Chebyshev inequality and $Q$-Function

Question

Prove that the Gaussian $Q$ function is bounded on the top by $1/2x^2$, i.e. $Q(x)\le 1/2x^2$. for $x\ge 0$ using the Chebyshev inequality and the Nakagami $m$ distribution with $m=0.5$ that reduces it to half normal distribution.

Answer 1

Note that $Q(x)$ is equal to $\mathbb P(X\geq x)$ where $X$ is a Gaussian RV with zero mean and variance 1. Now for $x>0$ we have: $$ \mathbb P(X\geq x)=\frac{1}{2}\mathbb P(|X|\geq x)\leq \frac{1}{2x^2} $$ where the last inequality is due to Chebyshev inequality.