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. $\endgroup$ May 9, 2018 at 1:28
  • $\begingroup$ @Cm7F7Bb You're right. I think I dropped the squares in transcribing the problem. $\endgroup$
    – MTyson
    May 9, 2018 at 12:31

1 Answer 1


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, 2018 at 11:42
  • $\begingroup$ Actually not necessarily hence I had to use dyadics in the revision. $\endgroup$ May 9, 2018 at 12:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.