Distribution whose PDF is proportional to the product of a PDF and a CDF Let $\phi$ be a symmetrical probability distribution function, that is nonnegative, that means  $\int_{-\infty}^\infty \phi(x) dx = 1$ and $\phi(-x) = \phi(x) \forall x$ and $\phi(x) \geq 0 \forall x$. Let $\Phi$ be it's cumulative distribution function, that means $\Phi(x) = \int_{-\infty}^x \phi(t) dt$.
For the normal distribution we know that $f(x) := \frac{\phi(x)\Phi(ax)}{\Phi(0)}$ is again a pdf for every $a\in \mathbb R$. (Note that $\Phi(0) > 0$ since $\phi$ is symmetrical. In case of the normal distribution $\Phi(0) = 0.5$.) Is this also true for any other distribution that meets the requirements above? Does it hold if we assume further restrictions?
It is known to work for the normal distribution (see skewed normal distribution), but I also made some numerical calculations that seem to support this idea for the Cauchy, Slash  and student-t distribution.
 A: 
Let $a$ denote some real number, $g$ any PDF, $G$ the corresponding CDF, $h$ the function defined by $h(x)=g(x)G(ax)$ for every $x$, and $(X,Y)$ some i.i.d. random variables with PDF $g$. 
Then the distribution of $X$ conditionally on the event $A=[Y\leqslant aX]$ has PDF  $h/P(A)$, for every $(g,a)$ are such that $P(A)\ne0$ (in particular, if $P(A)\ne0$ then $h/P(A)$ is a PDF).

To show this, note that for any bounded measurable $u$, $Y$ has CDF $G$ and is independent of $X$ hence $$E(u(X)\mathbf 1_A\mid X)=u(X)P(Y\leqslant aX\mid X)=u(X)G(aX),$$ which implies that $$E(u(X)\mid A)=\frac{E(u(X)G(aX))}{P(A)}=\frac1{P(A)}\int_\mathbb R u(x)G(ax)g(x)\mathrm dx.$$
Since this holds for every bounded measurable function $u$, the distribution of $X$ conditionally on $A$ has density $f$ defined by $$f(x)=\frac1{P(A)}G(ax)g(x)=\frac1{P(A)}h(x).$$ Finally, $$P(A)=\int_\mathbb R h(x)\mathrm dx=\int_\mathbb R G(ax)g(x)\mathrm dx.$$ That $g$ is even is not needed but if $g$ is even then $P(A)=P(-Y\leqslant-aX)=P(Y\geqslant aX)$ by symmetry hence $P(A)=\frac12$, which guarantees that $P(A)\ne0$. Likewise, if $a=1$ then $P(A)=\frac12$ for every $g$. Thus the identity $P(A)=G(0)$ is not always valid.
A: A proof can be found in A Class of Distributions Which Includes the Normal Ones by A. Azzalini:

Lemma 1: Let $f$ be a density function symmetric about $0$, and $G$ an absolutely continuous [cumulative] distribution function such that $G'$ is symmetric about $0$. Then
$$2G(\lambda y)f(y) $$
is a density function for any real $\lambda$.
Proof: Let $Y$ and $X$ be independent random variables with density $f$ and $G'$ respectively. Then
$$\frac{1}{2} = P(X-\lambda Y <0) = E_y(P(X<\lambda y \mid Y=y)) = \int_{-\infty}^\infty G(\lambda y) f(y) dy$$
q.e.d.

