# What is the lower bound of $\Phi(x) - x\Phi(x)\varphi(x) - \varphi(x)^2$ for $x\in(-\infty,\infty)$. [closed]

• $$\Phi(x)$$ is the cumulative distribution function of the standard normal distribution.
• $$\varphi(x)$$ is the probability density function of the standard normal distribution.

I'm an engineer rather than a mathematical researcher. During my work I need to know that if $$\Phi(x) - x\Phi(x)\varphi(x) - \varphi(x)^2$$ is always non-negative for any $$x\in(-\infty,\infty)$$. I have tried to calculate the values of the above formula in a wide range of $$x$$, and it shows that the values are certainly larger than $$0$$. But I don't know how to strictly prove that the lower bound of $$\Phi(x) - x\Phi(x)\varphi(x) - \varphi(x)^2$$ is $$0$$ or not. Please help.

• Please provide more context, including your initial approaches into answering your question. Isolated questions such as this one (completely lacking context) are not what Math Stack Exchange is designed for. Wither proper contextualization and details, your question may be closed soon. Commented Jan 8 at 3:39
• Welcome to MSC. You can simply plot the function to see its behavior. Adding some details may motivate the members to help to prove your findings.
– Amir
Commented Jan 8 at 21:35

I presume that $$\varphi(x)=\frac 1{\sqrt{2\pi}}e^{-x^2/2}$$ and $$\Phi(x)=\int_{-\infty}^x\varphi(y)\,dy$$. If not, please clarify what you mean by standard normal.
We'll just show that the derivative of your expression is positive everywhere. Since at $$-\infty$$ it tends to $$0$$, that will establish the claim.
Differentiating and using $$\Phi'=\varphi, \varphi'=-x\varphi$$, we get the full derivative equal to $$\varphi-\varphi\Phi+x^2\varphi\Phi-x\varphi^2+2x\varphi^2=\varphi[1-\Phi+x\varphi+x^2\varphi]$$ Since $$\varphi>0$$ and $$x^2\varphi>0$$, it suffices to show that $$1-\Phi+x\varphi>0$$. When $$x\ge 0$$, it is trivial because $$1-\Phi>0$$ and $$x\varphi\ge 0$$. When $$x<0$$, we have $$1-\Phi>\frac 12$$ and $$|x|\varphi(x)=\frac 1{\sqrt{2\pi}}\sqrt{x^2e^{-x^2}}\le \frac 1{\sqrt{2\pi}}\le \frac 12$$ just because $$e^s>1+s$$, so $$se^{-s}\le 1$$ for $$s\ge 0$$ (you can do better, but you don't need to).