TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Levajkovic)/Übungen 2023W/HW03.3

Uniform (0, 1)-distribution and its related distributions

Let $X$ be a Uniform(0,1) random variable.

(a) Find the cumulative distribution function of $V=e^{X}$ and its expectation.

(b) Show that $W=-2\ln {X}$ is $exp({\frac {1}{2}})$-distributed.

(c) Let $F$ be the cumulative distribution function of a continuous random variable. Find the cumulative distribution function of $Y=F^{-1}(X)$.

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Lösungsvorschlag von Lessi

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

Let $X$ be a Uniform(0, 1) random variable, e.g. $X\sim {\mathcal {U}}(0,1)$

a)

Let $V=e^{X}$ be a random variable. To determine its cdf we first define the pdf and cdf of X:

${\begin{aligned}f_{X}(x)&={\begin{cases}1,&x\in (0,1)\\0,&{\text{otherwise}}\end{cases}}\\F_{X}(x)&={\begin{cases}0,&{\text{for }}x\leq 0,\\x,&{\text{for }}0<x<1,\\1&{\text{for }}x\geq 1\end{cases}}\end{aligned}}$

Now we can determine the cdf of $V$:

$F_{V}(v)=P(e^{X}\leq v)=F_{X}(\ln v)={\begin{cases}0,&{\text{for }}v\leq 1,\\\ln v,&{\text{for }}1<v<e,\\1,&{\text{for }}v\geq e\end{cases}}$

To get the expected value we first determine the pdf of $V$ by taking the derivative of $F_{V}$: $f_{V}(v)={\frac {1}{v}},\ {\text{for }}v\in (1,e)$ So finally the expected value is computed $\mathbb {E} (V)=\int _{1}^{e}v*{\frac {1}{v}}dv=e-1$

b)

Let $W=-2\ln X$ be a random variable. Is $W\sim exp({\frac {1}{2}})$? We want to show that

$f_{W}(w)={\begin{cases}{\frac {1}{2}}e^{-{\frac {1}{2}}w},&{\text{for }}w\geq 0,\\0,&{\text{for }}x<0\end{cases}}$

So again we transform cdf of $X$ to get

$F_{W}(w)=1-F_{X}(e^{-{\frac {1}{2}}w})={\begin{cases}1-e^{-{\frac {1}{2}}w},&{\text{for }}x\geq 0,\\1,&{\text{for }}x<0\end{cases}}$

If we now integrate $F_{X}$ we have shown that $W\sim exp({\frac {1}{2}})$.

c)

Let F be a cumulative distribution function of a continuous random variable, and $Y=F^{-1}(X)$. Again we use $F_{X}$ as in any other transformation:

$F_{Y}(y)=P(Y\leq y)=P(F^{-1}(X)\leq y)=P(X\leq F(y))=F_{X}(F(y))$

Now since $F$ is a cdf (e.g. $F:\mathbb {R} \mapsto [0,1]$), take it as a argument of $F_{X}$ without any distinctions of its range, which means that in turn $F_{Y}(y)=F_{X}(F(y))=F(y)$.