When approaching this other question I came out with the inequality:
$$\frac{1}{4+x^2}e^{-x^2/2} \leq\Phi(x)\Phi(-x)\leq \frac{1}{4}e^{-x^2/2},\tag{1}$$
where $\Phi(x)$ is the cdf of the standard normal distribution:
$$\Phi(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}e^{-u^2/2}\,du = \frac{1}{2}\left(1+\operatorname{erf}\frac{x}{\sqrt{2}}\right).$$
$(1)$ can be proved by bounding the probability for a normal distributed random variable in $\mathbb{R}^2$ to take values outside a rectangle. $(1)$ is tight enough to solve many problems, but when playing with the Taylor series of $\log(4\Phi(x)\Phi(-x))$ I discovered that
$$\Phi(x)\Phi(-x)\geq\frac{1}{4}\exp\left(-\frac{2x^2}{\pi}\right)\tag{2}$$
in an impressive approximation, way better than the LHS in $(1)$, just (at most) 0.003 apart from the truth. My question now is: how much can we improve the RHS and LHS in $(1)$?
I know that there are quite good continued-fraction approximations for the Mills ratios, and I am wondering if we can provide extremely good approximations for $\Phi(x)\Phi(-x)$ or the logarithm of such a product.
Best Answer
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: