Sorry if this is to simple; I have not done any Math in a while.

The scenario is , we have two participants/Runners: A,B both headed

randomly towards the same place P, and staying in P for a random interval

then heading out. We want to know the probability of

B arriving right after A leaves.

So we normalize meeting at $[0,1]$ and A arrives at $a_1$ and stays

until $a_2$; and similar pair $(b_1, b_2)$ for B; $0 \geq a_1,a_2,b_1,b_2 \leq 1 $.

So we want to compute the probability that $a_2=b_1$

Working on $[0,1] \times [0,1] $. The probability of the event $a_2=b_1$

is the measure of all rectangles with base $(a_1,a_2)$ and side $(a_2,1)$,

with total area $(a_2- a_1)(1-a_2)$. WE can compute the total area by the

double integral:

$\int_0^1 (a_2-a_1)da_1 \int_0^1 (1-a_2)da_2 $

Getting 1/12 , which seems somewhat right intuitively. But I have been told

(non-rigorously) that the actual answer is 0. Can someone check and

criticque my work, please?

  • $\begingroup$ It seems that A and B are continuous random variables. Then $P(A-B=c)=0$. The probabilities are positive only for inequalities. $\endgroup$ – callculus Oct 11 '18 at 17:29
  • $\begingroup$ But what is wrong with my answer then? $\endgroup$ – gary Oct 11 '18 at 17:32
  • $\begingroup$ @callculus: I am not saying I am not wrong, I would just like to know what false assumption I am making in my post. $\endgroup$ – gary Oct 11 '18 at 17:52

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