Let $n(x) := \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}$, and $N(x) := \int_{-\infty}^x n(t)dt$. I have plotted the curves of the both sides of the following inequality. The graph shows that the following inequality may be true.
$$f(x):= (x^2+1)N + xn-(xN+n)^2 > N^2$$
where the dependency of $n$ and $N$ on $x$ are absorbed into the function symbols. However, I have not succeeded in providing a full proof except for $x$ above some positive number, with the help of various Mill's Ratio $\frac{m}{n}$ bounds.
I am asking for help in proving the above inequality or providing an $x$ that violates the above inequality. Judging from the aforementioned plot I am pretty confident the validity of the inequality, though.
The left hand side is actually the variance of a truncated normal distribution. I am trying to give it a lower bound. More explicitly,
$$f(x):=\int_0^\infty t^2n(t+x)dt-\Big(\int_0^\infty t\,n(t+x)dt\Big)^2>\Big(\int_0^\infty n(t-x)dt\Big)^2.$$
The form of the inequality is probably more transparent if we set $m=1-N$ and the inequality is equivalent to
$$g(x)\equiv m[(x^2+1)(1-m)+2xn]-n(x+n) > 0.$$
$N$ is the upper bound of $f$, i.e. $$(x^2+1)N + xn-(xN+n)^2 < N$$ or
$$h(x)\equiv x^2 m(1-m)-n[x(1-2m)+n]<0$$
Proof: $h$ is an even function and $h(0)<0$, so we only need to consider $x>0$.
From the integration by part of $m(x)$ and dropping a negative term, we have $$xm<n, \forall x>0.$$
The first term of $h(x)$ is then bounded and
\begin{eqnarray}
h(x)&<&x(1-m)n-n[x(1-2m)+n] \\
&=& n(xm-n) \\
&<& 0,
\end{eqnarray}
where last inequality is obtained by using $xm<n$ again.
The lower bound of $f(x)$ appears to be more difficult since it requires tighter approximation of $m$ without singularity at $x=0$. I can prove the lower bound for $x$ greater than some positive number. I know I need to stitch the small and large regions of positive $x$ together, but I have not carried the detailed computation out yet. Does anyone have more clever trick to accomplish this task?
$g(x)>0, \forall x\ge\sqrt{\frac{4}{3}}$
Proof:
\begin{align}
\frac{dg}{dx} &= 2n[xr(1-m)-2(0.5-m)] \\
&= 2n^2[(xr-1)n^{-1}+(2-xr)r]
\end{align}
where $r:=\frac{m}{n}$. In what follows we will use the first expression. The second expression is an alternative which I keep just for maybe future reference.
Since
$$r<\frac{1}{x}\Big(1-\frac{1}{x^2+3}\Big), \forall x>0,$$
\begin{align}
\frac{dg}{dx} &< \frac{2n^2}{x^2+3}(-n^{-1}+(x^2+4)r) \\
&<\frac{2n^2}{x^2+3}\Big(-n^{-1}+x\Big(1+\frac{4}{x^2}\Big)\Big),
\end{align}
where on the last line we apply the $r$ bound again.
Choose $x\ge x_0:=\sqrt{\frac{4}{3}}$,
$$n^{-1}-x\Big(1+\frac{4}{x^2}\Big)>n^{-1}-4x.$$
It can be shown that $n^{-1}-4x$ is positive at $x=x_0$ and its derivative is always positive for $x\ge x_0$. We thus have
$$\frac{dg}{dx}<0, \forall x\ge x_0.$$
It is easy to see that $g(x)>0$ for sufficiently large $x$. Therefore, $g(x)>0, \forall x\ge x_0$.
Best Answer
Here is a complete solution. The idea is to kill the entries of $N$ in two steps, by applying two appropriately constructed first-order differential operators, which will result in a simple elementary expression:
Let $b:=f-N^2$. As noted by cardinal, $b$ is an even function. So, it is enough to show that $b>0$ on $[0,\infty)$. Let $$ b_0(x):=\frac{b(x)}{x^2+1} $$ and $$ b_1(x)=\pi\, \left(x^2+1\right)^2 e^{x^2/2}\, b_0'(x). $$ Then $b_1'(x)=-e^{-\frac{x^2}{2}} \left(x^2+1\right)<0$, so that $b_1$ is decreasing. Also, $b_1(0)=0$. Hence, $b_1(x)<0$ for $x>0$, and so, $b_0$ is decreasing on $[0,\infty)$. Moreover, $b_0(x)\to0$ as $x\to\infty$. So, $b_0>0$ and hence $b>0$.