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# 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.

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)

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!