Let us suppose that $X\_1,\ldots,X\_n$ with $n\ge1$ are iid random variables such that $\operatorname EX\_1=0$ and $\operatorname E|X\_1|^s<\infty$ with some $s>2$ and define the DFT of $X\_1,\ldots,X\_n$ by setting
$$
D\_n(\omega)=n^{-1/2}\sum\_{t=1}^nX\_te^{-it\omega}
$$
for $n\ge1$ and $\omega\in[-\pi,\pi]$, where $i=\sqrt{-1}$. I am interested in the asymptotic behaviour of the expected value of the maximum of the periodogram given by
$$
\operatorname E\max\_{1\le j\le q}|D\_n(\omega\_j)|^2,
$$
where $q=\lfloor(n-1)/2\rfloor$ and $\omega\_j=2\pi j/n$ for $1\le j\le q$. If we also assume that $X\_1,\ldots,X\_n$ are Gaussian, then $\operatorname E\max\_{1\le j\le q}|D\_n(\omega\_j)|^2=O(\log n)$ as $n\to\infty$ since $|D\_n(\omega\_1)|^2,\ldots,|D\_n(\omega\_q)|^2$ are iid standard exponential random variables (we can use the idea from [this answer](https://math.stackexchange.com/a/89147/23249) to establish the growth rate). I suspect that this might be true even if we do not assume Gaussianity. Intuitively, for large values of $n$, the distribution of $D\_n(\omega)$ should be close to the Gaussian distribution.



> 
> Is it possible to establish that $\operatorname E\max\_{1\le j\le q}|D\_n(\omega\_j)|^2=O(\log n)$ as $n
> \to\infty$ if $X\_1,\ldots,X\_n$ are iid random variables with zero means and finite moments of order $s>2$?
> 
> 
> 


Any help is much appreciated!



Expected value of the maximum of the periodogram
Here is a sketch. Feel free to ask for clarifications if my writing gets too terse or confusing in places :-).


First recall the Bernstein (a.k.a. Hoeffding, Chernov, etc.) bound. If $Y\_m$ are mean $0$ 
independent random variables bounded by $s$, then for $Y=\sum\_{m=1}^n Y\_m$, we have for every 
positive $t$,
$$
P(|Y|\ge t)\le Ce^{-c\frac{t^2}{ns^2}}\,.
$$
with some $C,c>0$. The proof goes via the consideration of $Ee^{\beta Y}$ with appropriately 
chosen $\beta$, as usual.


We want a small refinement of this bound. Suppose that we know in addition that each $Y\_m$ is 
non-zero with probability at most $p\in(0,1)$. Then, conditioning upon the events that some 
$k$ of $Y\_m$ have any chance to be non-zero, we get the bound 
$$
P(|Y|\ge t)\le C\sum\_{k=1}^n e^{-c\frac{t^2}{ks^2}}{n\choose k}p^k(1-p)^{n-k}\,.
$$
Using the inequality $\frac 1k\ge 2\beta-\beta^2 k$, we can estimate the RHS by
$$
Ce^{-2c\frac{t^2}{s^2}\beta}\sum\_{k=1}^n {n\choose k}e^{c\beta^2\frac{t^2}{s^2}k}p^k(1-p)^{n-k}
=Ce^{-2c\frac{t^2}{s^2}\beta}\left[1+p(e^{c\beta^2\frac{t^2}{s^2}}-1)\right]^n\,.
$$
for any $\beta>0$ we want.


Now let us look at the distribution of our random variable $X$ assuming that $E|X|^q=1$. For 
every $p$, we can take the set $F$ of probability $p$ on which it attains the largest values 
and split $X$ as $X'+X''$ where $X'=X$ outside $F$ and $X'=\frac 1p E(X\chi\_F)$ on $F$. If 
$EX=0$, then $EX'=EX''=0$, $E|X'|^q\le E|X|^q$, $E|X''|^q\le CE|X|^q$ but $X''$ is not zero 
only with probability $\le p$ and $|X'|\le s$ where $s^qp=1$ (by Jensen). We can apply this 
trick successively with $p=2^{-r}, \log\_2 n\ge r\ge 0$ and get the decomposition of $X$ into 
the sum of mean $0$ random variables $Z+\sum\_{r=0}^{\log\_2n}{X\_r}$ where $Z$ is different from 
$0$ with probability about $\frac 1n$ and $E|Z|^q\le C$, while for each $r$, we have $|X\_r|\le 
s\_r=2^{r/q}$ and $X\_r$ is not zero only with probability $p\_r=2^{-r+1}$.


