TITLE: Average probability that a random cosine polynomial with bernoulli coefficients is small
QUESTION [5 upvotes]: Let $P_{n}(t)=\sum_{k=0}^{n}\varepsilon_{k}\cos(kt)$ where $\varepsilon_{i}$ are independent random variables taking values in $\left\{-1,1\right\}$ with equal probability. Is is true that for any $\delta\rightarrow 0$ we have 
\begin{equation}
\int_{-\pi}^{\pi}\mathbb{P}(|P_{n}(t)|<\delta)\,dt \approx \frac{\delta}{\sqrt{n}}?
\end{equation}
For Gaussian random variables this is trivial, but for discrete ones it becomes hard when $\delta \ll n^{-1/2}$. Can the averaging effect smooth things out so that it is still true in the discrete case?

REPLY [6 votes]: In general this is false, if $\delta$ is small enough. Indeed, let $Q_n:=\sum_{k=0}^{n}\varepsilon_{k}$ and 
$R_{n}(t):=P_{n}(t)-Q_n$. Then $\mathsf{E}R_{n}(t)=0$ and $\mathsf{Var}R_{n}(t)\le\frac1{20}(n+1)^5t^4$. 
Suppose now that $n$ is odd. Then
$$\mathsf{P}(|P_{n}(t)|<\delta)\ge\mathsf{P}(Q_n=0)-\mathsf{P}(|R_{n}(t)|\ge\delta)
\ge\sqrt{\frac2\pi}\,\frac1{\sqrt{n+2}}-\frac{(n+1)^5t^4}{20\delta^2}
\ge p_*$$ 
if $|t|\le t_*:=\Big(\frac{20}{\sqrt{2\pi}}\,\frac{\delta^2}{(n+2)^{11/2}}\Big)^{1/4}$, where $p_*:=\frac1{\sqrt{2\pi}}\,\frac1{\sqrt{n+2}}$. 
So, if (say) $20\delta^2<1$, then 
$$\int_{-\pi}^{\pi}\mathsf{P}(|P_{n}(t)|<\delta)\,dt
\ge2t_*p_*
>>\frac{\delta}{\sqrt{n}}
$$
if $\delta<< n^{-11/4}$.