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Feb 16th, 2005 10:55
Knud van Eeden,
  Knud van Eeden  25 December 2004  08:05 pm  Math: Probability: Distribution: Poisson: Can you describe Poisson distribution L^k . e^(L) / k!?  The Poisson distribution is a special case of the Binomial distribution, and can be derived from it.   It is a discrete distribution.   Both can be seen as variations on N throws, after each other, with a coin with a certain (larger in the case of the binomial distribution, or very small in the case of the Poisson distribution) probability.  Binomial or Bernouilli distribution: ... 0 1 2 3 4 N1 N > number of throws Poisson: ... 0 1 2 3 4 N1 N > number of intervals   Each of the N intervals has a probability p of success.   The difference is that the total amount of throws with a Poisson distribution is by definition very large, thus N very large, while the chance that you have success in each interval (e.g. head), thus p, is very small.   The difference between these two is further that the Binomial distribution requires knowledge of the probability that some event will happen (that is 'p'), and also about the probability that this event will not happen (that is '1p' usually). This probability p is often derived from finding the proportion of successes in N trials (e.g. throwing a coin N times, and counting the total amount of heads, and dividing this by the total amount of throws. This will give p). As usually if p is known, the probability that this event will not happen (that is '1p') is then also known (because the probability together must equal 1, or 100%, so you can subtract p from 1 to get the probability '1p'). Both 'p' and '1p' are needed in the binomial distribution formula. Now, there are many situations where we can count the occurrences of some event, but no figure of nonoccurrences is meaningful.  For example, A telephone center receives per agent 16 calls per hour on average. How many calls do NOT occur? Obviously, this is not known.  For example, On a road you count 16 cars passing per hour on average. How many cars did NOT occur? Obviously, this is not known.  To such situations the Binomial distribution formula is not applicable. Instead you can use the Poisson distribution.  The Binomial distribution works thus with the constant and known probability p.  The Poisson distribution works instead with the constant and known expectation L (this 'L' stands for the Greek character 'L'ambda, which is often used in this context to indicate the average rate)  This expectation of the Poisson distribution can be written as L = N . p  And you see that choosing this expectation L is a smart move, as it turns out to be equal to the average, something which you can find out experimentally by measuring via e.g. counting. And you will see that N and p drop out of the endresult of the calculations all together, so you do not need them anymore.  How can you possibly understand this expression L = N . p? If you divide your interval in N subintervals, with each subinterval having a probability of success p, and there are thus totally N of this subintervals, then you can say that this expression states: expectation = (total number of trials) times (probability of success at any trial)  as there are in the interval N subintervals, and each subinterval has a probability p of success, you are saying thus here (imagine that you, in each of this subintervals, in a serial sequence, throw a dice with a probability p, then you throw so N times after each other this dice. You are asking yourself what your expectation is after N of this throws) expectation = N . p  e.g. This is similar to doing throws, one after the other, and at each throw you have a probability of on average 1 times in 6 times that you get a 5. Thus after you have thrown 600 times, you can expect on average that you get 600 . 1/6, or thus 100 times the 5 as a result. So your expectation will be 100 occurrences of the 5.  e.g. If you play 100 times in the lottery, and the probability that you win something is on average 1 times in 5 times playing, you are expected on average to win 100 . 1/5, or thus 20 times. So your expectation will be 20 occurrences of a win.  e.g. If your football team should play 60 matches, and the probability that they win is on average 1 times in 2 times playing, then it is expected on average that they should win 60 . 1/2, or thus 30 times. So your expectation will be 30 occurrences of a win.  It can be shown that the average value of the Binomial distribution equals to N . p, or thus the average value of the Binomial distribution equals the expectation (you calculate this using the mean value theorem in calculus, applied to this curve)  (as the Poisson distribution is just a special case of the binomial distribution, so it should have the same properties, in this case the average) It can be shown that the average value of the Poisson distribution equals to N . p, or thus the average value of the Poisson distribution equals the expectation. (you calculate this using the mean value theorem in calculus, applied to this curve)  Thus N . p, the expectation, equals the average in this special case.  