Now it will be enough to treat each $X\_r$ and $Y$ separately. The exact nature of the discrete 
Fourier transform does not matter. All we need to know is that we are interested in the 
maximum of $n$ linear forms of $n$ iid copies of our random variables with coefficients not 
exceeding $\frac 1{\sqrt n}$. 


Let's start with $n$ iid copies of $Z$. Let $N$ be the number of non-zero values among $Z\_1,
\dots, Z\_n$. Notice that it is the sum of $n$ iid Bernoulli random variables each of which is 
$1$ with probability $\frac 1n$, so $Ee^N=(1+\frac en)^n\le e^e$. In particular, any fixed 
moment of $N$ is bounded by some constant.


We have for $q>2$, 
$$
\frac 1n \left(\sum\_m|Z\_m|\right)^2\le \frac 1n 
N^{2-\frac 2q}\left[\sum\_m |Z\_m|^q\right]^{2/q}
$$
so, using this crude bound and the Holder inequality, we get 
$$
E(\max(Z-\text{forms})^2)\le\frac 1n (EN^{\text{something}})^{1-\frac 2q}\left[E\sum\_m |Z\_m|
^q\right]^{2/q}\le Cn^{\frac 2q-1}\to 0
$$
as $n\to\infty$.
Thus this part is negligible for large $n$.


Now let us fix $r$ and consider $X\_r$. Notice that we can find $\delta>0$ such that $\frac 1q
+\delta<\frac 12$. Then if we change the notation $X\_r$ to $2^{-\delta r}X$, $p\_r=2^{-r+1}$ to 
$p$, and $s\_r=2^{r/q}$ to $2^{-\delta r}s$, we shall still have $s\le n^{\frac 12-
\varepsilon}$ with some $\varepsilon>0$ and $p\le s^{-2}$ for all $r\le\log\_2n$. The extra 
exponential factor $2^{-\delta r}$ we introduced is strong enough to enable us to consider 
each such $X$ separately and just to get a uniform bound of order $\log n$ for the 
expectations of the squared maximum of $n$ linear $X$-forms $L\_j$.


Now comes the trick: in order to show that $E(\max\_{1\le j\le n}|L\_j|)^2\le C^2\log n+O(1)$, 
it suffices to show that for each individual $j$, we have
$$
E(|L\_j|^2-C^2\log n)\_+=\int\_{C\sqrt{\log n}}^\infty 2tP(|L\_j|>t)\,dt\le \frac Cn\,.
$$  

However, we have the refined bound for the probability in question and, taking into account 
the $\sqrt n$ in the denominator, changing $\beta$ to $\beta/n$, and using the bound $p\le s^
{-2}$, we can rewrite it as 
$$
P(|L\_j|>t)\le Ce^{-2c\frac{t^2}{s^2}\beta}\left[1+s^{-2}(e^{c\beta^2\frac{t^2}{ns^2}}-
1)\right]^n
$$ 


We have to choose the optimal $\beta=\beta(t)$. There are two cases to consider:


*Case 1:* $t^2s^2\le n$. 


In this case we can take $\beta$ a small multiple of $s^2$ (the difference of the exponent and $1$ can be treated like a linear function in this range) and get the estimate $e^{-c't^2}$, which is as good as if we were dealing with (sub)Gaussian variables.


*Case 2:* $t^2s^2>n$. In this case we still want to stay in the linear range for the exponent in the parentheses, so we are forced to take $\beta$ a small multiple of $\frac{s\sqrt n}t$. Fortunately, it still gives the bound $e^{-c'\frac{\sqrt n}{s}t}$ and even if we integrate it against $t$ from $1$, we still obtain something like $e^{-c'\frac{\sqrt n}s}\le e^{-c'n^\varepsilon}$, which is much smaller than what we need for large $n$.


