# TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Bura)/Übungen 2019W/2.3

Density and distribution function

Let X be a continuous random variable with the density

${\displaystyle f_{x}(x)={\begin{cases}{\frac {x^{2}}{a}},&x\in (0,3)\\0,&{\text{otherwise}}\end{cases}}}$

where a is a nonzero constant.

(a) Determine the value of a.
(b) Find the cumulative distribution function of the random variable X.
(c) Compute the expectation of ${\displaystyle Z=2X^{3}+5}$.

## Lösungsvorschlag von Gittenburg

--Gittenburg 18:39, 21. Okt. 2019 (CEST)

(a)

Von einer Dichtefunktion wissen wir, dass ${\displaystyle \int _{-\infty }^{+\infty }f(x)dx=1}$.

${\displaystyle \int _{0}^{3}{\frac {x^{2}}{a}}\ dx={\frac {1}{a}}({\frac {x^{3}}{3}}){\Big |}_{0}^{3}={\frac {1}{a}}{\frac {3^{3}}{3}}={\frac {3^{2}}{a}}={\frac {9}{a}}=1\implies a=9}$.

(b)

${\displaystyle F_{X}={\begin{cases}0&x\leq 0\\\int _{0}^{x}{\frac {x^{2}}{a}}\ dx&x\in (0,3)\\1&x\geq 3\end{cases}}}$

(c)

{\displaystyle {\begin{aligned}E(Z)&=E(2X^{3}+5)\\&=2E(X^{3})+5\\&=2\int _{-\infty }^{+\infty }x^{3}\cdot f(x)dx+5\\&=2\int _{0}^{3}x^{3}\cdot {\frac {x^{2}}{9}}dx+5\\&={\frac {2}{9}}\int _{0}^{3}x^{5}dx+5\\&={\frac {2}{9}}({\frac {x^{6}}{6}}){\Big |}_{0}^{3}+5\\&={\frac {2}{9}}({\frac {3^{6}}{6}})+5=32\\\end{aligned}}}