Let $f_X(x) = \frac{1}{4}\cdot |X|$ a probability density function.
$-2<X<2$;
Find the Cumulative distribution function of $f_X(x)$.
What I thought was the answer is $F_X = \frac{1}2 + \frac{x}8$. Since a CDF is a function which tells us the probability of a $P(X<a)$, given certain.
Drawing the PDF generats a graph of a symetric function, which gives us a probability of a least $\frac{1}2$.
However I found out it is $F_x = P\left\{X\le x\right\}=\frac{1}2-\frac{1}2\cdot |x|\cdot |\frac{x}4|= \frac{1}2 - \frac{x^2}8$, for $-2 < x < 0$.
I can't understand why.
With the help above, I made it -
I will show the answer for the interval $-2 < x < 0$. The other one is symetric.
$$\int_{-2}^{x} \frac{1}{4}\cdot |t| dt = \int_{-2}^{x} \frac{1}{4}\cdot (-t) dt $$ (since t is negative). $$=\frac{1}4\cdot \int_{-2}^x (t) dt =- \frac{1}4\cdot \frac{t^2}2|_{-2}^{x}=-\frac{1}4\cdot \left( \frac{x^2}2\ - 2 \right) =\frac{1}2 - \frac{x^2}{8} $$.
So at the end we get -
$${F_x} = \left\{ {\begin{array}{*{20}{l}}0&{X < - 2}\\\begin{array}{l}\frac{1}{2} - \frac{{{x^2}}}{8}\\\frac{1}{2} + \frac{{{x^2}}}{8}\\1\end{array}&\begin{array}{l} - 2 < x < 0\\0 < x < 2\\x > 2\end{array}\end{array}} \right. % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbdfgBPj % MCPbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B % TfMBaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY-Hhbbf9v8 % qqaqFr0xc9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9 % q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaake % aaqaaaaaaaaaWdbiaadAeapaWaaSbaaSqaa8qacaWG4baapaqabaGc % peGaeyypa0Zaaiqaa8aabaqbaeaabiGaaaqaaiaaicdaaeaacaWGyb % GaeyipaWJaeyOeI0IaaGOmaaabaeqabaWaaSaaaeaacaaIXaaabaGa % aGOmaaaacqGHsisldaWcaaqaaiaadIhadaahaaWcbeqaaiaaikdaaa % aakeaacaaI4aaaaaqaamaalaaabaGaaGymaaqaaiaaikdaaaGaey4k % aSYaaSaaaeaacaWG4bWaaWbaaSqabeaacaaIYaaaaaGcbaGaaGioaa % aaaeaacaaIXaaaaqaabeqaaiabgkHiTiaaikdacqGH8aapcaWG4bGa % eyipaWJaaGimaaqaaiaaicdacqGH8aapcaWG4bGaeyipaWJaaGOmaa % qaaiaadIhacqGH+aGpcaaIYaaaaaaapeGaay5Eaaaaaa!5984! $$