How do I prove that a Skew-Normal or Skew-Logistic Distribution is a Distribution?

Question

How do I prove that a Skew-Normal or Skew-Logistic Distribution is a Distribution?

In the literature, I found that for $\phi(x)$ being Gaussian or Logistic or any other univariate distribution, and $\Phi(x)=\int_{-\infty}^{x} \phi(x) dx$, we obtain again a distribution by

$p(x)=2\phi(x)\Phi(ax)$

for any real valued $a$.

Answer 1

It seems $\phi(x)\geq 0$ implies $\Phi(x)\geq 0$. Thus, positivity is given. We also need $\Phi(x)+\Phi(-x)=\int_{-\infty}^\infty \phi(x) dx = 1$ and can show $$ \begin{align} & \int_{-\infty}^\infty 2 \phi(x) \Phi(ax) dx\\ &= \int_{-\infty}^0 2\phi(x) \Phi(ax) dx + \int_{0}^\infty 2 \phi(x) \Phi(ax) dx\\ &= -\int_0^{\infty} 2\phi(-x) \Phi(-ax) dx + \int_{0}^\infty 2 \phi(x) \Phi(ax) dx\\ &= \int_0^{\infty} 2\phi(x) \Phi(-ax) dx + \int_{0}^\infty 2 \phi(x) \Phi(ax) dx\\ &= \int_0^{\infty} 2\phi(x) (\Phi(-ax) + \Phi(ax)) dx\\ &= \int_0^{\infty} 2\phi(x) dx\\ &= 2\int_0^{\infty} \phi(x) dx\\ &=1 \end{align} $$