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
On
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
$$ 1 + \lim_{x \to -\infty}\int_{-\infty}^x xe^{-(u^2 - x^2)/2}\, du = 1 - \lim_{x \to \infty}\int_{-\infty}^x xe^{-(u^2 - x^2)/2}\, du = 1 \geq 0$$
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 $x$.