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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?


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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What do you mean by *? – M.B. May 18 '12 at 0:01
@M.B. Thats a habit I have for multiplication – Greg May 18 '12 at 0:10
up vote 3 down vote accepted

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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But, I also have that when $x=z$ then $U_x(z)=0$ and so does the RHS of (*), so wouldn't it then suffice to show that $U_x'(0)>0$ and then something about the second derivative? – Greg May 18 '12 at 17:14
$U_x(x)=0$ is used in the proof. On the other hand, $U_x'(0)$ is not defined (except if $x=0$). – Did May 18 '12 at 19:15
I'm still unclear, all you've done is established a lower bound and said that the lower bound is below the point we need. This does not imply that the inequality is false, which you seem to have claimed. Also, 2x-z is not greater than 0 everywhere. – Greg May 18 '12 at 20:31
The problem is that you are essentially making z and x 0 to get that $\Phi = 0.5$ uniformly on the interval, but in that case the RHS of $ (*) $ becomes 0. – Greg May 18 '12 at 23:18
The asymptotics is made more explicit and the proof somewhat shortened. The counterexample is not $U_0(0)=0$, naturally, but $U_x(2x)$, say, when $x\to0$. One shows that $U_x(2x)\sim2x^2$ when $x\to0$ instead of the prediction $U_x(2x)\geqslant3x^2$ uniformly. – Did May 19 '12 at 10:25

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