Why $\frac{1}{\sqrt{2\pi}} \int_x^\infty e^{-t^2/2} dt<e^{-x^2/2}$ when $x>0$?

Question

Why does the following inequality hold for $x > 0$? $$ Q(x)<e^{-x^2/2}, \tag{1} $$ where $$ Q(x) = \frac{1}{\sqrt{2\pi}} \int_x^\infty e^{-t^2/2} dt. \tag{2} $$

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The following is my attempt: Define $$ \begin{align} f(x) & \triangleq e^{-x^2/2} - Q(x) \\ & = e^{-x^2/2} - \frac{1}{\sqrt{2\pi}} \int_x^\infty e^{-t^2/2} dt \\ & = e^{-x^2/2} - \left( 1 - \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-t^2/2} dt \right) \\ & = e^{-x^2/2} - 1 + \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-t^2/2} dt. \tag{3} \end{align} $$ Then $$ \begin{align} f'(x) & = -\frac{2x}{2} e^{-x^2/2} + \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \\ & = \left(-x+\frac{1}{\sqrt{2\pi}} \right) e^{-x^2/2}. \tag{4} \end{align} $$ When $0<x<\frac{1}{\sqrt{2\pi}}$, $f'(x)>0 \Rightarrow f(x)$ is a monotonically increasing function. When $x=\frac{1}{\sqrt{2\pi}}$, $f'(x)=0$. When $x>\frac{1}{\sqrt{2\pi}}$, $f'(x)<0 \Rightarrow f(x)$ is a monotonically decreasing function. So $f(x)$ has a local maximum at $x = \frac{1}{\sqrt{2\pi}}$. $$ \lim_{x \rightarrow \infty} f(x) = 0 - 0 =0. $$ $$ f(0) = e^{-0^2/2} - Q(0) = 1 - \frac{1}{2} = \frac{1}{2}. $$ Then I don't know how to continue. Any comments and answers are welcome. Thanks in advance.

Answer 1

Your idea is quite good: if $$ f(x)=e^{-x^2/2}-\frac{1}{\sqrt{2\pi}}\int_x^{\infty} e^{-t^2/2}\,dt $$ then $f(0)=1/2$ and $\lim_{x\to\infty}f(x)=0$. On the other hand, $$ f'(x)=-xe^{-x^2/2}+\frac{1}{\sqrt{2\pi}}e^{-x^2/2} $$ so $f$ is increasing in the interval $[0,1/\sqrt{2\pi}]$ and decreasing in the interval $[1/\sqrt{2\pi},\infty)$.

Hence $f(x)>0$ for every $x>0$.

When a function is decreasing over an interval $[a,b)$ (where $b$ can be $\infty$), then $$ \lim_{x\to b}f(x)=\inf_{x\in[a,b)}f(x) $$ Thus $\inf_{x\in[1/2\pi,\infty)}f(x)=0$. The infimum cannot be reached, or the function would be constant in some interval $[c,\infty)$, which it isn't, because the derivative is not zero over $(1/\sqrt{2\pi},\infty)$.

Answer 2

$$e^{-x^2/2}=\int^{\infty}_xe^{-t^2/2}t\,dt$$ Therefore $$Q(x)-e^{-x^2/2}=\frac{1}{\sqrt{2\pi}}\int^{\infty}_xe^{-t^2/2}\,dt-\int^{\infty}_xe^{-t^2/2}t\,dt=\int^{\infty}_x\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt$$ If $x\geqslant1/\sqrt{2\pi}$ then $$Q(x)-e^{-x^2/2}=\int^{\infty}_x\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt<0$$ otherwise split the integral into two parts $$\int^{\infty}_x\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt=\int^{1/\sqrt{2\pi}}_x\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt+\int^{\infty}_{1/\sqrt{2\pi}}\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt$$ The first is positive and the second part negative. You can estimate each part $$\int^{1/\sqrt{2\pi}}_x\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt\leqslant\int^{1/\sqrt{2\pi}}_0\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt<0.08$$ and $$\Big|\int^{\infty}_{1/\sqrt{2\pi}}\Big(\frac{1}{\sqrt{2\pi}}-t\Big)e^{-t^2/2}\,dt\Big|>0.5$$ The negative part outweights the positive part. So whenever $x>0$ the inequality holds.