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Let $a$,$b$ be positive real numbers such that $a. 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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    @AntonioVargas: Yes, that is what I meant. I should have read the whole question, instead of just the title!2012-05-02

1 Answers 1

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Since $a 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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    @Greg, I have edited my answer to reflect your new question. Is it satisfactory?2012-05-02