Die Dichtefunktion einer Zufallsgröße X sei gegeben durch
Berechnen Sie die Quantile xq{\displaystyle x_{q}} der Ordnung q=0.25{\displaystyle q=0.25}, 0.5{\displaystyle 0.5} (Median) bzw. 0.75{\displaystyle 0.75}.
f(x)={0fuer x<−11+xfuer −1≤x≤01−xfuer 0<x<10fuer x≥1{\displaystyle f(x)={\begin{cases}0&{\text{fuer }}x<-1\\1+x&{\text{fuer }}-1\leq x\leq 0\\1-x&{\text{fuer }}0<x<1\\0&{\text{fuer }}x\geq 1\end{cases}}}
∫−1x1+xdx=x+x22|−1x=x+x22+12{\displaystyle \int _{-1}^{x}1+x\,dx=\left.x+{\frac {x^{2}}{2}}\right|_{-1}^{x}=x+{\frac {x^{2}}{2}}+{\frac {1}{2}}}
∫−101+xdx=12{\displaystyle \int _{-1}^{0}1+x\,dx={\frac {1}{2}}}
∫0x1−xdx=x−x22|0x=x−x22{\displaystyle \int _{0}^{x}1-x\,dx=\left.x-{\frac {x^{2}}{2}}\right|_{0}^{x}=x-{\frac {x^{2}}{2}}}
F(x)={0fuer x<−1x+x22+12fuer −1≤x≤0x−x22+12fuer 0<x<11fuer x≥1{\displaystyle F(x)={\begin{cases}0&{\text{fuer }}x<-1\\x+{\frac {x^{2}}{2}}+{\frac {1}{2}}&{\text{fuer }}-1\leq x\leq 0\\x-{\frac {x^{2}}{2}}+{\frac {1}{2}}&{\text{fuer }}0<x<1\\1&{\text{fuer }}x\geq 1\end{cases}}}
F(x)=0.25x+x22+12=0.25x2+2x+12=0x=−1+1−12=−1+12{\displaystyle {\begin{aligned}F(x)&=0.25\\x+{\frac {x^{2}}{2}}+{\frac {1}{2}}&=0.25\\x^{2}+2x+{\frac {1}{2}}&=0\\x&=-1+{\sqrt {1-{\frac {1}{2}}}}=-1+{\frac {1}{\sqrt {2}}}\end{aligned}}}
F(x)=0.5x+x22+12=0.5x2+2x+0=0x=−1+1=0{\displaystyle {\begin{aligned}F(x)&=0.5\\x+{\frac {x^{2}}{2}}+{\frac {1}{2}}&=0.5\\x^{2}+2x+0&=0\\x&=-1+{\sqrt {1}}=0\end{aligned}}}
F(x)=0.75x−x22+12=0.75x2−2x+12=0x=1−1−12=1−12{\displaystyle {\begin{aligned}F(x)&=0.75\\x-{\frac {x^{2}}{2}}+{\frac {1}{2}}&=0.75\\x^{2}-2x+{\frac {1}{2}}&=0\\x&=1-{\sqrt {1-{\frac {1}{2}}}}=1-{\frac {1}{\sqrt {2}}}\end{aligned}}}
Beispiel 92
