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On a uni-directional circular road where all vehicles travel in the same direction with varying speeds and overtaking is allowed on the entire road, what is the expected number of times (a) a particular vehicle will overtake others? (b) a particular vehicle will be overtaken by others?

In particular, what I want to find out: Is (a) = (b)? Thanks,

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    @Alvi: please do not use answers to make comments.2011-03-10

2 Answers 2

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Certainly not unless the speed of the vehicle is the mean speed of the traffic. I think in that case you can make a symmetry argument-go to a coordinate system that rotates at the mean speed.

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    +1 Because you sound a little tentative I provided the details ;-).2011-03-08
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Let the speeds be $s_1 \le s_2 \le \cdots \le s_n$ (in laps per unit time). Relative to vehicle $k$, vehicle $j$ moves at a rate of $s_j - s_k$. Assuming the vehicles are presently uniformly and independently distributed around the road, the expected number of overtakings in a period of time $t$ therefore is $t|s_j - s_k|$. Vehicles $1, 2, \ldots, k-1$ will be overtaken and vehicles $k+1, k+2, \ldots, n$ will overtake vehicle $k$. Therefore:

(a) expected number of times vehicle $k$ overtakes others = $t\sum_{j=1}^{k-1}(s_k-s_j)$;

(b) expected number of times vehicle $k$ is overtaken = $t\sum_{j=k+1}^{n}(s_j-s_k)$.

(Empty sums equal zero, of course.)

When, as Ross Millikan suggests, vehicle $k$ moves at the mean speed of traffic, the sums (a) and (b) must be equal, because $s_k = \frac{1}{n}\sum_{j=1}^{n}s_j$ is algebraically equivalent to (b) - (a) equalling zero. Clearly the implication works in the other direction, too.