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First, sorry for the debacle that was my previous question...hopefully this one works out better....

First some notation for simplicity:

Let $ h(x)=\phi(x)+x\Phi(x)$

Let $ g(x)=xh(x) $

where $\phi$ is the standard normal PDF and $\Phi$ is the standard normal CDF

Let $F(x)=[12b^2(g(\frac{1+ax}{b})-g(\frac{ax}{b})+g(\frac{2ax-1}{b})-g(\frac{2ax}{b}))] + [12b^2(\Phi(\frac{1+ax}{b})-\Phi(\frac{ax}{b})+\Phi(\frac{2ax-1}{b})-\Phi(\frac{2ax}{b})] +12(a^2x^2-1)$

where a,b are positive constants and $a<1$

I want to show F(x) is positive for $ x \in (0,\frac{1}{a})$

Any thoughts?

Edit:

One possibility is to replace $\Phi(x)$ with $1-\Phi(-x)$ in $h$ so that the extra terms plus the last term yields $12(1-ax)^2$ which is obviously positive, then I only have to worry about the remaining term, which just just like above except now has $...-x\Phi(-x) $ in the definition of $h$ instead of whats above.

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    @M.B. Thats a habit I have for multiplication2012-05-18

1 Answers 1

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Summary: The inequality $F(x)\geqslant0$ does not hold for every admissible values of the parameters $x$, $a$ and $b$, but replacing the last $12$ by $6$, or $a^2x^2$ by $ax$, both make it true.

1. The change of variables $ax/b\to s$, $1/b\to z$ shows that the inequality to prove is equivalent to $ U_s(z)\geqslant z^2-s^2\tag{$\ast$} $ for every $0\lt s\lt z$, where $ U_s(z)=u(s+z)-u(s)+u(2s-z)-u(2s), $ and $ u(t)=t\phi(t)+(1+t^2)\Phi(t). $ 2. Since $\Phi'(t)=\phi(t)$ and $\phi'(t)=-t\phi(t)$, one has $u''(t)=2\Phi(t)$ and $ U_s'(z)=u'(s+z)-u'(2s-z)=\int_{2s-z}^{s+z}u''(t)\mathrm dt=\int_{2s-z}^{s+z}2\Phi(t)\mathrm dt. $ 3. Consider the asymptotics $s=o(1)$ and $z=rs$, with $r\gt1$ fixed. Then $2\Phi(0)=1$ and $(s+z)-(2s-z)=(2r-1)s$ hence $U_s'(rs)=(2r-1)s+o(s)$ when $s\to0$, for every fixed $r\gt1$. Since $U_s(s)=0$, one gets $ U_s(rs)=\int_1^rsU_s'(ws)\mathrm dw=s^2\int_1^r(2w-1)\mathrm dw+o(s^2)=(r^2-r)s^2+o(s^2). $ Since $z^2-s^2=(r^2-1)s^2$ and $r^2-1\gt r^2-r$, one sees that $(\ast)$ is wrong in the regime $z=rs$, $r\gt1$ fixed, for $s$ small enough.

4. On the other hand, at least in the regime $z=rs$, $r\gt1$ fixed, $U_s(z)\geqslant z(z-s)+o(s^2)$ hence $U_s(z)\geqslant\frac12(z^2-s^2)+o(s^2)$. Reformulating everything as in the question, this shows that one can still hope that, for every $b\gt0$, $0\lt a\lt 1$ and $0\lt x\lt1/a$,

$\qquad[12b^2(g(\frac{1+ax}{b})-g(\frac{ax}{b})+g(\frac{2ax-1}{b})-g(\frac{2ax}{b}))] +$ $\qquad\qquad\qquad +{}[12b^2(\Phi(\frac{1+ax}{b})-\Phi(\frac{ax}{b})+\Phi(\frac{2ax-1}{b})-\Phi(\frac{2ax}{b})] \geqslant\mathbf{6}(1-a^2x^2), $

5. Indeed, $U_s(z)\geqslant z(z-s)$ holds true for every $0\lt s\lt z$. To show this, start from the identity $\Phi(-t)=1-\Phi(t)$ and the change of variable $\tau=-t$ in the formula of $U_s'(z)$ above, which yield $ U_s'(z)=\int_{-s-z}^{-2s+z}2(1-\Phi(\tau))\mathrm d\tau=2(2z-s)-\int_{-s-z}^{-2s+z}2\Phi(t)\mathrm dt, $ hence $ U_s'(z)=2(2z-s)-\int_{2s-z}^{s+z}2\Phi(t-3s)\mathrm dt. $ Since $\Phi(t-3s)\leqslant\Phi(t)$ for every $t$, this yields $ U_s'(z)\geqslant2(2z-s)-\int_{2s-z}^{s+z}2\Phi(t)\mathrm dt=2(2z-s)-U_s'(z), $ hence $U_s'(z)\geqslant2z-s$ and, finally, $ U_s(z)\geqslant z^2-zs. $ In the notations of the question, this proves that

$[12b^2(g(\frac{1+ax}{b})-g(\frac{ax}{b})+g(\frac{2ax-1}{b})-g(\frac{2ax}{b}))] +$ $\qquad\qquad\qquad +{}[12b^2(\Phi(\frac{1+ax}{b})-\Phi(\frac{ax}{b})+\Phi(\frac{2ax-1}{b})-\Phi(\frac{2ax}{b})] \geqslant\mathbf{12}(1-ax)\geqslant\mathbf{6}(1-a^2x^2). $

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    Ok I see the logic now, thanks. For the record, I don't think my concerns were petty. I will now focus on figuring out the parameters I need to make it work. Presumably, I need z sufficiently large.2012-05-19