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?

  • $\begingroup$ isn't this integral just $2\pi \varepsilon_0$ ? $\endgroup$ Mar 29, 2015 at 10:44
  • $\begingroup$ I am sorry, I posted the wrong thing. Corrected. $\endgroup$
    – TOM
    Mar 29, 2015 at 10:46
  • $\begingroup$ Have you tried using the CLT. The variance of $P_n(t)$ is $s(t)^2=\sum_{k=0}^n \cos^2kt$. Then, for fixed $t$, $\frac{1}{s(t)}P_n(t)$ converges weakly to $N(0,1)$ $\endgroup$ Mar 29, 2015 at 11:49
  • $\begingroup$ Liviu Nicoaescu: it was exactly the point when I said that it is hard for delta small - using explicit bounds in the CLT the magnitude of approximation is like $n^{-1/2}$, so it will not work for shorter intervals ($\delta<<n^{-1/2}$). $\endgroup$
    – TOM
    Mar 29, 2015 at 13:54

1 Answer 1


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 $\delta<< n^{-11/4}$.


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