TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Levajkovic)/Übungen 2023W/HW02.3

The Cauchy cummulative distribution function

Let

$F(y)={\frac {1}{2}}+{\frac {1}{\pi }}\arctan y,\ y\in \mathbb {R}$.

(a) Show that $F$ is a cumulative distribution function of a certain random variable $Y$.

(b) Find the density function and use R to sketch its graph.

(c) Find $a\in \mathbb {R}$ such that $P(Y>a)=0.1$.

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

a

Let $F(y)={\frac {1}{2}}+{\frac {1}{\pi }}\arctan y,\ y\in \mathbb {R}$ be a cumulative distribution function

1) $0\leq F(y)\leq 1,\forall y\in \mathbb {R}$: Given that $arctan:\mathbb {R} \rightarrow ({\frac {-\pi }{2}},{\frac {\pi }{2}})$,

$0<{\frac {1}{2}}-{\frac {\pi }{2\pi }}<F(y)<{\frac {1}{2}}+{\frac {\pi }{2\pi }}<1$

2) F is monotonically increasing: ${\frac {d}{dy}}tan^{-1}y={\frac {1}{1+y^{2}}}>0{\text{ for all }}y\in \mathbb {R}$ implies arctan is monotonically increasing. Therefore F is also monotonically increasing

3) $\lim _{x\rightarrow -\infty }F(x)=0{\text{ and }}\lim _{x\rightarrow \infty }F(x)=1$

$\lim _{y\rightarrow -\infty }{\frac {1}{2}}+{\frac {1}{\pi }}\arctan y=\lim _{y\rightarrow -\infty }{\frac {1}{2}}+\lim _{y\rightarrow -\infty }{\frac {1}{\pi }}*\lim _{x\rightarrow -\infty }\arctan y={\frac {1}{2}}+{\frac {1}{\pi }}*({\frac {-\pi }{2}})=0$

$\lim _{x\rightarrow \infty }F(x)=1$ is shown similarly.

4) F is right continuous

$\lim _{h\searrow 0}F(y+h)=F(h)\iff \lim _{h\searrow 0}\arctan(y+h)=arctan(y)$

Since $\arctan$ is coninous over the entirity of $\mathbb {R}$ it is also right-continuous for all $x\in \mathbb {R}$

b

The cdf of a continuous random variable is defined as the integral over the pdf.

${\begin{aligned}F(y)&=\int _{-\infty }^{y}f_{Y}(t)dt\\F'(y)&=f_{Y}(y)\\{\frac {1}{\pi }}{\frac {1}{1+y^{2}}}&=f_{Y}(y)\end{aligned}}$

fY <- function(x) {
  return(1 / (pi * (1 + x^2)))
}

curve(fY(x), from=-10, to=10, col = "red", xlab = "x-axis", ylab = "y-axis", main = "Plot of fY")

c

${\begin{aligned}0.1&=P(Y>a)=1-P(X\leq a)=1-F(a)=1-{\frac {1}{2}}-{\frac {1}{\pi }}\arctan {a}\\{\frac {1}{2}}-{\frac {1}{10}}&={\frac {1}{\pi }}\arctan {a}\\{\frac {4\pi }{10}}&=\arctan a\\a&=\tan({\frac {4\pi }{10}})\approx 3.078\end{aligned}}$

F <- function(y) {
  return(1 / 2 + atan(y) / pi)
}

a <- tan(4 * pi / 10)

1 - F(a)