inequalitynormal distribution
I am trying to show that $$P(X>t)\leq \frac{1}{2}e^\frac{-t^2}{2}$$ for $t>0$ where $X$ is a standard normal random variable. Perhaps this is simple. I have been starting with
$$ \int_{t}^{\infty} \frac{1}{\sqrt{2\pi}} e^\frac{-x^2}{2}dx \leq \int_{t}^{\infty}\frac{x}{t}\frac{1}{\sqrt{2\pi}}e^\frac{-x^2}{2}dx = \frac{1}{t\sqrt{2\pi}}e^\frac{-t^2}{2}$$
but this is not what I want…I am looking for a much stronger inequality. Any help in the direction of getting the $1/2$?
I will use Mill's ratio (one of them) for that:
$$\frac{1}{x+\sqrt{x^{2}+2}}<\mathop{\mathsf{M}}\nolimits\!\left(x\right)\leq%
\frac{1}{x+\sqrt{x^{2}+(4/\pi)}},$$
where $\mathop{\mathsf{M}}(x)=\frac{\int_{x}^{\infty}e^{-t^{2%
}}\mathrm dt}{e^{-x^{2}}}=e^{x^{2}}\int_{x}^{\infty}e^{-t^{2}}\mathrm dt=e^{x^{2}}\frac{\sqrt{\pi}}{2}\mathbb{erfc}(x)$
One can define $\Phi(x)\Phi(-x)=\frac{1}{4}\mathbb{erfc}\left(\frac{x}{\sqrt{2}}\right)\left(2-\mathbb{erfc}\left(\frac{x}{\sqrt{2}}\right)\right)=\frac{1}{2}\mathbb{erfc}\left(\frac{x}{\sqrt{2}}\right)-\frac{1}{4}\mathbb{erfc}^2\left(\frac{x}{\sqrt{2}}\right)$.
And obtain
$$\frac{\frac{2}{\sqrt{\pi}}e^{x^{-2}}}{x+\sqrt{x^{2}+2}}<\mathbb{erfc}(x)\leq%
\frac{\frac{2}{\sqrt{\pi}}e^{x^{-2}}}{x+\sqrt{x^{2}+(4/\pi)}}$$
$$-\left(\frac{\frac{2}{\sqrt{\pi}}e^{x^{-2}}}{x+\sqrt{x^{2}+(4/\pi)}}\right)^2<-\mathbb{erfc}^2(x)\leq%
-\left(
\frac{\frac{2}{\sqrt{\pi}}e^{x^{-2}}}{x+\sqrt{x^{2}+2}}\right)^2$$
Finally, combining them and changing the variable to $\frac{x}{\sqrt{2}}$ one will get
$$\!\sqrt{\frac{2}{\pi }}\!\!\frac{e^{-\frac{x^2}{2}}}{x+\sqrt{x^2+4}}\!-\!\frac{2}{\pi}\!\!\!\left(\!\frac{e^{-\frac{x^2}{2}}}{\sqrt{x^2+\frac{8}{\pi }}+x}\!\right)^2\!\!\!<\Phi(x)\Phi(-x)\!\leq
\sqrt{\frac{2}{\pi }}\!\!\frac{e^{-\frac{x^2}{2}}}{x\!+\!\sqrt{x^2\!+\!\frac{8}{\pi }}}\!-\!\frac{2}{\pi}\!\left(\!\!\frac{e^{-\frac{x^2}{2}}}{x\!+\!\sqrt{x^2\!+\!4}}\!\!\right)^2
$$
Here is a little log-plot of new (dashed) and old (yours)(solid) bounds (black line is the multiplacation of cdf's $\Phi(x)\Phi(-x)$).
Moreover it is worth noticing that new bounds are much looser then the initial ones for $0<x<1$ and work only for positive $x$.
Comparing the obtained bounds with the proposed by OP: $$-\frac{2x^2}{\pi}\leq\log(4\Phi(x)\Phi(-x))\leq -\frac{2x^2}{\pi+(\pi/3-1)x^2}$$ one can split all the range of $x$ into $3$ regions:
Best Answer
For every $t\gt0$, $$P(X\gt t)=\int_0^\infty\frac1{\sqrt{2\pi}}\mathrm e^{-(x+t)^2/2}\mathrm dx,$$ and, for every $x\geqslant0$, $$\mathrm e^{-(x+t)^2/2}\leqslant\mathrm e^{-t^2/2}\mathrm e^{-x^2/2},$$ hence $$P(X\gt t)\leqslant\mathrm e^{-t^2/2}\int_0^\infty\frac1{\sqrt{2\pi}}\mathrm e^{-x^2/2}\mathrm dx=\mathrm e^{-t^2/2}\,P(X\gt0).$$