Is the Inverse Mills Ratio Strictly Decreasing?

Question

As far as I know, the Inverse Mills ratio, $\lambda(x)=\phi(x)/\Phi(x)$, is decreasing in $x$. Thus, I am curious now whether $\lambda(x)$ is in fact strictly decreasing in $x$.

To see this, I derived the derivative of the inverse Mills ratio: \begin{align*}\lambda(x)&=\phi(x)/\Phi(x) \\[2pt] \Rightarrow \lambda'(x)&=\frac{\phi'(x)\Phi(x)-\phi(x)^2}{\Phi(x)^2} \\[2pt] &=-x\lambda(x)-\lambda(x)^2\quad\because \phi(x)'=-x\phi(x)\end{align*} Here, I have no idea how to show whether $\lambda(x)<0$ or $\lambda(x)\leq0$.

Thus, is the inverse Mills ratio in fact "strictly" decreasing in $x$?

Answer 1

Theorem $1.$ (Sampford) If $\lambda(x):=\frac{\varphi(x)}{\Phi(x)},$ then $\lambda^\prime(x)\in(-1,0).$

$\frac{e^{-\frac{1}{2} z^2}}{\int^{x}_{-\infty}e^{-\frac{1}{2} z^2}\, \mathrm dz}$ is nothing but the density of a truncated normal with support $z\in (-\infty,x]$. Its variance is $1-x\lambda(x) -\lambda^2(x). $ This implies $\lambda^\prime(x) >-1.$ And by definition, $\lambda(x) > 0; ~x+\lambda(x)> 0 $ for $x\geq 0.$ This is also true if $x< 0.$ The result follows.

$\blacksquare$

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Reference:

$\rm [I]$ Some Inequalities on Mill's Ratio and Related Functions, M. R. Sampford, The Annals of Mathematical Statistics, Vol. $24, $ No. $1$ (Mar., $1953$), pp. $130-132.$