I am working on this exercise:
Suppose $X_{n}$ is standard normal. Prove that $S_{n}/\sqrt{n}\stackrel{d}{=}X_{1}$, and that for every $x>0$, $$(x^{-1}-x^{-3})e^{-x^{2}/2}\leq\int_{x}^{\infty}e^{-t^{2}/2}dt\leq x^{-1}e^{-x^{2}/2}.$$ Deduce that, for some absolute constant $C>0$, and every sufficient large $n$, we have $$P\Big(|S_{n}|\geq 2\sqrt{n\log n}\Big)\leq Cn^{-2}.$$
I have proved everything except for the last part. I had some attempts but I was stuck in the end.
Before writing down my attempt, I have a question which is essential to this exercise. In this exercise, I only proved that $S_{n}/\sqrt{n}$ has the same distribution as $X_{1}$, which implies $$P(S_{n}/\sqrt{n}\leq a)=P(X_{1}\leq a)$$ for all $a$, and the same can be applied to $\geq$.
But this exercise asks me about the distribution of $|S_{n}|$, so it seems that I need to split the probability into $$P\Big(|S_{n}|\geq 2\sqrt{n\log n}\Big)=P\Big(S_{n}/\sqrt{n}\geq 2\sqrt{\log n}\ \text{or}\ S_{n}/\sqrt{n}\leq -2\sqrt{\log n}\Big),$$ but how could I split out this probability? Is that true we have $$P\Big(|S_{n}|\geq 2\sqrt{n\log n}\Big)=P\Big(S_{n}/\sqrt{n}\geq 2\sqrt{\log n}\Big)+P\Big(S_{n}/\sqrt{n}\leq -2\sqrt{\log n}\Big)?$$
I have some attempt of $P\Big(S_{n}/\sqrt{n}\geq 2\sqrt{\log n}\Big)$.
Here is my attempt:
Since $S_{n}/\sqrt{n}$ has the same distribution as $X_{1}$, we have $$P\Big(S_{n}/\sqrt{n}\geq 2\sqrt{\log n}\Big)=P\Big(X_{1}\geq 2\sqrt{\log n}\Big)=(2\pi)^{-1/2}\int_{2\sqrt{\log n}}^{\infty}e^{-t^{2}/2}dt,$$ where the last equality is using the density function of standard normal distribution.
Then, by inequality in the exercise, we have \begin{align*} (2\pi)^{-1/2}\int_{2\sqrt{\log n}}^{\infty}e^{-t^{2}/2}dt&\leq(2\pi)^{-1/2}\dfrac{1}{2\sqrt{\log n}}\exp(-2\log n)\\ &=\dfrac{(2\pi)^{-1/2}}{2}n^{-2}\dfrac{1}{\sqrt{\log n}}. \end{align*}
Now, for $n$ large enough, $\sqrt{\log n}\leq \sqrt{n^{2}}=n$, and thus $$\text{the above}\leq \dfrac{(2\pi)^{-1/2}}{2}n^{-3}\leq \dfrac{(2\pi)^{-1/2}}{2}n^{-2}.$$
Is my attempt correct? and what should I do with $|S_{n}|$?
Thank you!
Edit 1:
The above has a mistake that while $\sqrt{\log n}\leq \sqrt{n^{2}}=n$ is true, it it in the denominator.
So I will just use $\log n\geq 1$ for all $n\geq e$, and thus $$P\Big(S_{n}/\sqrt{n}\geq 2\sqrt{\log n}\Big)\leq \dfrac{(2\pi)^{-1/2}}{2}n^{-2}\dfrac{1}{\sqrt{\log n}} \leq \dfrac{(2\pi)^{-1/2}}{2}n^{-2}.$$
Therefore, $$P\Big(|S_{n}|\geq 2\sqrt{n\log n}\Big)=2P\Big(S_{n}/\sqrt{n}\geq 2\sqrt{\log n}\Big)\leq \dfrac{(2\pi)^{-1/2}}{4}n^{-2}.$$
Yes, your attempt seem correct.
So is your splitting, since the events $\left\{S_{n}/\sqrt{n}\geq 2\sqrt{\log n}\right\}$ and $\left\{ S_{n}/\sqrt{n}\leq -2\sqrt{\log n}\right\}$ are disjoint. Then by symmetry of $S_n/\sqrt n$, $$P\Big(\left\lvert S_{n}/\sqrt{n}\right\rvert\geq 2\sqrt{\log n}\Big)=2P\Big(S_{n}/\sqrt{n}\geq 2\sqrt{\log n}\Big).$$