Inverse Uniform Distribution | ISI MStat 2007 PSB Problem 4

This problem is an interesting application of the inverse uniform distribution family, which has infinite mean. This problem is from ISI MStat 2007. The problem is verified by simulation.

Problem

The unit interval (0,1) is divided into two sub-intervals by picking a point at random from inside the interval. Denoting by $Y$ and $Z$ the
lengths of the long and the shorter sub-intervals respectively show that $\frac{Y}{Z}$ does not have a finite expectation.

This is the 4th Problem ISI MStat 2008. Enjoy it.

Prerequisites

Distribution Function
Distribution Function of $\frac{1}{X}$ in terms of the Distribution Function of $X$ .

Solution

$\frac{Y}{Z} + 1 = \frac{Y+Z}{Z} = \frac{1}{Z}$ , where $Z$ is the shorter length of the broken stick.

So, $E( \frac{Y}{Z}) = E(\frac{1}{Z}) - 1$ .

Let's try to find the distribution of $\frac{1}{Z}$ .

Let $U$ ~ Unif $(0,1)$ whcih denotes the random uniform cut.

Number line - Inverse Uniform Distribution

The shorter stick of length smaller than $x$ can be achieved if the stick is cut either before $x$ or it is cut after $1-x$ .

Observe that $P( Z \leq x) = P ( U \leq x ) + P ( U \geq 1 - x) = x + 1 - (1-x) = 2x$ . This answer is natural since, the total valid length is $2x$ .

$P( \frac{1}{Z} \leq z) = P ( Z \geq \frac{1}{z} ) = 1 - \frac{2}{z} \Rightarrow F_{\frac{1}{Z}}(z) = 1 - \frac{2}{z}$ if $2 \leq z < \infty$ .

Therefore, $f_{\frac{1}{Z}}(z) = \frac{2}{z^2}$ if $2 \leq z < \infty$ .

Hence, $E( \frac{Y}{Z}) = E(\frac{1}{Z}) - 1 = (\int_{2}^{\infty} \frac{2}{z} dz) - 1 = \infty$

Simulation and Verification

Exercise: Prove that $F_{\frac{Y}{Z}}(x) = \frac{(x-1)}{(x+1)}$ if $1 \leq x < \infty$ .

 u = runif(1000,0,1)
  w = 1 - u
  Z = pmin(u,w)
  Y = pmax(u,w)
  YbyZ = Y/Z
plot(ecdf(YbyZ), xlim = c(0,50))
x = seq(0, 50, 0.01)
curve((x - 1)/(x+1), from = 0, col = "red", add = TRUE)

Graph - Inverse Uniform Distribution — Red = Actual Distribution Function, Black = Simulated ECDF

The Mean moves really slowly to infinity ~ logx. Hence it is really hard to show it is going to $\infty$ . Also, the probability of occurrence of high value is almost 0. Hence, it really hard to show my simulation that the mean is $\infty$ . But, we can show that the mean of the maximum values is really large.

v = rep(0,200)
m = NULL
for ( i in 1:200)
{
  #v[i] = 100*i
  u = runif(10000,0,1)
  w = 1 - u
  Z = pmin(u,w)
  Y = pmax(u,w)
  YbyZ = Y/Z
  m = c(m, max(YbyZ))
}
mean(m) = 79079.43

Beware of the simulation, it can be totally counterintuitive. This is really enjoyable though.

Stay Tuned!

This problem is an interesting application of the inverse uniform distribution family, which has infinite mean. This problem is from ISI MStat 2007. The problem is verified by simulation.

Problem

The unit interval (0,1) is divided into two sub-intervals by picking a point at random from inside the interval. Denoting by $Y$ and $Z$ the
lengths of the long and the shorter sub-intervals respectively show that $\frac{Y}{Z}$ does not have a finite expectation.

This is the 4th Problem ISI MStat 2008. Enjoy it.

Prerequisites

Distribution Function
Distribution Function of $\frac{1}{X}$ in terms of the Distribution Function of $X$ .

Solution

$\frac{Y}{Z} + 1 = \frac{Y+Z}{Z} = \frac{1}{Z}$ , where $Z$ is the shorter length of the broken stick.

So, $E( \frac{Y}{Z}) = E(\frac{1}{Z}) - 1$ .

Let's try to find the distribution of $\frac{1}{Z}$ .

Let $U$ ~ Unif $(0,1)$ whcih denotes the random uniform cut.

Number line - Inverse Uniform Distribution

The shorter stick of length smaller than $x$ can be achieved if the stick is cut either before $x$ or it is cut after $1-x$ .

Observe that $P( Z \leq x) = P ( U \leq x ) + P ( U \geq 1 - x) = x + 1 - (1-x) = 2x$ . This answer is natural since, the total valid length is $2x$ .

$P( \frac{1}{Z} \leq z) = P ( Z \geq \frac{1}{z} ) = 1 - \frac{2}{z} \Rightarrow F_{\frac{1}{Z}}(z) = 1 - \frac{2}{z}$ if $2 \leq z < \infty$ .

Therefore, $f_{\frac{1}{Z}}(z) = \frac{2}{z^2}$ if $2 \leq z < \infty$ .

Hence, $E( \frac{Y}{Z}) = E(\frac{1}{Z}) - 1 = (\int_{2}^{\infty} \frac{2}{z} dz) - 1 = \infty$

Simulation and Verification

Exercise: Prove that $F_{\frac{Y}{Z}}(x) = \frac{(x-1)}{(x+1)}$ if $1 \leq x < \infty$ .

 u = runif(1000,0,1)
  w = 1 - u
  Z = pmin(u,w)
  Y = pmax(u,w)
  YbyZ = Y/Z
plot(ecdf(YbyZ), xlim = c(0,50))
x = seq(0, 50, 0.01)
curve((x - 1)/(x+1), from = 0, col = "red", add = TRUE)

Graph - Inverse Uniform Distribution — Red = Actual Distribution Function, Black = Simulated ECDF

The Mean moves really slowly to infinity ~ logx. Hence it is really hard to show it is going to $\infty$ . Also, the probability of occurrence of high value is almost 0. Hence, it really hard to show my simulation that the mean is $\infty$ . But, we can show that the mean of the maximum values is really large.

v = rep(0,200)
m = NULL
for ( i in 1:200)
{
  #v[i] = 100*i
  u = runif(10000,0,1)
  w = 1 - u
  Z = pmin(u,w)
  Y = pmax(u,w)
  YbyZ = Y/Z
  m = c(m, max(YbyZ))
}
mean(m) = 79079.43

Beware of the simulation, it can be totally counterintuitive. This is really enjoyable though.

Stay Tuned!

Leave a Reply Cancel reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

HALL OF FAME BLOG OFFLINE CLASS

CHEENTA ON DEMAND BOSE OLYMPIAD

Cheenta Academy Private Limited |

support@cheenta.com

Menu

ISI Entrance Solutions

ISI CMI Program

ISI Entrance Solutions

ISI CMI Self Paced