Suppose that $Z,Z_1,Z_2,\ldots$ are iid random variables such that $\operatorname EZ=0$, $\operatorname EZ^2=1$ and $\operatorname E|Z|^s<\infty$ with some $s>2$. Let $\tilde Z_t=Z_tI_{\{|Z_t|\le n^{1/s} \}}$ with $1\le t\le n$ be the triangular array of the truncated random variables, where $I_A$ is the indicator function of a set $A$. Denote the periodograms $$ I_{n,Z}(\lambda)=\Bigl|n^{-1/2}\sum_{t=1}^n\exp(-i\lambda t)Z_t\Bigr|^2 \quad\text{and}\quad I_{n,\tilde Z}(\lambda)=\Bigl|n^{-1/2}\sum_{t=1}^n\exp(-i\lambda t)\tilde Z_t\Bigr|^2 $$ for $\lambda\in[0,\pi]$.

I would like to show that \begin{equation}\label{eq1}\tag{#} \max_{1\le j\le q}I_{n,Z}(\omega_j)-\max_{1\le j\le q}I_{n,\tilde Z^{(n)}}(\omega_j)\to0 \end{equation} a.s. as $n\to\infty$, where $\omega_j=2\pi j/n$ with $j=1,\ldots q$ and $q=\lfloor(n-1)/2\rfloor$. Intuitively speaking, this means that the truncation does not affect the asymptotic behaviour of the maximum of a periodogram.

I am trying to verify a proof of this statement, which is given here (see Lemma 3.3). The proof establishes that \begin{equation}\label{eq2}\tag{*} \sum_{t=1}^n|Z_t-\tilde Z_t|=\sum_{t=1}^n|Z_t|I_{\{|Z_t|>n^{1/s}\}} \end{equation} converges to $0$ almost surely as $n\to\infty$ and then it is claimed that the periodograms of the sequences $Z_1,\ldots,Z_n$ and $\tilde Z_1,\ldots,\tilde Z_n$ have to be identical identical a.s. for all $n$ sufficiently large. However, I am struggling to see why \eqref{eq2} implies \eqref{eq1}. In my opinion, \eqref{eq2} implies that $$ \max_{1\le j\le q}\Bigl|n^{-1/2}\sum_{t=1}^n\exp(-i\omega_jt)Z_t-n^{-1/2}\sum_{t=1}^n\exp(-i\omega_jt)\tilde Z_t\Bigr|^2 \le\Bigl|n^{-1/2}\sum_{t=1}^n|Z_t-\tilde Z_t|\Bigr|^2\to0 $$ a.s. as $n\to\infty$ and I do not see a way to conlcude that \eqref{eq1} holds. Maybe I am missing something trivial or maybe I do not understand the proof properly. So does \eqref{eq2} imply that \eqref{eq1} holds?

Any help is greatly appreciated!

  • $\begingroup$ If (*) converges to 0, that means for a.e. realization of $(Z_t)$, for all sufficiently large $n$, $|Z_t|<n^{1/s}$ for all $t\le n$. That is $Z_t=\tilde Z_t$ for all $t\le n$. Since they are equal for large $n$, they have equal periodograms. $\endgroup$ – Anthony Quas May 9 '18 at 1:28
  • $\begingroup$ @Cm7F7Bb You're right. I think I dropped the squares in transcribing the problem. $\endgroup$ – MTyson May 9 '18 at 12:31

Let $A_N$ be the event defined by $$ A_N:=\bigcup_{n=2^{N-1}+1}^{2^N}\left\{\max_{1\le j\le q}I_{n,Z}(\omega_j)\neq \max_{1\le j\le q}I_{n,\tilde Z^{(n)}}(\omega_j)\right\}. $$ Then the following inclusion holds $$ A_N\subset \bigcup_{n=2^{N-1}+1}^{2^N}\bigcup_{t=1}^n\left\{ Z_t\neq \widetilde{Z_t}^{(n)} \right\}= \bigcup_{n=2^{N-1}+1}^{2^N}\bigcup_{t=1}^n\left\{ \left\lvert Z_t\right\rvert\gt n^{1/s} \right\}=\bigcup_{t=1}^{2^N}\left\{ \left\lvert Z_t\right\rvert\gt \left(\max\left\{2^{N-1}+1,t\right\}\right)^{1/s} \right\}. $$ Using the fact that the random variables $Z_t,1\leqslant t\leqslant n$, have the same distribution, we infer that $$ \Pr\left(A_N\right)\leqslant 2^N\Pr\left\{ \left\lvert Z_1\right\rvert\gt 2^{\frac{N-1}s} \right\} $$ and since $\mathbb E\left\lvert Z\right\rvert^s$ is finite, we derive finiteness of $\sum_{N=1}^{+\infty}\Pr\left(A_N\right)$ and the Borel-Cantelli lemma allows to conclude.

  • $\begingroup$ Thank you very much (+1)! Could you please explain how we derive finiteness of $\sum_{n=1}^{+\infty}\operatorname{Pr}(A_n)$? We have that $\sum_{n=1}^\infty\operatorname{Pr}(|Z_1|>n^{1/s})\le\operatorname E|Z_1|^s<\infty$. Is it true that $\sum_{n=1}^\infty n\operatorname{Pr}(|Z_1|>n^{1/s})<\infty$ as well? $\endgroup$ – Cm7F7Bb May 9 '18 at 11:42
  • $\begingroup$ Actually not necessarily hence I had to use dyadics in the revision. $\endgroup$ – Davide Giraudo May 9 '18 at 12:04

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