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Apr 2nd, 2009 07:46
engatoo engatoo, Knud van Eeden,
  Knud van Eeden  17 February 2005  00:40 am  Math: Probability:Distribution: Exponential: Can you derive Exponential distribution L . e^(L . t)?   The Exponential distribution is mainly used to model Poisson processes. E.g. when calculating a Poisson distribution with rate of change L, and looking for the distribution of waiting times between successive changes.  Waiting time distributions are based on the exponential distribution, which in turn is derived from the Poisson distribution.  While the Poisson distribution governs the occurrence of random events in time, space, volume, ..., the intervals between such events is controlled by the Exponential distribution.  Consider a problem such as the arrival of telephone calls.  The total number of occurrences during any particular unit of time is controlled by the Poisson distribution with mean L per 1 unit time.  Here is tried to find the distribution of time intervals between successive arrivals of calls and find the probability that there is a time interval of length t between successive arrivals of calls.   You divide the time interval t in subintervals of length dt such that there is a small probability p of the occurrence of a call during this subinterval dt.  Poisson: ... 0 1 2 3 4 N1 N > number of intervals   Each of the N intervals has a probability p of success. p p p p p p p p ... 0 1 2 3 4 N1 N > number of intervals   Here q is the probability of no occurrence of a call in that subinterval dt, where p + q = 1 (or thus in other words together they cover 100%). Thus q = 1  p (1)  The subinterval dt is assumed small enough to make the probability of more than one occurrence of a call in this subinterval dt negligible.  The probability, denoted by dP, that there are N incremental intervals between successive arrivals of a call is given by: dP = (probability of a call not arriving) . + (probability of a call not arriving) .  (probability of a call not arriving) . > N intervals ...  (probability of a call not arriving) . + (probability of a call arriving) . = q . q . q . q . q ... q . p   + N intervals +  That is N intervals with no arrivals, and in the (N+1)th interval there is an arrival  N So dP = q . p (2)  figure: The probability of a call coming in after waiting N intervals <interval with length t until call arrives > q q q q q q q q p ... 0 1 2 3 4 N N+1 ^  > number of intervals  here a call arrives  Note: The right side of this structure (2) is called the 'geometric waiting time distribution', which is in turn a special case of the 'binomial waiting time distribution'.   Note: This process is the same as throwing a dice, and asking yourself what the probability is that you throw it N times until say a 5 occurs. The probability that this occurs is (probability of not a 5) . + (probability of not a 5) .  (probability of not a 5) . > N throws ...  (probability of not a 5) . + (probability of a 5) . Thus for example, what is the probability that you throw 7 times a dice and that no 5 occurs. Now you know that a 5 occurs on average 1 times in 6 times throwing. Thus its probability is 1 / 6. Similarly the probability that no 5 occurs is thus 5 times in 6 times throwing, or thus 5/6, or thus 1  1/6. Thus if you throw a dice 7 times after each other the probability that no 5 occurs is thus 5/6 . 5/6 . 5/6 . 5/6 . 5/6 . 5/6 . 5/6 or thus 7 (5/6) The probability that a 5 occurs after throwing 7 + 1, or 8 times is thus (5/6 . 5/6 . 5/6 . 5/6 . 5/6 . 5/6 . 5/6) . (1/6) or thus 7 (5/6) . (1/6)   Putting (1) in (2) gives N dP = (1  p) . p (3)  What follows will replace p by L, N and dt, from which a differential equation follows.  Now if there are on average L arrivals per 1 unit time, correspondingly the (mean number of arrivals in subinterval dt) follows by linearly scaling down from (old arrivals) (new arrivals)  =  (4) (old interval) (new interval) or thus here: L x (5)  =  1 dt from which follows that x = dt . L  1 Thus the (mean number of arrivals in subinterval dt) equals here dt . L (6)  Now using that the mean number of arrivals in any interval equals here (number of trials) times (probability of success at any trial), thus if you have only only 1 interval dt, then you can write (mean number of arrivals in subinterval dt) = ( 1 subinterval ) . ( probability of success at any trial) = 1 . p (7) As (6) and (7) say the same, you can equate them: dt . L = 1 . p (8) As you want to eliminate p from (3), you express (8) in p: p = L . dt (9) As you divided here your time interval in N