Expected length of unit segments

Question

Suppose we draw $x, y, z$ from the uniform distribution on $(0, 5)$. How would we calculate the expected value of $|(x, x + 1) \cup (y, y + 1) \cup (z, z+1)|$?

Answer 1

Evan's answer to this question might be of some interest to you. I'll use a similar method to talk about the points covered by at least one segment. Let $p_1(t)$ be the probability that a point $t\in[0, 6]$ is covered by a particular segment. If the segment's start is placed at position $x$, $p_1(t)$ is the ratio of the length of the interval of $x$ values that result in $t$ being covered and the total interval from which $x$ is drawn. From this, we can obtain a piecewise expression for $p_1(t)$:

$$ p_1(t) = \begin{cases} \frac {t}{5} & 0 \le t < 1 \newline \frac {1}{5} & 1 \le t < 5 \newline \frac {6-t}{5} & 5 \le t \le 6 \newline \end{cases} $$

If we integrate $p_1(t)$ on $[0, 6]$, we obtain the expected size of the interval of points covered, which is of course $1$. To obtain the expected length covered by 3 segments, we need to obtain the probability that a point be covered by at least one of 3 such segments, chosen independently, which I'll call $q(t)$.

To do this, we can note that the probability of not being covered by a particular segment is $1-p_1(t)$ Since the probability of being covered be a particular segment is independent of being covered by another segment, the probability of not being covered by any of the 3 segments is simply $(1-p_1(t))^3$. Thus, the probability of being covered by at least one segment is the compliment, $q(t) = 1-(1-p_1(t))^3$. We can plug our piecewise expression for $p_1(t)$ into this relation to obtain $q(t)$.

$$ q(t) = \begin{cases} 1 - (1 - \frac t 5)^3 & 0 \le t < 1 \newline 1 - (1 - \frac 1 5)^3 & 1 \le t < 5 \newline 1 - (1 - \frac{6-t}5)^3 & 5 \le t \le 6 \newline \end{cases} = \begin{cases} \frac{1}{125} t^3 - \frac{3}{25} t^2 + \frac{3}{5} t & 0 \le t < 1 \newline \frac{61}{125} & 1 \le t < 5 \newline \frac{1}{125} (6-t)^3 - \frac{3}{25} (6-t)^2 + \frac{3}{5} (6-t) & 5 \le t \le 6 \newline \end{cases} $$

To obtain the expected length, all we need to do is evaluate the integral:

$$\int_0^6q(t) dt = \int_0^1q(t) dt + \int_1^5q(t) dt + \int_5^6q(t) dt$$

Notice that the first and last integrals are equivalent due to symmetry. After a bit of calculus, we obtain the expected length of $\frac{619}{250}$.