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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


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
up vote 1 down vote accepted

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 $f(x) = \phi(a-x) - \phi(b+x) > 0$.

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I had deleted this answer since Robert's comment gets right to the point, but I figure I might as well leave it. – Antonio Vargas May 2 '12 at 1:43
There should not have been CDF's in the question. I'm sorry for the confusion. – Greg May 2 '12 at 1:48
@Greg, I have edited my answer to reflect your new question. Is it satisfactory? – Antonio Vargas May 2 '12 at 3:59

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