Upper Bound Estimation, Product of 2 Normal Distributions I want to show that $(1-\Phi(x))(1-\Phi(y))\le\frac{xy}{xy-1}$ with  $x>0>y$. 
I thought about using that $(1-\Phi(x))\le \frac{1}{\sqrt{2\pi}}\frac{1}{x}e^{-\frac{-x^2}{2}}$ if $x>0$, but than i am getting problems with the fact that $y<0$ and the inequality is not holding anymore if i am using this estimate for x and -y.
 A: The symmetry gives $(1-\Phi(x))(1-\Phi(y)) = \Phi(-x)\Phi(|y|)$ for $y < 0 <x$. However, this doesn't matter, because the proposed inequality doesn't hold. 
When the magnitudes of $x$ and $y$ are small, the right hand side is also small, while the left hand side is very close to $1/4$.
For example, $x = \dfrac1{20}$ and $y = -\dfrac1{10}$, we have 
$\qquad$the left hand side $(1-\Phi(x))(1-\Phi(y)) \approx 0.239261$, 
$\qquad$the right hand side $\dfrac{x y }{ xy - 1} = \dfrac1{201} \approx 0.00497512$.
How about a modified version $(1-\Phi(x))(1-\Phi(y))\overset{?}{\leq}\dfrac14 - \dfrac{x y }{ xy - 1}$. Too bad, this doesn't hold either. At $x = \dfrac1{10}$ and $y = -\dfrac1{10}$, 
$\qquad$the left hand side is $(1-\Phi(x))(1-\Phi(y)) \approx 0.248414$
$\qquad$the new right hand side is $\dfrac14 - \dfrac1{101} \approx 0.240099$

Could it be that by $\Phi(x)$ you actually mean the error function $\mathrm{Erf}(x)$? Note that the CDF of standard Normal is $\Phi(x) = \dfrac12 \left( 1 + \mathrm{Erf}\bigl( \dfrac{x}{\sqrt{2}} \bigr) \right)$
Please let me know if I misunderstood your question.
