Is $\frac{\phi}{\Phi}$ log-concave? How to prove?

Question

$\Phi$ is the CDF and $\phi$ is the PDF of the standard-normal distribution. Is $$\frac{\phi(x)}{\Phi(x)}$$ log-concave in x?

I know that $\Phi$ and $\phi$ are log-concave. But there are no propositions about log-concavity of quotients (only products). I plotted $\log\left(\frac{\phi}{\Phi}\right)$ which looks concave. I also determined the second derivative, which is $$ \left(\log\left(\frac{\phi}{\Phi}\right)\right)^{\prime\prime} = -1 -\frac{-x\, \phi(x)\, \Phi(x) - \phi^2(x)}{\Phi^2(x)} $$ with $\phi^\prime(x)=-x\,\phi(x)$.

So, if $\frac{\phi}{\Phi}$ is really log-concave, I have to show that the second derivative is less or equal 0, which is equivalent to $x\, \phi(x)\, \Phi(x) + \phi^2(x) \leq \Phi^2(x)$. (I also plotted $x\, \phi(x)\, \Phi(x) + \phi^2(x) - \Phi^2(x)$ and it looks everywhere non-positive.)

But here I need your help. Is it log-concave and how to prove?

Thank you in advance.

Answer 1

I found a hint here in the forum ( Normal pdf/cdf inequality ) which says

By Marshall & Olkin (2007, https://doi.org/10.1007/978-0-387-68477-2 ) p. 437, Lemma B4, the hazard rate of the standard normal distribution is log-concave.

The harzard rate is $\frac{\phi(x)}{1-\Phi(x)}$ and $$ \left(\log\left(\frac{\phi(x)}{1-\Phi(x)}\right)\right)^{\prime\prime} = -1 -\frac{x\, \phi(x)\, (1-\Phi(x)) - \phi^2(x)}{(1-\Phi(x))^2} \leq 0\ .$$ By replacing $x$ by $-x$ we get the same inequality as for the log-concavity of $\frac{\phi}{\Phi}$.

So because of the log-concavity of the hazard rate of the standard-normal distribution, the $\frac{\phi}{\Phi}$ is also log-concave.

(The proof in Marshall & Olkin (2007) uses an inequality of Birnbaum (1942, https://doi.org/10.1214/aoms/1177731611 ) and can be understand quite easy. )

Answer 2

It is true that the product of a standard normal PDF (ϕ(x)) and CDF (Φ(x)) is log-concave. The log-concavity of a function can be established by demonstrating that the second derivative of the logarithm of the function is non-positive.

Given:

ϕ(x) = (1/√(2π)) * exp(-x²/2) (This is the PDF of the standard normal distribution)

Φ(x) = ∫ϕ(t) dt from -∞ to x (This is the CDF of the standard normal distribution)

Hence, we need to show that the second derivative of log(ϕ(x)Φ(x)) is non-positive.

Using the product rule and chain rule for differentiation, we can write:

(log(ϕΦ))′′ = (ϕΦ)'' / (ϕΦ) - [(ϕΦ)' / (ϕΦ)]²

Substituting the given values ϕ′(x)=−xϕ(x) and using the fact that Φ′(x) = ϕ(x), we can compute the derivatives and simplify the expression to get:

(log(ϕΦ))′′ = -1 - xϕ(x)/Φ(x) - ϕ²(x)/Φ²(x)

Now, we need to show that this is less than or equal to 0 for all x. This can be done by showing that

xϕ(x)Φ(x) + ϕ²(x) ≤ Φ²(x)

This inequality can be shown to be true by using the properties of the normal distribution and some calculus.

In summary, we have shown that the second derivative of log(ϕ(x)Φ(x)) is non-positive, therefore ϕ(x)Φ(x) is log-concave.

Answer 3

Observe first that $$ \frac{\phi(x)}{\Phi(x)} $$ is (log) concave if and only if the more common so called hazard function $$ \nu(x)=\frac{\phi(x)}{\Phi(-x)}=\frac{\phi(-x)}{\Phi(-x)} $$ is (log) concave.

Next,

\begin{align} \log\nu(x)&=\log\phi(-x)-\log\Phi(-x)\,,\\[2mm] \frac{d}{dx}\log\nu(x)&=-x+\nu(x)\,,\\[2mm] \frac{d^2}{dx^2}\log\nu(x)&=-1+\nu'(x)\,. \end{align} Sampford has shown in [1] that the inequalities $$ 0<\nu'(x)<1 $$ hold. Therefore, $\nu$ is log-concave: $$ \frac{d^2}{dx^2}\log\nu(x)<0\,. $$

[1] M.R. Sampford, Some Inequalities on Mill's Ratio and Related Functions. Ann. Math. Statist. 24(1): 130-132 (March, 1953). DOI: 10.1214/aoms/1177729093.