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Feb 19th, 2006 10:19
Knud van Eeden,
  Knud van Eeden  25 December 2004  10:53 pm  Math: Probability: Distribution: Can you derive the Erlang distribution C^k.e^(C.T).T^(k1)/(k1)!?  Suppose that the oocurrence of a certain type of events in time follow a Poisson distribution. In other words, if k is the total amount of events (e.g. the occurrence of an agent getting 5 telephone calls within an hour), that you can calculate the probability of it by using the Poisson distribution formula: k L L . e (1) =  k! Now you can write L as follows: Given is (see the description of the Poisson distribution formula) for the average rate L: L = N . p (2) Let dt = T / N (3) (where t is your total (time) interval, which you divide in N small subinterval, thus each subinterval dt has a length T divided by N, thus T/N). From (3) follows you can find N by dividing your total interval by the length of the subinterval: N = T / dt (4) Given is that the probability p is proportional with the subinterval length, so you get. E.g. if you have an average rate per hour of 60 calls, with a certain probability, then you have a proportionally smaller probability if you look for the probability in an interval 60 times smaller, thus in an interval of 1 minute. p = constant . dt or thus abbreviating constant to C gives p = C . dt Putting (4) and (5) in (2) gives: L = (T / dt) . (C . dt) (6) worked out (elimination again the subinterval dt) gives L = C . T (7) Putting (7) in (1) gives the Poisson distribution formula expressed in the total time interval length instead. k (C . T) (C . T) . e (8) =  k! For example if you have 60 calls per hour, than is your total time interval T equal to 1 hour. The constant C will then be equal to 60.   Now the Erlang distribution follows from for example the question, how much time is passed until (but not included) the agent is getting his 5th call?  The method used to find this, is to first use the cumulative Poisson probability function, and then to find the corresponding probability function via the derivative of it. That will lead to the Erlang formula.  OK, introduce a variable t which presents this (time) interval until the agent is getting his 5th call, or thus in general his kth call.  Now before he is getting his 5th call, he is getting his 0th call, 1st call, 2nd call, 3th call and 4th call. Thus in general before receiving his kth call he has received his 0th, 1st, 2nd, 3rd, 4th, 5th, ..., (k  1) calls.  You split your time interval until the 5th call up in time between (0th and 1th call) + (1th and 2nd call) + (2nd and 3th call) + (3rd and 4th call) + (4th and 5th call)  That this happens can be found via the cumulative Poisson distribution, or thus the sum (you add all this time events before the kth call together so): 0 (C . T) 1 (C . T) (C . T) . e (C . T) . e  +  + ... (9) 0! 1! (k1) (C . T) (C . T) . e ... +  (k  1)! or thus written otherwise as: k (C . T) (C . T) . e = SUM( k = 0 to N1 ) (  ) (10) k! Now as usual, similar to e.g. finding the 'not' situation in the binomial distribution, you use the complement of this, or thus the subtraction of 1, to give the looked for expression. Thus the cumulative probability that an agent is getting his 5th call is equal to 1 minus the probability that the agent is getting his 4th or lower call. (11) Combining (11) and (10) gives thus for this probability in general: k (C . T) (C . T) . e 1  ( SUM( k = 0 to N1 ) (  ) ) (12) k! As you are interested in the probability function itself, you determine the derivative of (12) to T. ' (C . T) . e ' 1  ( SUM( k = 0 to N1 ) (  ) ) (13) k! = (C . T) . e ' ( SUM( k = 0 to N1 ) (  ) ) 0  k! That gives, using the product rule (u . v)' = u' . v + u . v' for each term of the above sum (where thus C and k are considered constants for this differentation): k (C . T) (C . T) . e  k! gives for each term: (k  1) (C . T) k (C . T) k . (C . T) . C . e + (C . T) . C . e (13)  k! by summing all the terms, you get: (k1) (C.T) k (C.T) k.(C.T) .C.e (C.T) .C.e (14)  SUM(k=0 to N1) () k! (C . T) Now put the common factor C . e in front of the sum k (k1) (C.T) (C.T)  k.(C.T) = C.e . SUM(k=0 to N1) () (15) k! You can further work this eliminate this sum, by writing and working out (15) as follows (write the sum as 2 sums) k (C.T) (C.T) = C.e . SUM(k=0 to N1) () + k! (k1) (C.T) k.(C.T)  C.e . SUM(k=0 to N1) () (16) k! You can simplify the second term, by using the fact that k / k! equals per definition k / ( k . (k1) . (k2) . ... 2. 1) or thus there remains, after dividing numerator and denominator both by k 1 / ((k1) . (k2) . ... 2. 1) which equals per definition 1 / (k1)! (17) To avoid making this equal to 1 / (0  1)! when you choose k=0 in the sum, you split the second term of (16) in 2 parts (basically removing the k=0 term, as this gives 0 as a result anyhow. + this term equals 0  V (C.T) (k1) (k1) C.e 0.(C.T) k.(C.T) . ( + SUM(k=1 to N1) ()) (18) 0! k! Putting the (17) in (18), and the endresult of this back in (16) leaves thus: k (k1) (C.T) (C.T) (C.T) =C.e .