I've been trying to prove the following: $$\phi'(x)(1-\Phi(x))+2\phi(x)^2>0$$ where $\phi$ is the standard Normal pdf and $\Phi$ its cdf. I've tried many simulations and I believe it's true in general, how would one go proving it formally?
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1$\begingroup$ it's true indeed $\endgroup$– AksakalCommented Nov 17, 2018 at 16:22
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1$\begingroup$ Try using the fact that $\phi^\prime(x) = -x\phi(x)$ and cancel out a $\phi(x)$ which makes the inequality trivial when $x<0$. $\endgroup$– Dilip SarwateCommented Nov 17, 2018 at 17:03
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$\begingroup$ when $x<0$ it's already trivial, becuase $\phi'>0$ $\endgroup$– AksakalCommented Nov 17, 2018 at 17:07
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$\begingroup$ obviously, I need it for x>0 $\endgroup$– andrea m.Commented Nov 17, 2018 at 17:10
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$\begingroup$ you really need to worry only about $x\to\infty$, btw, is this a home work? $\endgroup$– AksakalCommented Nov 17, 2018 at 17:11
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2 Answers
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$$\phi(x) = \frac1{\sqrt{2\pi}}\exp\left( -\frac{x^2}{2}\right)$$
$$\phi'(x) = \frac{-x}{\sqrt{2\pi}}\exp\left( -\frac{x^2}{2}\right)=-x\phi(x)$$
Hence the problem is equivalent to
$$-x(1-\Phi(x))+2\phi(x) >0$$
As discussed in the comment, when $x \le 0$, the problem is trivial and I will only focuses on when $x>0$.
We want to show that $$ (1-\Phi(x))< \frac{2\phi(x)}{x} $$
for $x>0$.
\begin{align} 1-\Phi(x) &= \int_{x}^\infty \phi(t) \, dt\\ &< \int_x^\infty \frac{t}{x} \phi(t) \, dt \\ &= \int_x^\infty \frac{-\phi'(t)}{x} \, dt \\ &= \frac{\phi(x)}{x} \\ &< \frac{2\phi(x)}{x} \end{align}
It seems that a stronger statement is true:
$$\phi'(x)(1-\Phi(x))+\phi(x)^2 >0$$
answered Nov 17, 2018 at 17:31
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Consider that for a random variable with any density above $x$, $E(X|X>x) > x$
For a standard normal variate, you should be able to show that $E(X|X>x) = \phi(x)/[1-\Phi(x)]$
(e.g. for the integral in the numerator use the fact that $x\phi(x)=-\phi'(x)$)
Hence $\phi(x)/[1-\Phi(x)]>x$ or $\phi(x)>x[1-\Phi(x)]$ which is stronger than your result.
