TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Levajkovic)/Übungen 2023W/HW02.4

Transformations

Let

$f_{X}(x)={\begin{cases}{\frac {1}{2}}(x+1),\ x\in (-1,1),\\0,\ {\text{otherwise}}\end{cases}}$

be the probability density function of a random variable $X$. Find the cumulative distribution function of

$Y={\frac {9}{(X-2)^{2}}}$.

Dieses Beispiel ist als solved markiert. Ist dies falsch oder ungenau? Aktualisiere den Lösungsstatus (Details: Vorlage:Beispiel)

Lösungsvorschlag von Lessi

--Lessi 2024-02-07T13:04:11Z

First we determine the cdf of $X$:

${\begin{aligned}F(x)&=\int _{-\infty }^{x}{\frac {1}{2}}(t+1)dt=\int _{-1}^{x}{\frac {1}{2}}(t+1)dt={\frac {t^{2}}{4}}+{\frac {t}{2}}{\Big |}_{-1}^{x}\\&={\frac {x^{2}+2x+1}{4}}={\frac {(x+1)^{2}}{4}}\end{aligned}}$

So the cdf is $F_{X}(x)={\begin{cases}0,&{\text{for }}x\leq -1,\\{\frac {(x+1)^{2}}{4}},&{\text{for }}-1<x<1,\\0,&{\text{for }}x\geq 1\end{cases}}$

Now we can find the cdf of $Y={\frac {9}{(X-2)^{2}}}$:

${\begin{aligned}F_{Y}(y)=P(Y\leq y)&=P{\Big (}{\frac {9}{(X-2)^{2}}}\leq y{\Big )}\\&=P{\Big (}-{\sqrt {\frac {9}{y}}}\leq X-2\leq {\sqrt {\frac {9}{y}}}{\Big )}\\&=P{\Big (}2-{\frac {3}{\sqrt {y}}}\leq X\leq 2+{\frac {3}{\sqrt {y}}}{\Big )}\\&=F{\Big (}2+{\frac {3}{\sqrt {y}}}{\Big )}-F{\Big (}2-{\frac {3}{\sqrt {y}}}{\Big )}\end{aligned}}$

Since for any $y>0,\ 2+{\frac {3}{\sqrt {y}}}>1$ we get $F{\Big (}2+{\frac {3}{\sqrt {y}}}{\Big )}=1$. For $F{\Big (}2-{\frac {3}{\sqrt {y}}}{\Big )}$ we get ${\frac {9}{4}}(1-{\frac {1}{\sqrt {y}}})^{2}$, for $1<y<9$ so finally

$F_{Y}(y)={\begin{cases}1,&{\text{for }}1\leq y,\\1-{\frac {9}{4}}(1-{\frac {1}{\sqrt {y}}})^{2},&{\text{for }}1<y<9,\\0,&{\text{for }}y\geq 9\end{cases}}$