I am wondering whether there is some known family $c(x,a)$ of CDFs (cumulative distribution functions), where $a$ are some parameters and $x \geq 0 $, such that
$c(x,a)$ is continuous $\forall x,a$.
The PDF $c'(x,a)$ is continuous $\forall x$ and for almost all $a$, apart for some "special value" $a=a_0$.
For some particular limit of the parameters, say formally $a = a_0$, we have that $c(x, a_0) =0$ if $x<1$ and that $c'(1^+, a_0) = +\infty$. This is the red curve sketched in the image below.
For every $a$, including $a=a_0$, we have that $c(0,a)=0$ and $c'(0,a)=0$.
If no such a family is known, is there a strategy to construct an explicit example (maybe a minimal one, where $a$ is just a single parameter)? Partial answers and hints are appreciated.
Visual sketch: For more clarity it may be convenient to give a visual sketch of the problem. We have to construct a family of functions $c(x,a)$ that behave as the blue curves in the image below:
$c(x,a)$ and $c'(x,a)$ are both continuous.
$c'(x,a)=0$ for $x \rightarrow 0^+$.
$c(x,a)$ tends to the red curve $r(x)$ for $a \rightarrow a_0$, where the red curve is such that $r'(1^+) = \infty$. This red curve is the function $c(x, a_0)$ mentioned in the first part of this post.
Possibly, $c(x,a)$ tends to the green curve $g(x)$ in some other limit, where the green curve is, for example, $ g(x)=H(1-x) x^n+H(x-1)$, with $n>0$ and $H$ the Heaviside step function. This curve is there just to say that there could be another non-smooth limit of the blue curves, but that in general we want all the curves of the family to be zero (and possibly have zero derivative) at the origin.
Clearly the "trivial" solution $$ c(x,a) = (1-a)g(x) + a r(x) $$
does not work because it gives rise to a PDF $c'(x,a)$ that is not continuous. Stated in other words, I am looking for an "example of homotopy" between the "red CDF" and the "green CDF" such that the "blue CDFs" are at least of class $C^1$ (i.e. the "blue PDFs" are at least of class $C^0$).
Loosely speaking: For example, the Weibull, the Log-logistic, the Chi-squared, the Gamma and many other distributions all seem to be able to reproduce "smooth shapes" for most of the parameters, as well as shapes "where there is angle at the origin" ...the problem is that I want to "move the angle" to some $x>0$. Maybe some clever re-parametrization of the three parameters of the Generalized extreme value distribution in terms of just one parameter could work?
Does something like this work?
Take any $K:[0,\infty) \rightarrow [0,1]$ that is differentiable.
Let $g_n:[0,1) \rightarrow [0,a_n]$ be a sequence of functions that are continuous, differentiable, non-decreasing, strictly decreasing in $n$ pointwise, and satisfy $\lim_{x\uparrow 1}g_n(x)=a_n$ and $\lim_{x\uparrow 1}g_n'(x)=a_n$. For example, $a_n \dfrac{1}{n} x^{n} \rightarrow 0$ on $[0,1)$, with derivative $a_n x^{n-1}$, and at $x=1$, $a_n$.
Let $h_n:[1,\infty) \rightarrow [a_n,\infty)$ be a sequence of functions that are continuous, differentiable, strictly increasing, and satisfy $\lim_{x \downarrow 1}h_n(x)=a_n$ and $\lim_{x \downarrow 1}h_n'(x)=a_n$. For example, $ a_n (1-\frac{1}{\lambda}e^{-\lambda (x-1)})$ with derivative $a_n e^{-\lambda(x-1)}$, and at $x=1$, $a_n$. Or even $h_n(x) = a_n x$ would work.
Consider $$F(x)= \begin{cases} K(g_n(x)), & 0 \le x < 1\\ K(h_n(x)), & x\ge 1. \end{cases} $$ then $$ f(x) = \begin{cases} k(g_n(x))g_n'(x), & 0 \le x < 1\\ k(h_n(x))h_n'(x), & x \ge 1. \end{cases} $$
At $x=1$, you need $k(g_n(1))g_n'(1)=k(h_n(1))h_n'(1)$, but both of those terms equal $a_n$ since $g_n(1) = h_n(1)=a_n$, and $g_n'(1)=h_n'(1)=a_n$.
You just need to smoothly paste the derivative across $x=1$, if I'm not mistaken?