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.

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    $\begingroup$ As a point of reference for other readers, here's a WolframAlpha logplot showing the log of the CDF product and the three approximations. (I've scaled the variables and flipped things to make plotting simpler.) $\endgroup$ Commented Aug 6, 2014 at 4:54
  • $\begingroup$ If I'm not mistaken, the bound in (2) is just the (exponentiation of) the Taylor series bound for the log of the product. (The strength of the bound being due to exp. turning absolute differences into fractional differences.) $\endgroup$ Commented Aug 6, 2014 at 5:09
  • $\begingroup$ @Semiclassical: absolutely yes. $\endgroup$ Commented Aug 6, 2014 at 5:10
  • $\begingroup$ What is a bit interesting is that adding more terms to the Taylor series further improves the agreement in the middle (as it should) but it appears to do so at the cost of making the approximation break down more quickly in the tails. $\endgroup$ Commented Aug 6, 2014 at 5:13
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    $\begingroup$ @Semiclassical: however, dealing with the Taylor series in a continued-fraction-fashion, we have that: $$\log(4\Phi(x)\Phi(-x))\leq -\frac{2x^2}{\pi+(\pi/3-1)x^2},$$ an even tighter upper bound. $\endgroup$ Commented Aug 6, 2014 at 5:16

1 Answer 1


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 $$\!\sqrt{\frac{2}{\pi }}\!\!\frac{e^{-\frac{x^2}{2}}}{x+\sqrt{x^2+4}}\!-\!\frac{2}{\pi}\!\!\!\left(\!\frac{e^{-\frac{x^2}{2}}}{\sqrt{x^2+\frac{8}{\pi }}+x}\!\right)^2\!\!\!<\Phi(x)\Phi(-x)\!\leq \sqrt{\frac{2}{\pi }}\!\!\frac{e^{-\frac{x^2}{2}}}{x\!+\!\sqrt{x^2\!+\!\frac{8}{\pi }}}\!-\!\frac{2}{\pi}\!\left(\!\!\frac{e^{-\frac{x^2}{2}}}{x\!+\!\sqrt{x^2\!+\!4}}\!\!\right)^2 $$
enter image description here
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:

  • where both of the proposed bounds are worse than OP's (somewhere from $0$ to $1.5$)

  • where the proposed lower bound is tighter, but the upper bound is looser (somewhere from $1.5$ to $2.28$)

enter image description here

  • where both proposed are tighter (starting somewhere from $2.28$)

enter image description here

  • $\begingroup$ Interesting. Can I ask you to compare your bounds with the improved ones: $$-\frac{2x^2}{\pi}\leq\log(4\Phi(x)\Phi(-x))\leq -\frac{2x^2}{\pi+(\pi/3-1)x^2}$$? $\endgroup$ Commented Aug 6, 2014 at 6:04
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    $\begingroup$ @JackD'Aurizio I've updated a little bit the answer :) $\endgroup$ Commented Aug 6, 2014 at 11:49

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