Finding density function for a given distribution

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.

• It simply reduces to $F_X(t)=P(X\leq t)=\int_{-\infty}^t f_X(x)\mathrm{d}x$ and we interpret this integral as a sum, when the distribution is discrete – Fakemistake Nov 18 '18 at 18:49
• @Fakemistake But if one changes a finite number of points in a function then the integral stays the same, and I can redefine $f_X(0)=f_X(1)=f_X(2)=0$ and get that the integral equals $0\ne F_X(t)$. – J. Doe Nov 18 '18 at 18:58

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.

• Thanks! What I don't understand is the fact that if I change a finite number of points in any finction $f$ (and let $f'$ be the new function after I changed the points) then $\int f(x) dx= \int f'(x) dx$. In our case, I can redefine the points $f_X(0):=f_X(1):=f_X(2):=0$, thus, the redefined function $f_X\equiv 0$, thus $\int_{-\infty}^{t} f_X(s)ds=0\ne F_X(t)$ which doesn't settle with the fact that $\int_{-\infty}^{t} f_X(s)ds=F_X(t)$$. – J. Doe Nov 19 '18 at 15:26 • @J.Doe If$\int_{-\infty}^{t} f_X(s)ds=0\ \forall \ \ t\in \mathbb R$then your redefined pdf$f_X(s)$is not valid. One property for a valid pdf is$\int_{-\infty}^{\infty} f_X(s)ds=1\$. Thus there is no surprise that an ill-defined function does not fulfill other necessary properties. – callculus Nov 19 '18 at 17:12