Log-convexity of a function defined by an integral (Normal Mills ratio)

Question

Let's define $f(x)$, for all $x>0$ by :

$$f(x)=e^{x^2/2}\int_x^{+\infty}e^{-t^2/2}dt$$

I would like to prove that $f$ is log-convex, which is equivalent to the following condition :

$$\forall x>0,f''(x)f(x)-f'(x)^2\ge0\tag{$\star$}$$

Simple calculations lead to :

$f'(x)=xf(x)-1$

$f''(x)=(x^2+1)f(x)-x$

$f''(x)f(x)-f'(x)^2=f(x)^2+xf(x)-1$

Hence, in order to prove $(\star)$, it is sufficient to prove that :

$$\forall x>0,f(x)\ge\frac12\left(\sqrt{x^2+4}-x\right)$$

and I didn't find any way to show this last inequality.

Answer 1

We need to prove that, for all $x> 0$, $$\mathrm{e}^{x^2/2}\int_x^{\infty}\mathrm{e}^{-t^2/2}\, \mathrm{d} t \ge \frac12\left(\sqrt{x^2+4}-x\right)$$ or $$g(x) := \int_x^{\infty}\mathrm{e}^{-t^2/2}\, \mathrm{d} t - \mathrm{e}^{-x^2/2}\cdot \frac12\left(\sqrt{x^2+4}-x\right) \ge 0.$$

We have $$g'(x) = \mathrm{e}^{-x^2/2}\left(\frac{x^3 + 3x}{2\sqrt{x^2+4}} - \frac{1 + x^2}{2}\right) < 0, \quad \forall x > 0.$$ Also, we have $\lim_{x\to \infty} g(x) = 0$. Thus, $g(x) \ge 0$ for all $x \ge 0$.

We are done.

Answer 2

OP's function $f(x)$ is not some "function defined by an integral" but the well-known Mills ratio for the Normal distribution:

$$f(x) =e^{x^2/2}\int_x^{+\infty}e^{-t^2/2}dt = \frac{\int_x^{+\infty}\frac{1}{\sqrt{2 \pi}}e^{-t^2/2}dt}{\frac{1}{\sqrt{2 \pi}}e^{-x^2/2}}$$ $$=\frac{1-\Phi(x)}{\phi(x)}=\frac{\Phi(-x)}{\phi(x)} \equiv m(x),$$

where $\Phi$ is the Normal distribution function and $\phi$ is the Normal density, $\Phi'(x) = \phi(x)$.

Various results have been established over the years for the Normal Mills ratio. In particular,

Birnbaum, Z. W. (1942). "An inequality for Mill's ratio". The Annals of Mathematical Statistics, 13(2), 245-246.

proved that

$$m(x) \geq \frac{\sqrt{x^2+4} - x}{2},$$

...which is exactly what is needed for log-convexity of $m(x)$.

Mills ratio is a good illustration that the ratio of two log-concave functions (as $\Phi$ and $\phi$ are), may not be log-concave. In general, this is because the logarithm of such a ratio is the difference of two concave functions, which is equivalent to the sum of a concave and a convex function (since "minus-concave" is convex).

Answer 3

This is essentially the solution from

• Z. W. Birnbaum. "An Inequality for Mill's Ratio." The Annals of Mathematical Statistics, 13(2) 245-246 June, 1942. https://doi.org/10.1214/aoms/1177731611

(already cited in Alecos Papadopoulos' answer). Instead of Jensen's inequality we apply the Cauchy-Schwarz inequality to the functions $g(t) = e^{-t^2/4}$ and $h(t) = te^{-t^2/4}$:

$$ \left(\int_x^\infty t e^{-t^2/2} \, dt \right)^2 < \int_x^\infty e^{-t^2/2} \, dt \cdot \int_x^\infty t^2 e^{-t^2/2} \, dt \, . $$ The inequality is strict because $g$ and $h$ are not constant multiples of each other.

Evaluating the first integral and integration by part for the last integral gives $$ \left(e^{-x^2/2} \right)^2 < \int_x^\infty e^{-t^2/2} \, dt \cdot \left( x e^{-x^2/2} + \int_x^\infty e^{-t^2/2} \, dt\right) $$ or $$ 1 < f(x) \cdot ( x + f(x)) \, , $$ so that $$ f''(x)f(x)-f'(x)^2=f(x)^2+xf(x)-1 > 0 \, . $$ This proves that $f$ is (strictly) log-convex.