Let $X$ be a random variable with distribution function $$ \\ F_X(t)=\begin{cases} 0, & t<0 \\ 2/11, & 0\leqslant t<1 \\7/11, & 1 \leqslant t<2 \\1, & 2 \leqslant t \end{cases} \ $$ Find the density function of $X$.
I know that the density is $$ \\ f_X(t)=\begin{cases} 0, &t\notin\{0,1,2\} \\2/11, &t=0 \\ 5/11, &t=1 \\ 4/11, & t=2 \end{cases} \ $$
but I don't know how or why this solution is true, hope for an explanation.
I have made a sketch of the cdf. The distance between the circles and the red bullets show the jump of the cdf from $t-\epsilon$ to $t$ where $t=0,1,2$ and $\epsilon\to 0$.
At a cdf the horicontal lines shows that the pdf is $0$ at that interval. I hope the sketch make it easier to understand the connection of the cdf and the pdf here. Feel free to ask, if something is still not comprehensible.