Comments on: Subsampling weakly dependent time series and application to extremes

Appendix

Here we provide an outline of the proof of the Theorem. First, note that by the continuity of the limit law ℍ(⋅) and by a subsequence argument, it is enough to show that for each fixed x,

$$\hat{\mathbb{H}}_{m,n}(x) - \mathbb{H}(x)\rightarrow_p 0.$$

Fix x∈ℝ. Write M _i,b =maxY _i,b and $M_{b}^{*}= \max\{X_{1}^{*},\ldots,X_{b}^{*}\}$, the maximum over a single resampled block. Also, let $\mathbb{H}^{\dagger}_{b}(x) =P_{*}([M_{b}^{*}-{v}_{m}]/{u}_{m} \leq x)$. Then, by (1) and the independence of the resampled blocks, it is easy to check that

which, by Theorem 5 and Condition (i), converges to ℍ(x) in probability provided $k^{2}\operatorname{Var}({\mathbb{H}}^{\dagger}_{b}(x)) \rightarrow0$. Note that, by stationarity, with $w_{m}(x) = u_{m}^{-1} x+v_{m}$,

where C ₁,C ₂,…∈(0,∞) are constants. By Conditions (i) and (iii), it follows that I _1n (p)=O(m/n)=o(1) for every fixed p≥1. And, by retracing the arguments in the proof of Theorem 5, one can show that for any j≥pb, p>1, and α _jn >0,

Setting α _jn =[mu _m η(j−b)]^1/2 and noting that m=kb, we have

which, by Condition (iii), goes to zero by first letting n→∞ and then p→∞. This completes the proof of the theorem.