Show that $\log \phi(z;\lambda)$ is a concave function

Question

Let $Z\sim SN(\lambda)$ where SN means skew-normal. Then a random variable $Z$ have density function given by $$\phi(z;\lambda)=2\phi(z)\Phi(\lambda z)\qquad -\infty<z<\infty$$

where $\phi$ and $\Phi$ are the standard normal density and function. One of the properties of this density is

Property: The density of $Z$ is strongly unimodal, i.e log $\phi(z;\lambda)$ is a concave function.

I want to make a proof of this property

Calling $f= \log \phi(z;\lambda)$ I want to show that for any $x$,$y$ and $a$,$b\in(0,1)$ with $a+b=1$ that $$f(ax+by)\leq a f(x)+b f(y)$$

I have that $$\Phi(z;\lambda)=\int_{-\infty}^z\int_{-\infty}^{\lambda t}\phi(t)\phi(u)du dt$$

and $$\phi(z;\lambda)=2\frac{1}{\sqrt{2\pi}} \exp \Big(-\frac{1}{2}z^2\Big)\int_{-\infty}^{\lambda z} \frac{1}{\sqrt{2\pi}} \exp\Big(-\frac{1}{2}u^2\Big)du$$

Making some research I found this remark

Remark: Log-concavity of a function $g$ on $(a,b)$ is equivalent to each of the following two conditions.

i) $\frac{g'(x)}{g(x)}$ is monotone decreasing on $(a,b)$.

ii) $ \log (g(x))''<0$

But I don't know how I can use it.

Answer 1

First write

$$f=\ln(2)+\ln(\phi(z))+\ln(\Phi(\lambda z)).$$

Concavity is linear: the sum of concave functions is also concave. You can easily prove this from the definition. Thus we just need to show each piece above is concave.

1) $\ln(\phi(z))=\ln(C)-z^2/2,$ for $C$ constant. Then taking two derivatives gives $-1<0$ which implies $\ln(\phi(z))$ is concave.

2) $\ln(\Phi(\lambda z))$: one derivative gives $\lambda\frac{\phi(\lambda z)}{\Phi(\lambda z)}$,. One more derivative gives:

$$\lambda \frac{-z\lambda^2\phi(\lambda z)\Phi(\lambda z)-\lambda \phi^2(\lambda z)}{\Phi^2(\lambda z)}.$$

The denominator is always positive. For the numerator, write it as:

$$-\lambda \phi(\lambda z)[z\lambda^2 \Phi(\lambda z)+\lambda \phi(\lambda z)].$$

There are two cases: $\lambda<0$ and $\lambda\geq 0$. If $\lambda\geq0$, the numerator is clearly negative.

Otherwise, $\lambda<0$, in which case we need to show $z\lambda^2\Phi(\lambda z)+\lambda \phi(\lambda z)\leq 0$, or equivalently $z\lambda \Phi(\lambda z)+\phi(\lambda z)\geq 0$. This is true if $z<0$, so lets prove it for $z\geq 0$. First check it for $z=0$, where it's clearly true. Dividing through by $z$ implies we need to show:

$$\lambda \Phi(\lambda z)+\phi(z)/z\geq 0$$

Differentiating gives:

$\lambda^2 \phi(\lambda z)-\lambda^2 \phi(\lambda z)+\phi(\lambda z)/z^2=\phi(\lambda z)/z^2\geq 0$.

So since the derivative is positive and the statement is true at $z=0$, it's true for all $z>0$.