subintervals of size dt, you can write t = N . dt (10) thus dt = t / N (11) Putting (11) in (9) gives p = L . t (12)  N Putting (9) and (12) in (3) gives: N dP = (1  L . t) . L . dt (13)  N  You know maybe that N (another) definition of e is: N 1 1 (1 + ) = e (14) N lim N > +infinity An analog expression is: N 1 1 (1  ) = e (15) N lim N > +infinity and similarly: p N +p (1 + ) = e (16) N lim N > +infinity and similarly: p N p (1  ) = e (17) N lim N > +infinity  Thus here using (17), if N > infinity, then in (13) you have that N (1  L . t)  N tends to (L . t) e (18) Putting (18) in (13) gives (L . t) dP = L . e . dt (19) Dividing both sides of (19) by dt gives: (L . t) dP = L . e (20)  dt The righthand side is the Exponential distribution, and is the probability density function for the interval between successive events, i.e. it is the probability that the time interval lies between t and t+dt. The expectation is, similar to an archimedes lever, where you multiply the arm by the mass at that distance, given by in general by: t>infinity    (independent variable).(dependent variable).d(independent variable)    t=0 thus here the expectation is given by: t>infinity    (L . t)  t . L . e . dt    t=0 Integrating by parts gives + t>infinity (L . t) (L . t)  t . L . e  e    L  + t=0 Filling in the begin and end points gives: (0 . t) (0 . t) 0  ( 0 . L . e  e  ) L = 0  ( 0  1 )  L or thus expectation = 1 (21)  L This result is what you should expect. If L is the expected number of events in unit time (from the original Poisson distribution), then you could also expect that the mean interval between events is 1/L. In other words if L, the average total amount of occurrences or thus calls in the given interval gets larger and larger, the average waiting time for an occurrence of a call will get smaller and smaller. Thus here they are inversely proportional.   To further find the total waiting time between 2 events, you integrate both sides of (19), giving       (L . t)  dP =  L . e . dt (22)       or thus the probability P for a waiting time equals:    (L . t) P =  L . e . dt (22)    to be more precise: t>infinity    (L . t) P =  L . e . dt (22)    t=t end   For example: In a telephone call center, incoming telephone calls can be modeled as a Poisson process with a mean of 25 incoming telephone calls per hour. What is the probability that there are no telephone calls in an interval of 6 minutes? First convert units. Now 25 calls per hour is thus 25 calls per 60 minutes. Thus L = 25  60 The probability that there are no telephone calls in the first 6 minutes can be calculated as the probability of getting telephone calls between 6 minutes and infinity. t>infinity    (25 . t)    60  P =  (25) . e . dt (23)    60  t=6 working out, by integrating the right side of (23) gives: + t>infinity (25 . t)    60   P = e  (24) + t=6 (25 . 6)  60 P = 0  (  e ) (25) (25 . 6)  60 P = e (26) P = 0.082 (27) or thus in about 8 times in 100 times you will have to wait 6 minutes until the next call arrives.   For example: At a bus stop, arriving buses can be modeled as a Poisson process with a mean of 2 incoming buses per hour. You arrive at the busstop, just seeing the current bus leave. What is the probability that there is no bus coming in an interval of 15 minutes? First convert units. Now 2 buses per hour is thus 2 busses per 60 minutes. Thus L = 2  60 The probability that there are no buses arriving in the first 15 minutes can be calculated as the probability of buses arriving between 15 minutes and infinity. t>infinity    (2 . t)    60  P =  (2 ) . e . dt (28)    60  t=15 working out, by integrating the right side of (23) gives: + t>infinity (2 . t)    60   P = e  (29) + t=15 (2 . 15)  60 P = 0  (  e ) (30) (2 . 15)  60 P = e (31) P = 0.607... (32) or thus in about 6 times in 10 times you will have to wait 15 minutes until the next bus arrives.   Book: see also:  [book: Montgomery, Douglas, C. / Runger, George C. / Hubele, Norma, F.  engineering statistics  publisher: John Wiley  ISBN 047117026 7  p. 100 'Exponential distribution', 'in a large corporate network, user logons to the system can be modeled as a Poisson process with a mean of 25 logons per hour. What is the probability that there are no logons in an interval of 6 minutes?]  [book: Hastings, Kevin J.  Probability and statistics  ISBN 0201 592789  p. 135 'Exponential density' (suppose that the time interval between arrivals of buses to a particular stop is a random variable T (in hours) with the exp(2) distribution. You arrive, panting, to the stop just after the bus leaves. What is the probability that you will have to wait at least 15 minutes for the next bus?']   