Prove that the Q function is bounded such that $Q(x)\le\frac{1}{2x^2}$

Question

Prove that the gaussian Q function is bounded on the top by $\frac{1}{2x^2 }$, i.e. $Q(x)\le\frac{1}{2x^2}$ for $x\ge0$, using the chebyshev inequality and the nakagami-m distribution with m=0.5(that reduces it to half normal distribution). This is also known as the chebyshev bound I think. Can't get too far with it, any help/hints appreciated. The Chebyshev inequality to used here is: Pr{|r-$\mu|$$\ge$x}$\le$$\frac{\sigma^2}{x^2}$

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:

$$ Q(x)\ = Pr(X\geq x)=\frac{1}{2}\Pr(|X|\geq x)\leq \frac{1}{2x^2} $$ where the last inequality is due to Chebyshev inequality. And $Pr${|$r$-$\mu|$$\ge$x}=Pr{|$X|\le x$}

Ref: Chebyshev inequality and $Q$-Function