Question on part of a proof using Chebyshev's inequality

Question

This is a question concerning a specific part of a proof, namely, the proof of Prediction $8$ in Tao's note: https://terrytao.wordpress.com/2015/01/04/254a-supplement-4-probabilistic-models-and-heuristics-for-the-primes-optional/.

We have a sequence of random variables $X_n$ for $2 \leq n \leq x$, where each ${X_n}$ has mean zero, is bounded by ${O(1)}$, and ${X_n}$ is independent of ${X_m}$ unless ${|n-m| \leq 2}$, from which we easily calculate that

$\displaystyle \mathop{\bf E} (\sum_{2 \leq n \leq x} X_n)^2 = O(x)$.

The proof then proceeds to claim the following:

A direct application of Chebyshev’s inequality then gives the bound that $\displaystyle \sum_{2 \leq n \leq x} X_n = o( \frac{x}{\log^2 x} )$ with failure probability of ${O( x^{-1+o(1)})}$.

Question: By Chebyshev's inequality, ${\bf P}(|\sum_{2 \leq n \leq x} X_n| \geq o(\frac{x}{\log^2 x})) \leq \frac{O(x)}{o(\frac{x^2}{\log^4 x})}$, are we supposed to show that the RHS equals ${O( x^{-1+o(1)})}$?

Answer 1

Let $Y_n=X_2+X_3+\dots+X_n$, then $\mathbb{E}(Y_n)=0$ and $Var(Y)=\mathbb{E}(Y_n^2)=O(x)$ indeed. Hence by Chebyshev's inequality $$ \mathbb{P}(|Y_n|>\epsilon)\le \frac{Var(Y_n)}{\epsilon^2} $$ If you set $\epsilon=\frac{x}{\log^2 x}\alpha_n$ where $\alpha_n\to 0$, then $$ \mathbb{P}(|Y_n|>\epsilon)\le \frac{O(x)\log^4 x}{\alpha_n^2 x^2} =\frac{O\left(\frac{\log^4 x}x\right)}{\alpha_n^2} $$ so if you set for example $\alpha_n=\frac 1{{\log x}}\to 0$ then $$ \mathbb{P}\left(|Y_n|>\frac x{\log^2 x}\right)\le \frac{O(x)\log^4 x}{\alpha_n^2 x^2} =O\left(x^{-1}\log^6 x\right) $$ and $x^{-1}\log^6 x=o(x^{-1+\beta_n})$ for e.g. $\beta_n=\frac1{\sqrt{\log x}}$ as $$ \log (\log^6 x)=6\log\log x \ll \sqrt{\log x}=\log\left(x^{\frac1{\sqrt{\log x}}}\right). $$