Approximation in Normal distribution random variable

Question

Let ${X_n : n \geq 1}$ be independent $\mathcal{N}(0,1)$ random variables.

How do we get the following approximation?

Answer 1

Let $X \sim N(0,1)$ be a normal distributed random variable with mean $0$ and variance $1$. Then,

$$\begin{align*} \mathbb{P}(X \geq r) &= \frac{1}{\sqrt{2\pi}} \int_r^{\infty} \exp \left(- \frac{x^2}{2} \right) \, dx \\ &\leq \frac{1}{\sqrt{2\pi}} \int_r^{\infty} \frac{x}{r} \exp \left(- \frac{x^2}{2} \right) \, dx \\ &= \frac{1}{\sqrt{2\pi} r} \exp \left(- \frac{r^2}{2} \right). \end{align*}$$

for any $r>0$. Using symmetry, we conclude

$$\mathbb{P}(|X| \geq r) \leq \sqrt{\frac{2}{\pi}} \frac{1}{r} \exp \left(- \frac{r^2}{2} \right).$$

Apply this inequality for $r := \sqrt{2 \log n}(1+\varepsilon)$.