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.

  • $\begingroup$ 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 $\endgroup$ – Fakemistake Nov 18 '18 at 18:49
  • $\begingroup$ @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 $0\ne F_X(t)$. $\endgroup$ – 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.

enter image description here

  • $\begingroup$ 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)$$. $\endgroup$ – J. Doe Nov 19 '18 at 15:26
  • 1
    $\begingroup$ @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. $\endgroup$ – callculus Nov 19 '18 at 17:12

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