the statement of Proposition 3. Additionally, note that we need the second part of Condition 5∗ to justify that the sets {Bk (u)}k≥0 appearing below satisfy the assumptions of Theorem 1 and to apply the standard argument for evaluating the double sum (see the relation (9) below). It might seem that we can replace the second part of Condition 5∗ with the following weaker assumption in spirit of the condition E4, Piterbarg and Rodionov (2020), Moreover, if h1 (e) = 0 for all e ∈ Sd−1 , then Eq. 2 holds uniformly in t from any closed set and there exist c > 0 and K < ∞ such that for every x ∈ [0, c], t0 ∈ Rd and e ∈ Sd−1 , the number of roots of the equation 1 − σ 2 (t0 + ye) = x with respect to y does not exceed K. The analogue of this assumption was used to examine the stationary-like case in the proof of Theorem 1, Piterbarg and Rodionov (2020). The exact value of K does not play a role, but its finiteness guarantees the existence of “large” subintervals Bk (u) satisfying the assumptions of Theorem
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