URBAN OPERATIONS RESEARCH PROFESSOR ARNOLD I BARNETT LECTURE NOTE OF 9/24/2001 PREPARED BY JAMES S. KANG Crofton's Method Let XI and X2 be independent random variables that are uniformly distributed over the interval [0,a]. We are interested in computing EllX1-X2l. For instance, in an urban setting, X1 and X2 may denote the location of an accident and the location where an emergency vehicle is currently parked in a road segment of length a, respectively. In this case, we want to know the distance(or the travel time) on average between the two locations, EllX1-X2l. We may solve this question using a joint distribution of X1 and X2, but Crofton's method is quite useful for the question Let G(a)=ElX1-X2l. Now consider the following question: If the interval were [ 0, a+e where e is small, what would G(a+e) be? Table 1 summarizes G(a+a) depending on the locations of X1 and X Table 1: G(a+e) Probability of a case G(a+e)given a case a=(a a<X1≤a+E,0≤X2≤a (a+e) a 0≤X1≤ X2≤a+E <X1<a+E,a<X2≤a+E We do not care Figure 1: Case where0≤X1< a and a<X2≤a+e Note that we did not specify G(a +a) for the case where a< Xi <a+E and a< X2 <a+E, because the probability of that case,(_2)2, is negligible when e is small ("If e is small, e2 To compute G(a+a) from Table 1, we invoke the total expectation theorem(or the conditional xpectation rule). When a sample space is divided into A1, A2, .. An that are mutually exclusive and collectively exhaustive events, the expected value of a random variable Z is computed byUrban Operations Research Professor Arnold I. Barnett Lecture Note of 9/24/2001 Prepared by James S. Kang Crofton’s Method Let X1 and X2 be independent random variables that are uniformly distributed over the interval [0, a]. We are interested in computing E[|X1 − X2|]. For instance, in an urban setting, X1 and X2 may denote the location of an accident and the location where an emergency vehicle is currently parked in a road segment of length a, respectively. In this case, we want to know the distance (or the travel time) on average between the two locations, E[|X1 − X2|]. We may solve this question using a joint distribution of X1 and X2, but Crofton’s method is quite useful for the question. Let G(a) ≡ E[|X1 − X2|]. Now consider the following question: If the interval were [0, a + ε] where ε is small, what would G(a+ε) be? Table 1 summarizes G(a+ε) depending on the locations of X1 and X2. Table 1: G(a + ε) Case Probability of a case G(a + ε) given a case 0 ≤ X1 ≤ a, 0 ≤ X2 ≤ a a a+ε · a a+ε = ( a a+ε )2 G(a) a<X1 ≤ a + ε, 0 ≤ X2 ≤ a ε a+ε · a a+ε = εa (a+ε)2 a + ε 2 − a 2 = a+ε 2 0 ≤ X1 ≤ a, a<X2 ≤ a + ε a a+ε · ε a+ε = εa (a+ε)2 a + ε 2 − a 2 = a+ε 2 a<X1 ≤ a + ε, a<X2 ≤ a + ε ε a+ε · ε a+ε = ( ε a+ε )2 We do not care. 0 a a+ε X1 X2 Figure 1: Case where 0 ≤ X1 ≤ a and a<X2 ≤ a + ε Note that we did not specify G(a + ε) for the case where a<X1 ≤ a + ε and a<X2 ≤ a + ε, because the probability of that case, ( ε a+ε )2, is negligible when ε is small (“If ε is small, ε2 is pathetic.”). To compute G(a+ ε) from Table 1, we invoke the total expectation theorem (or the conditional expectation rule). When a sample space is divided into A1, A2,... ,An that are mutually exclusive and collectively exhaustive events, the expected value of a random variable Z is computed by 1