(SUM(k=0 to N1) ()SUM(k=1 to N1) ()) (19) k! (k1)! Now if you look carefully to this sum, you see that this are 2 sums with the same common term (C . T) which are subtracted from each other. You split out the highest term of this, and are left with 2 sums, which are equal to each other and subtracted of each other. Thus 0, thus eliminated. (k1) (C.T) (C.T) =C.e .() + (k1)! k (k1) (C.T) (C.T) (C.T) C.e .(SUM(k=0 to N2) ()SUM(k=1 to N1) ()) (20) k! (k1)! (k1) (C.T) (C.T) =C.e .() + (k1)! (k1) (k1) (C.T) (C.T) (C.T) C.e .(SUM(k=1 to N1) ()SUM(k=1 to N1) ()) (21) (k1)! (k1)! (k1) (C.T) (C.T) =C.e .() + (k1)! (C.T) C.e .( 0 ) (22) (k1) (C.T) (C.T) =C.e .() (k1)! Adding the exponents of C gives finally: k (C.T) (k1) = C . e . T (23)  (k1)! And this function is the Erlang probability function.   Note: This Erlang probability function is a special case of the Gamma probability function. Because you can write (this follows from the definition of Gamma( k ), which is basically a generalization of the faculty for noninteger numbers): (k1)! as Gamma( k ) (24) if k is an integer greater than 0. Putting (24) in (23) gives: k (C.T) (k1) = C . e . T (25)  Gamma( k ) which is the definition of the Gamma probability function. From this follows that you can use examples using the Gamma probability function to further understand the Erlang probability function.   Note: The exponential probability function is a special case of this Erlang probability function (and thus a special case of the gamma probability function). Because you can write if you choose k=1 and put this in (23): 1 (C.T) (11) = C . e . T  (1  1)! 1 (C.T) 0 = C . e . T  0! (C.T) = C . e and this is the definition for the exponential probability function.   Example: If an agent receives 16 calls per day on average, what is the time until his 5th call? Thus C = 16 and T = 1 (day) and k = 5 Filling this in in (23) gives 5 (16 . 1) (5  1) = 16 . e . 1  (5  1)! = 16^5 * exp(16) / 4! = 0.00491673681 (I have to further check the meaning of this  is it after 0.004 part of a day, thus 0.004 of 8 hours, or thus after already 2 minutes, which is a bit short of course, or should I interpret it differently. And or the values of the parameters C and or T). The problem is the large value of C, probably you should make your interval shorter.   Example: Thus for example, if an agent receives 16 calls per day on average, thus 16 calls per 8 hours, thus 2 calls per hour. What is the time until his 2th call in the first hour? Thus C = 2 and T = 1 (hour) and k = 2 Filling this in in (23) gives 2 (2 . 1) (2  1) = 2 . e . 1  (2  1)! = ( 2^2 * exp(2) * 1^1 ) / 1! = 0.541341133 (this should thus probably mean after about 0.54 part of 1 hour, or thus about an half an hour?)  Note: It looks like that the problem is that you should very probably integrate this formula (23) between the begin and end of your time interval. You see a probable example of this in, using the Gamma probability function (note that the Erlang probability function is a special case of the Gamma probability function). Thus you should calculate T=T2   k (C.T) (k1)  C . e . T (25)   . dT  (k1)! T=T1  Using this example: Example: [book: source: Hastings, p. 139] Suppose that cars pass an expressway according to a Poisson process with a rate of 10 per minute. Find the probability that the second car passes the exit between 5 and 8 seconds Since 5 seconds is 1/12 minute, 8 seconds is 2/15 minute, and the time T2 of passage of the second car has the Gamma( 2, 10 ) distribution and given: C = 10 k = 2 you get after filling this in in (24): P[ 1/12 <= T2 <= 2/15 ] = T=2/15   2 (10.T) (21)  10 . e . T (26)   . dT  (21)! T=1/12 which after working out gives: = 0.18 In other words, the probability is about 1 in 5 times that a 2nd car passes the exit between 5 and 8 seconds.   Book: see also:  [book: Hastings, Kevin J.  Probability and statistics  ISBN 0201 592789  p. 136 'Gamma density']  [book: Montgomery, Douglas, C. / Runger, George C. / Hubele, Norma, F.  engineering statistics  publisher: John Wiley  ISBN 047117026 7  p. 54 'Random variables: you may use the Poisson distribution function with L = C . t (where C is a constant) / p. 152 'Poisson random process']  [book: Neeleman, D. / Bolhuis, van J.  Statistics and probability  Free University  p. 118 'Erlang distribution']  [book: Weisstein, Eric W.  CRC concise encyclopedia of mathematics  ISBN 0849396409  p. 694 'Gamma distribution']   Internet: see also:  Math: Probability: Distribution: Link: Overview: Can you give an overview of links? http://www.faqts.com/knowledge_base/view.phtml/aid/32917/fid/815  Erlang distribution http://mathworld.wolfram.com/ErlangDistribution.html  Gamma distribution (see (5) 'The corresponding probability function P (x) of waiting times until the hth Poisson event is then obtained by differentiating D(x)') http://mathworld.wolfram.com/GammaDistribution.html  Erlang distribution http://en.wikipedia.org/wiki/Erlang_distribution  Math: Traffic engineering: Can you give overview of links about the distribution of telephone calls? http://www.faqts.com/knowledge_base/view.phtml/aid/39736/fid/815 