Question

The Chernoff bound on a standard normal random variable$Z$gives$P\{Z>a\}\le e^{-a^2/2},a>0$. Show, by considering the density$Z$, that the right side of the inequality can be reduced by the factor$2$. That is, show that

$P\{Z>a\}\le \frac{1}{2}{e}^{-a^2/2}a>0$

Short answer

Therefore,

$$\begin{gathered} P\{Z>a\}={\int }_a^{\infty }\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}d{u}^{x=\underset{¯}{=}-a}{\int }_0^{\infty }\frac{1}{\sqrt{2\pi }}{e}^{-\frac{(x+{)}^2}{2}}dx= \\ {\int }_0^{\infty }\frac{1}{\sqrt{2\pi }}{e}^{-\frac{\left(x^2+3ax+a^2\right)}{2}}dx=\frac{e^{\frac{u^2}{2}}}{\sqrt{2\pi }}{\int }_0^{\infty }{e}^{-\frac{x^2}{2}}\underset{\le 1}{\underbrace{e^{-ax}}}dx\le \\ \frac{e^{\frac{{\pi }^2}{2}}}{\sqrt{2\pi }}\underset{=\sqrt{\pi /2}}{\underbrace{{\int }_0^{\infty }{e}^{-\frac{x^2}{2}}dx}}=\frac{1}{2}{e}^{-\frac{a^2}{2}},a>0 \end{gathered}$$

Step-by-step solution

Given Information.

The Chernoff bound on a standard normal random variable$Z$ gives$P\{Z>a\}\le e^{-a^2/2},a>0$.

Explanation.

Let $Z$be a standard normal random variable. Assume that$a>0$. Using the definition of the standard normal density function, we have that

$$\begin{gathered} P\{Z>a\}={\int }_a^{\infty }\frac{1}{\sqrt{2\pi }}{e}^{-\frac{y^2}{2}}d{u}^{x=u-a}={\int }_0^{\infty }\frac{1}{\sqrt{2\pi }}{e}^{-\frac{(x+\Delta {)}^2}{2}}dx= \\ {\int }_0^{\infty }\frac{1}{\sqrt{2\pi }}{e}^{-\frac{\left(x^2+2ux+a^2\right)}{2}}dx=\frac{e^{\frac{{\pi }^2}{2}}}{\sqrt{2\pi }}{\int }_0^{\infty }{e}^{-\frac{z^2}{2}}\underset{\le 1}{\underbrace{e^{-ax}}}dx\le \\ \frac{e^{\frac{z^2}{2}}}{\sqrt{2\pi }}\underset{=\sqrt{\pi /2}}{\underbrace{{\int }_0^{\infty }{e}^{-\frac{z^2}{2}}dx}}=\frac{1}{2}{e}^{-\frac{{\pi }^2}{2}} \end{gathered}$$.