[Math] Assuming the two people wait for each other, what is the expected waiting time

conditional-expectationexpectationprobabilityprobability distributionsuniform distribution

Two people agree to meet at a restaurant. Assume their arrival times are independent
and uniformly distributed on the one hour interval from 1:00–2:00 p.m. Assuming the two people wait for each other, what is the expected
waiting time?

$\textbf{Hint}$: Let X be the arrival time of one person and Y the arrival time of the other
person. Express the event of interest in terms of X and Y . S

I think the expected time of waiting should be 0 right? Since it's a uniform distribution, then EX = 30 and EY = 30. Therefore E(X-Y) = 0.

What do you guys think?

Best Answer

Because the joint distribution is so simple: $$f_{X,Y}(x,y) = 1, \quad 0 \le x, y \le 1,$$ it is easier to integrate directly: $$\operatorname{E}[|X-Y|] = \int_{x=0}^1 \int_{y=0}^1 |x-y| f_{X,Y}(x,y) \, dy \, dx = \int_{x=0}^1 \int_{y=0}^x x-y \, dy \, dx + \int_{x=0}^1 \int_{y=x}^1 y-x \, dy \, dx.$$

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[Math] Two people meeting, expected time of waiting

The random variable for $B$ has pdf $g(b)=1/2$ over the interval from $t=2$ to $t=3$, together with a point mass of $1/2$ at $t=2$. As you've pointed out, $A$ has pdf $f(a)=1$ on this interval. To compute the expected waiting time, it's convenient to break the calculation up into three disjoint pieces:

$B$ arrives at $t=2$
$B$ arrives later ($t>2$), but before $A$
$A$ arrives before $B$

The first case contributes $${\rm Pr}(B=2) \int_2^3 (a-2)f(a)\,da = 1/4$$ to the expectation. The second case contributes $$\int_2^3 f(a)\,da \int_2^a (a-b)g(b) \,db=1/12.$$ The third case also contributes $1/12$; I leave that integral to you. So the expected waiting time is $5/12$ hour, or $25$ minutes.

What are the expected values of the ordered arrival times

Some clues:

Standard formula for order statistics: Suppose $X_, X_2, \dots, X_n$ is a random sample of size $n$ from a continuous distribution with density $f(x),$ where $f(x) > 0$ for $a < x < b.$ Then the density function of the $k$th order statistic $Y$ is $$\frac{n!}{(k-1)!(n-k)!}[F(y)]^{k-1}[1-F(y)]^{n-k}f(y),$$ for $a < y < b,$ and $0$ otherwise.

This implies that order statistics of $h$ standard uniform random variables will be beta random variables, with means evenly spaced at $1/(h+1), \dots, h/(h+1).$ For example if there are three of your arrivals in an hour, they will occur on average at 15, 30, and 45 minutes into the hour.

Simulation in R for the smallest of three:

set.seed(2020)
y1 = replicate(10^6, min(runif(3)))
mean(y1)> set.seed(2020)
[1] 0.2501943  # aprx E(min of 3) = 1/4

hist(y1, prob=T, br=50, col="skyblue2")
 curve(dbeta(x, 1,3),add=T, col="red", lwd=2)

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[Math] Two people meeting, expected time of waiting

What are the expected values of the ordered arrival times

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