# Prove that a difference of normal probability densities is non-negative

Let $a$,$b$ be positive real numbers such that $a<b$. Show that $f(x)$ is non-negative for all $x>0$ where

$f(x)= \phi(a-x)-\phi(b+x)$ where $\phi$ is the standard normal PDF.

So far I have that $f(0)>0$ and that

$f'(x)=(a-x)\phi(a-x)+(b+x)\phi(b+x)$

and then I get stuck

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$a-x < a < b < b+x$ and $\Phi$ is an increasing function, so ... –  Robert Israel May 2 '12 at 1:25
Using the words "Equation is non-zero" is not right. –  Aryabhata May 2 '12 at 1:32
Thanks, I fixed the problem. Hopefully, it makes more sense. –  Greg May 2 '12 at 1:37
Greg, Aryabhata's point (which he could have made more clear) was that equations are neither negative nor positive nor zero, but functions can be any of these. The title should read: "Prove the following function is non-negative". –  Antonio Vargas May 2 '12 at 1:45
@Antonio Vargas Done –  Greg May 2 '12 at 1:51

Since $a<b$ and $a,b,x>0$ we have $-b-x < a - x < b + x$, or $|a-x| < |b+x|$, so that $-(b+x)^2/2 < -(a-x)^2/2$, and hence that
$$\sqrt{2\pi} \phi(b+x) = e^{-(b+x)^2/2} < e^{-(a-x)^2/2} = \sqrt{2\pi} \phi(a-x).$$
Thus $f(x) = \phi(a-x) - \phi(b+x) > 0$.