# Question on part of a proof using Chebyshev's inequality

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 $${O( x^{-1+o(1)})}$$?

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 $$\log (\log^6 x)=6\log\log x \ll \sqrt{\log x}=\log\left(x^{\frac1{\sqrt{\log x}}}\right).$$
• For the last step, I think it suffices to substitute $\log^6 x$ by $x^{o(1)}$. Commented Apr 9 at 17:36
• Yes, this is exactly what I am proving: $\log\log^6 x\ll \log(x^{1/\sqrt{\log x}})$ so $\log^6 x\ll x^{1/\sqrt{\log x}}$ and $x^{1/\sqrt{\log x}}=x^{o(1)}$. Commented Apr 10 at 6:37