0
$\begingroup$

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.

  • 0
    What do you mean by *?2012-05-18
  • 0
    @M.B. Thats a habit I have for multiplication2012-05-18

1 Answers 1

3

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). $

  • 0
    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?2012-05-18
  • 0
    $U_x(x)=0$ is used in the proof. On the other hand, $U_x'(0)$ is not defined (except if $x=0$).2012-05-18
  • 0
    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.2012-05-18
  • 0
    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 0.2012-05-18
  • 0
    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.2012-05-19
  • 0
    Thank you for all your help so far and for putting up with my questions, naturally I have more. For steps 3 and 4, you assume both a>1 and a range of values z=ax, both of which are outside the bounds Ive set... The reason I'm so skeptical is that Ive done many simulations and all of them work out. I realize though thats not a sufficient reason.2012-05-19
  • 0
    This parameter $a\gt1$ has nothing to do with your $a\lt1$, I thought this was obvious. Post modified. May I suggest that you cease to decree hastily that the answer does not solve your question for such petty concerns, it does. About simulations: you might want to try the values $a=0.5$, $x=1$, $b=100$.2012-05-19
  • 0
    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