Given random variable $X$ with cumulative distributive function
$$
F_X(x)=
\begin{cases}
0&x<0
\\
\dfrac{1}{4}x^2&0\leq x<1
\\
\dfrac{1}{2}&1\leq x<2
\\
\dfrac{1}{3}x&2\leq x<3
\\
1&x\geq 3
\end{cases}.
$$
Find the probability density function of $X$.
To find the p.d.f, I plot the graph of $F_X(x)$,as below.
Based on picture, I conclude $X$ is mixed random variable.
For continuous random variable, differentiating $F_X(x)$ we have
$$f_X(x)=
\begin{cases}
\dfrac{1}{2}x&0\leq x< 1\\
\dfrac{1}{3}&2\leq x<3
\end{cases}$$
For discrete random variable, we have
$$f(1)=F_X(1^+)-F_X(1^-)=\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{1}{4},$$
$$f(2)=F_X(2^+)-F_X(2^-)=\dfrac{2}{3}-\dfrac{1}{2}=\dfrac{1}{6}.$$
Now I want check $f(x)$ is p.d.f or not p.d.f.
\begin{eqnarray}
\int\limits_{0}^{1} \dfrac{1}{2}x dx + \int\limits_{2}^{3} \dfrac{1}{3} dx &=&
\left[\dfrac{1}{4}x^2\right]_0^1 + \left[\dfrac{1}{3}x\right]_2^3\\
&=& \dfrac{1}{4}+1-\dfrac{2}{3}\\
&=& \dfrac{7}{12}
\end{eqnarray}
total probability of discrete and continuous:
$$\dfrac{7}{12}+\dfrac{1}{4}+\dfrac{1}{6}=\dfrac{14+6+4}{24}= 1.$$
So, we have p.d.f. for continous variable
$$f_X(x)=
\begin{cases}
\dfrac{1}{2}x&0\leq x< 1\\
\dfrac{1}{3}&2\leq x<3
\end{cases}$$
and for discrete random variable
$f(1)=\dfrac{1}{4}$ and
$f(2)=\dfrac{1}{6}.$
Is my answer correct?
Best Answer
As you have correctly observed, $F_X$ is not continuous and thus $X$ cannot have an (absolutely) continuous distribtion. So there is no (measurable) positive function $f$ such that
$$F_X(x) = \int_{-\infty}^x f(t) dt, x \in \mathbb{R}$$ and thus no 'density' exists.