For the function $Q(x) := \mathbb{P}(Z>x)$ where $Z \sim \mathcal{N}(0,1)$
\begin{align}
Q(x) = \int_{x}^\infty \frac{1}{\sqrt{2\pi}} \exp \left(-\frac{u^2}{2} \right) \text{d}u,
\end{align}
for $x \geq 0$ the following bound is given in many communication systems textbooks:
\begin{align}
Q(x) \leq \frac{1}{2} \exp \left(-\frac{x^2}{2} \right).
\end{align}
The bound without the $\frac{1}{2}$ in front of the exponential can be proven directly by Chernoff bound on the Gaussian distribution. However, how do we show the case with the $\frac{1}{2}$ before the exponential?
Best Answer
Start from the inequality $$e^{x^2/2}Q(x)=\int_x^\infty\frac 1{\sqrt{2\pi}}\exp\left(-\frac{u^2-x^2}2\right)\mathrm du.$$ After the substitution $s+x=u$, this equality becomes
$$e^{x^2/2}Q(x)=\int_0^\infty\frac 1{\sqrt{2\pi}}\exp\left(-\frac{s^2+2sx}2\right)\mathrm ds=\int_0^\infty\frac 1{\sqrt{2\pi}}\color{red}{\exp\left(-sx\right)}\exp\left(-\frac{s^2}2\right)\mathrm ds.$$ Conclude noticing that the red term is smaller than $1$.