# Show that Y/Z does not have finite expectation

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 longer and the shorter sub-intervals respectively, show that Y/Z does not have finite expectation.

Let X~U(0,1) and U be the length of the shorter of the intervals (0,X) and (X,1); that is Z=$min$(X,1-X) and let Y=1-U.So,Y and Z be the lengths of the longer and the shorter sub-intervals respectively

Then to find Expectation I need to find P.D.F of Y/Z but I can't do that. Please Help.

• You've asked quite a few homework questions the past few days, @Ranabir. This is perfectly fine of course, but perhaps not advisable: if you actually want to learn statistics then it's often better to solve the problems on your own (this includes looking in lots of books)... Aug 4, 2012 at 8:04
• @MånsT I am a student of statistics honours(3rd year).I faced problem to solve some problem so I ask your help. I am not putting every homework question of mine. Aug 4, 2012 at 15:21
• See the related question: stats.stackexchange.com/questions/299722/… Oct 24, 2019 at 10:05

We can write $Y$ and $Z$ as functions of $X \sim \mathcal U (0, 1)$, i.e. $Y := Y(X)$ and $Z := Z(X)$. To show the expectation is infinite, all we have to do is show $\int_0 ^ 1 \frac{Y(x)}{Z(x)} \ dx = \infty$ (by the Law of the Unconscious Statistician). As you noted, we can write $Z = \min(X, 1 - X)$ and $Y = \max(X, 1 - X)$. But $$\int_0 ^ {1/2} \frac{Y(x)}{Z(x)} \ dx = \int_0 ^ {1/2} \frac{\max(x, 1 - x)}{\min(x, 1 -x)} \ dx = \int_0 ^ {1/2} \frac{1 - x}{x} \ dx = \infty$$ One gets the same thing on the interval $[1/2, 1]$ as well after a substitution, so the integral on $[0, 1]$ is $\infty$.
EDIT: Okay, we can get the pdf as well. Let $T(X) = \frac{\max(X, 1 - X)}{\min(X, 1 - X)}$. We calculate $P(T(X) \le t)$ for some $t$. If $t < 1$ then trivially we get $0$, otherwise $$P(T(X) \le t) = P\left(\frac{1 - X}{X} \le t, X \in (0, 1/2) \right) + P\left( \frac{X}{1 - X} \le t, X \in (1/2, 1)\right).$$ Careful reflection reveals that the two probabilities on the RHS are actually the same, so after some manipulation we get $$P(T(X) \le t) = 2 P\left(X \ge \frac 1 {t + 1}, X \in (0, 1/2)\right) =2 \int_{1 / (t + 1)} ^ {1/2} \ dx = 1 - \frac 2 {t + 1}.$$ Differentiating to get the pdf gives $$f(t) = \frac 2 {(t + 1) ^ 2}$$ for $t \ge 1$ and $f(t) = 0$ otherwise.
As a sanity check, $\int_1 ^ \infty \frac{2t}{(t + 1)^2} \ dt = \infty$ as expected.