Determine the expected time until the first arrival.

Question

Two people agree to meet at a specified place between 3:00 P.M. and 4:00 P.M. Suppose that you measure time to the nearest minute relative to 3:00 P.M. so that, for instance, time 40 represents 3:40 P.M. Further suppose that each person arrives according to the discrete uniform distribution on {0, 1, ..., 60} and that the two arrival times are independent. Determine the expected time until the first arrival.

Here I've tried setting : $X_i$=person i's arrival time with i=1,2

$$P_X(x_i)= \begin{cases} 1/61, & \text{if $x_i$ $\in[0,1,2,3...60]$} \\ 0, & \text{otherwise} \end{cases} $$

I create a function Z=min{x,y}

$$E(z)=\sum_{z=0}^{60}zP_Z(z)=\sum_{z=0}^{60}z[P(x_1=z,x_2>z)+P(x_1>z,x_2=z)+P(x_1=z,x_2=z)]$$ And that is where I'm kind of stuck because I've tried rewrittin $P(x_1=z,x_2>z)$ as $P(x_2>z)P(x_1=z|x_2>z)$ and drawing a little pmf table but still no luck.

If anybody by any chance knows a much easier way to solve this problem I would be happy to hear from them.

quasi · Accepted Answer · 2018-04-12T09:52:59.613

2

A reusable trick . . .

Since the values of $z$ are nonnegative integers, it follows that $$E(z) = p_1 + p_2 + p_3 + \cdots$$ where $p_k = P(z \ge k)$.

Hence we get $$E(z) = \sum_{k=1}^{60} \left(\frac{61-k}{61}\right)^2 = \frac{1210}{61}\approx 19.84$$

To explain the trick . . .

For each positive integer $k$, let $q_k=P(z=k)$.

Then we have \begin{align*} p_1 &= q_1 + q_2 + q_3 +\cdots\\[4pt] p_2 &= \phantom{q_1 +\; }q_2 + q_3 +\cdots\\[4pt] p_3 &= \phantom{q_1 + q_2 + \;}q_3 + \cdots\\[4pt] \vdots\\[4pt] \end{align*} hence, summing the columns, we get $$ p_1 + p_2 + p_3 + \cdots = 1q_1 + 2q_2 + 3q_3 + \cdots \qquad\qquad\;\;\;\;\; $$

edited Apr 12 '18 at 09:52

answered Apr 12 '18 at 09:26

quasi

58,772

1

I like this trick, but if you haven't seen it before it could be a little cryptic. Perhaps a short explanation or reference would be useful? This might do... – Theoretical Economist Apr 12 '18 at 09:35
@Theoretical Economist: I edited in an explanation of the trick. – quasi Apr 12 '18 at 09:54

score 1 · Answer 2 · answered Apr 12 '18 at 09:26

I think that I've found a way around it. $P(x_2>z|x_1=z)$=$ 60-z\over 61$

So I'm gonna get: (With the assumption that someone can arrive at time 0) $$E(z)=\sum_{z=0}^{60}zP_Z(z)=\sum_{z=1}^{60}z[P(x_1=z,x_2>z)+P(x_1>z,x_2=z)+P(x_1=z,x_2=z)]$$

$$=\sum_{z=1}^{60}z\frac{2(60-z)}{61*61}+\sum_{z=1}^{60}\frac{z}{61}\frac{1}{61}$$ The rest would be to compute that.

And my answer to my own question would be $E(z)=19.8360$

Determine the expected time until the first arrival.

2 Answers2