consider following piecewise cdf. It is continuous but not differentiable at just $2$ points. My question is it valid to take piecewise derivative of cdf and call it pdf.
$ G(x) = \begin{cases} \dfrac{x}{3 *0.1} . & 0 < x \leq 0.1 \\ \dfrac{x - 0.1}{3 *(0.6 - 0.1) } + \frac{1}{3} & 0.1 < x \leq 0.6\\ \dfrac{x - 0.6}{3* (0.9 - 0.6) } + \frac{2}{3} & 0.6 < x \leq 0.9\\ \end{cases} $
$ g(x) = \begin{cases} \dfrac{1}{3 *0.1} . & 0 < x \leq 0.1 \\ \dfrac{1}{3 *(0.6 - 0.1) } & 0.1 < x \leq 0.6\\ \dfrac{1 }{3* (0.9 - 0.6) } & 0.6 < x \leq 0.9\\ \end{cases} $
My guess is that it is okay because I think $g$ satisfies following $$\int_a^b g(x) = G(b) - G(a)$$ for all $0<a<b<0.9$ because Riemann Integral doesn't depend on values at finite number of points.I am asking because I haven't studied probability rigorously and I want to make sure my guess is correct.Also, Please don't involve any measure theory.
It is correct. For example, function $p(x) = \begin{cases} 1,\ x \in [0, 1]\\ 0,\ x \notin [0, 1] \end{cases}$ is density function of uniform distribution on $[0, 1]$ even though it's CDF is not-differentiable in $0$ or $1$.
(formally, PDF is defined up to measure zero, although we usually try to find "most continuous" one)