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

Question

• $\Phi(x)$ 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.

Answer 1

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).

That's it. Hope this helps a bit. Feel free to ask questions if something is unclear.