Internet: see also:  [Internet: source: Waiting Time Distributions: http://mathforum.org/library/drmath/view/52162.html]  Math:Probability:Distribution:Exponential: Can you describe Exponential distribution L . e^(L . t)? http://www.faqts.com/knowledge_base/view.phtml/aid/33099/fid/815  Math: Probability: Distribution: Link: Overview: Can you give an overview of links? http://www.faqts.com/knowledge_base/view.phtml/aid/32917/fid/815 http://www.roo7e.com http://www.roo7e.com http://www.al33ab.com http://ksacam.com http://chat.ksacam.com http://ksacam.info http://oo2o.com http://s4cam.com http://saudi4cam.com http://voice.ksacam.com http://سعوديكام.oo2o.com http://سعوديكام.com http://xnmgbply5cnrr.com http://www.3rabstarz.com/vb http://www.3rabstarz.com/vb http://www.3rabstarz.com/ http://www.3rabstarz.com/vb http://www.3rabstarz.com/ http://www.dir.3rabstarz.com/ http://www.up.3rabstarz.com http://www.3rabstarz.com/vb/forumdisplay.php?f=5 http://www.3rabstarz.com/vb/forumdisplay.php?f=116 http://www.3rabstarz.com/vb/forumdisplay.php?f=66 http://www.3rabstarz.com/vb/forumdisplay.php?f=78 http://www.3rabstarz.com/vb/forumdisplay.php?f=147 http://www.3rabstarz.com/vb/forumdisplay.php?f=2 http://www.3rabstarz.com/vb/forumdisplay.php?f=31 http://www.3rabstarz.com/vb/forumdisplay.php?f=136 http://www.3rabstarz.com/vb/forumdisplay.php?f=34 http://www.3rabstarz.com/vb/forumdisplay.php?f=89 http://www.3rabstarz.com/vb/forumdisplay.php?f=26 http://www.3rabstarz.com/vb/forumdisplay.php?f=72 http://www.3rabstarz.com/vb/forumdisplay.php?f=73 http://www.3rabstarz.com/vb/forumdisplay.php?f=149 http://www.3rabstarz.com/vb/forumdisplay.php?f=150 http://www.3rabstarz.com/vb/forumdisplay.php?f=151 http://www.3rabstarz.com/vb/forumdisplay.php?f=152 http://www.3rabstarz.com/vb/forumdisplay.php?f=67 http://www.3rabstarz.com/vb/forumdisplay.php?f=3 http://www.3rabstarz.com/vb/forumdisplay.php?f=37 http://www.3rabstarz.com/vb/forumdisplay.php?f=115 http://www.3rabstarz.com/vb/forumdisplay.php?f=76 http://www.3rabstarz.com/vb/forumdisplay.php?f=77 http://www.3rabstarz.com/vb/forumdisplay.php?f=36 http://www.3rabstarz.com/vb/forumdisplay.php?f=153 http://www.3rabstarz.com/vb/forumdisplay.php?f=122 http://www.3rabstarz.com/vb/forumdisplay.php?f=118 http://www.3rabstarz.com/vb/forumdisplay.php?f=119 http://www.3rabstarz.com/vb/forumdisplay.php?f=120 http://www.3rabstarz.com/vb/forumdisplay.php?f=121 http://www.3rabstarz.com/vb/forumdisplay.php?f=35 http://www.3rabstarz.com/vb/forumdisplay.php?f=65 http://www.3rabstarz.com/vb/forumdisplay.php?f=7 http://www.3rabstarz.com/vb/forumdisplay.php?f=9 http://www.3rabstarz.com/vb/forumdisplay.php?f=11 http://www.3rabstarz.com/vb/forumdisplay.php?f=142 http://www.3rabstarz.com/vb/forumdisplay.php?f=143 http://www.3rabstarz.com/vb/forumdisplay.php?f=146 http://www.3rabstarz.com/vb/forumdisplay.php?f=140 http://www.3rabstarz.com/vb/forumdisplay.php?f=137 http://www.3rabstarz.com/vb/forumdisplay.php?f=114 http://www.3rabstarz.com/vb/forumdisplay.php?f=117 http://www.3rabstarz.com/vb/forumdisplay.php?f=113 http://www.3rabstarz.com/vb/forumdisplay.php?f=51 http://www.3rabstarz.com/vb/forumdisplay.php?f=84 http://www.3rabstarz.com/vb/forumdisplay.php?f=138 http://www.3rabstarz.com/vb/forumdisplay.php?f=139 http://www.3rabstarz.com/vb/forumdisplay.php?f=15 http://www.3rabstarz.com/vb/forumdisplay.php?f=130 http://www.3rabstarz.com/vb/forumdisplay.php?f=97 http://www.3rabstarz.com/vb/forumdisplay.php?f=102 http://www.3rabstarz.com/vb/forumdisplay.php?f=101 http://www.3rabstarz.com/vb/forumdisplay.php?f=104 http://www.3rabstarz.com/vb/forumdisplay.php?f=131 http://www.3rabstarz.com/vb/forumdisplay.php?f=103 http://www.3rabstarz.com/vb/forumdisplay.php?f=99 http://www.3rabstarz.com/vb/forumdisplay.php?f=59 http://www.3rabstarz.com/vb/forumdisplay.php?f=60 http://www.3rabstarz.com/vb/forumdisplay.php?f=39 http://www.3rabstarz.com/vb/forumdisplay.php?f=40 http://www.3rabstarz.com/vb/forumdisplay.php?f=42 http://www.3rabstarz.com/vb/forumdisplay.php?f=4 http://www.3rabstarz.com/vb/forumdisplay.php?f=1 http://www.3rabstarz.com/vb/forumdisplay.php?f=88 http://www.3rabstarz.com/vb/forumdisplay.php?f=148 http://www.3rabstarz.com/vb/forumdisplay.php?f=62 http://www.3rabstarz.com/vb/forumdisplay.php?f=63 http://www.3rabstarz.com/vb/forumdisplay.php?f=64 http://www.3rabstarz.com/vb/forumdisplay.php?f=6 http://www.3rabstarz.com/vb/forumdisplay.php?f=12 http://www.3rabstarz.com/vb/forumdisplay.php?f=96 http://www.3rabstarz.com/vb/forumdisplay.php?f=58 http://www.3rabstarz.com/vb/forumdisplay.php?f=38 