Ratio of CDF to PDF increasing? Let $\Phi(x)$ be a cumulative normal distribution function and $\phi(x)$ the associated probability density function.
Is the ratio $\frac{\Phi(x)}{\phi(x)}$ increasing in x?
Numerically it seems to be true. Is there any ways to prove it analytically?
Thanks
 A: The inverse Mill's ratio is defined as
$$
\lambda(x)=\frac{\phi(x)}{\Phi(x)}.
$$
For answering the question, it suffices to show that $\lambda'(x)<0$. Note that the p.d.f. of standard normal $\phi$ is differentiable. Thus we can apply quotient rule to it
$$
\lambda'(x)=\frac{\phi'(x)\Phi(x)-\phi(x)^2}{\Phi(x)^2}=\frac{-x\phi(x)\Phi(x)-\phi(x)^2}{\Phi(x)^2}=-\lambda(x)(x+\lambda(x)).
$$
Observe that
$$
\phi'(x)=-x\frac{1}{\sqrt{2\pi}}\exp\left\lbrace -\frac{x^2}{2}\right\rbrace=-x\phi(x).
$$
It is clear that
$$
x+\lambda(x)>0,\quad \forall x\geq 0.
$$
The challenge is to show that
$$
-x+\lambda(-x)>0,\quad \forall x>0.
$$
Here we exploit two facts:


*

*normal distribution is symmetrical around $0$.

*the inverse Mill's ratio is the expectation of the truncated normal
$$
\frac{\phi(x)}{1-\Phi(x)}=E[X|X>x],\quad X\sim \mathcal{N}(0,1).
$$
Thus the condition we want to show can be written as
$$
E[X|X>x]=\frac{\phi(x)}{1-\Phi(x)}=\frac{\phi(-x)}{\Phi(-x)}=\lambda(-x)>x,\quad \forall x>0.
$$
If this is not immediate to you, invoke Chebychev's inequality to get
$$
E[X|X>x]\geq x.
$$
For strict inequality, argue by contradiction. Suppose $E[X|X>x]=x$, then it is necessary that
$$
\int_{x}^\infty (X-x)d\Phi(X)=0,
$$
which implies that $X=x$ except on a set of $\Phi$ measure zero. But this is clearly false that we can always find a set with strictly positive $\Phi$ measure for which $X>x$ given any $x$.
A: Maybe there is a simpler way, but here is a thought. Use the relations $ \Phi(x)' = \phi(x)$ and $\phi(x)'= -x\phi(x)$
$$\left(\frac{\Phi(x)}{\phi(x)}\right)' = \frac{\Phi(x)'\phi(x) - \Phi(x)\phi(x)'}{\phi(x)^2} = \frac{\phi^2(x) + x \Phi(x)\phi(x)}{\phi(x)^2} = 1 + x \frac{\Phi(x)}{\phi(x)}$$ 
If $\left(\frac{\Phi(x)}{\phi(x)}\right)' \geq 0$ we are done.
But note that since $x$ can take negative values, we can't be sure that $1 + x \frac{\Phi(x)}{\phi(x)} \geq 0$.
If $1 + x \frac{\Phi(x)}{\phi(x)} = 0$ then 
$$\left(\frac{\Phi(x)}{\phi(x)}\right)'' =  \left(1 + x \frac{\Phi(x)}{\phi(x)}\right)' = \frac{\Phi(x)}{\phi(x)} + x\left(1 + x \frac{\Phi(x)}{\phi(x)}\right) = \frac{\Phi(x)}{\phi(x)} \geq 0 $$ 
Therefore once we prove that $\lim_{x \to -\infty}\left(\frac{\Phi(x)}{\phi(x)}\right)' \geq 0$ then we will have proved that $\left(\frac{\Phi(x)}{\phi(x)}\right)' \geq 0$ holds for every $x\in \Bbb{R}$
Therefore we calculate:
$$\lim_{x \to -\infty}1 +  x \frac{\Phi(x)}{\phi(x)} = 1 + \lim_{x \to -\infty}x\frac{\int_{-\infty}^x e^{-u^2/2}\, du}{e^{-x^2/2}}  = 1 + \lim_{x \to -\infty}\int_{-\infty}^x xe^{-(u^2 - x^2)/2}\, du  = *$$
And 
\begin{align}
\int_{-\infty}^x xe^{-(u^2 - x^2)/2}\, du &= -\int_x^{\infty} xe^{-(u^2 - x^2)/2}\, du \\
&= -\int_0^{\infty} x e^{-((x+h)^2 - x^2)/2}\, dh  = \int_0^{\infty} -x e^{-h^2/2} e^{ - xh}\, dh \xrightarrow[x \to \infty]{} 0
\end{align}
Once $xe^{ - xh}\xrightarrow[x \to \infty]{} 0$ Now note that
$$ 1 + \lim_{x \to -\infty}\int_{-\infty}^x xe^{-(u^2 - x^2)/2}\, du = 1 - \lim_{x \to \infty}\int_{-\infty}^x xe^{-(u^2 - x^2)/2}\, du = 1 \geq 0$$
A: As in Conrado Costa's answer:
$$
\Bigl(\frac{\Phi(x)}{\phi(x)}\Bigr)' = 1 + x \frac{\Phi(x)}{\phi(x)} \tag{1}
$$
and the RHS is clearly positive for $x\ge0$.
For $x<0$ write
$$
1 + x \frac{\Phi(x)}{\phi(x)}
  =  \frac{\phi(x)}{x} \, \Psi(x) ,
$$
where
$$
\Psi(x) := \frac{\phi(x)}{x} + \Phi(x) .
$$
$\Psi(x)$ is negative for $x<0$
since $\lim_{x\to-\infty}\Psi(x) = 0$ and
$\Psi'(x) := -\frac{\phi(x)}{x^2} < 0$,
hence the RHS of (1) is positive for $x<0$ too.