Thus L, the expectation, equals the average value of the Poisson distribution.  It is assumed for the cases where you apply your Poisson distribution that this mean or average or expectation remains constant. Thus L = N . p = constant Thus if N gets much larger, then p must get much smaller, such that L remains constant.  This L is your average total number of occurrences. E.g. totally 16 calls (in your time interval of 1 hour, and per agent) on average.  Thus what you then do is to divide your interval (this can in general be any interval, be it time, space, volume, wire length, area, ...) in N smaller subintervals of equal length.  original interval [ ]  original interval divided in smaller equal subintervals. [                  ...  ]  Events (e.g. incoming telephone calls) should occur at random in your time, length, volume, ..., subinterval  Now if the following conditions are present, you could use the Poisson distribution: 1. The probability of more than one count in a subinterval is zero (e.g. in a 1 millisecond interval it is very unlikely that you get at the same moment 3 telephone calls) 2. The probability in each subinterval is the same for all (in other words no subinterval is more important than another subinterval) 3. The probability is proportional to the total length of the subinterval (if you wait 5 minutes, you should have a 5 times bigger chance of getting a call) 4. The probability of each subinterval is independent of another subinterval (what happens in another subinterval should have no influence on the probability in the subinterval)   How to calculate L? L is a positive real number, equal to the expected total number of occurrences that occur in the given interval.  Thus L follows from the nominator of (total number of occurrences)  (the given interval)  Note: So this expression can also be seen as a frequency, if you look at the units used, that is, a dimension less (integer) value divided by a dimensioned value, similar to e.g. a clock beats 10 times per minute, a wave oscillates 60 times per second.  For instance, if the events occur on average 1 time every 2 minutes, thus ( 1 time )  ( 2 minutes ) and you are interested in the number of events occcurring in a 30 minute interval, you would linearly scale it, by multiplying everything with a constant, as that does not change the quotient (if you multiply both nominator and denominator in a quotient with the same value, the end result remains the same, as multiplying with 1 does not change the endresult). Thus you can also write: (some constant) (total number of occurrences)  (some constant) (your given time interval)  Another way of saying this is that L, the total number of occurrences is proportional with the given time interval. Thus if the time interval is chosen 2 times as large, the total number of occurrences is going to be 2 times as large. Thus if the time interval is chosen 3 times as large, the total number of occurrences is going to be 3 times as large, and so on. e.g. 15 . ( 1 time )  15 . ( 2 minutes ) remains proportionally the same.  Given a total occurrences of 15 . 1 times in 15 . 2 minutes, or thus totally 15 occurrences in a 30 minutes interval. Thus L would in that case be chosen to be this new total number of occurrences, or thus 15.  For instance, if the events occur on average 24 times every 1 working day or thus ( 24 times )  ( 1 working day) and you are interested in the number of events occcurring in a 1 hour interval, or thus you would proportionally or linearly scale it, by multiplying by 1/8. or thus (1/8) . ( 24 times )  (1/8) . ( 1 working day) or thus (1/8) . ( 24 times )  (1/8) . ( 8 hours) or thus ( 3 times )  ( 1 hour ) Thus L, the total number of occurrences would in this case be chosen equal to 3.   So a possible algorithm to calculate the Poisson distribution is: 1. Choose a total time interval e.g. 8 hours 2. Count the total of occurrences in this interval e.g. 16 calls 3. Determine the interval in which you are interested e.g. 1 hour 4. Scale from the proportion: (new total occurrences) (old total occurrences)  =  (new time interval) (old time interval) Note: so you are basically saying that (new frequency) = (old frequency) follows: (new time interval) . (old total occurrences) (new total occurrences) =  (old time interval) thus you use as the scale factor (new time interval)  (old time interval) and you multiply the old total occurrences with this number. e.g. (1 hour ) . (16 total occurrences ) (new total occurrences) =  (8 hours ) = 2 total occurrences 5. Fill in 1. L, or this 'new total occurrences' 2. k, or thus the amount of times you are asking it to occur 3. in the formula k L L . e  k! e.g. choosing k equal to 3, gives 3 2 2 . e  3! 6. That will give a number between 0 and 1. e.g. this fraction equals in this example: 0.180447044 7. Interpret this result You can also use scaling to interpret this result yourself better. or thus it will occur 0.18 times in the 1 hour interval. Usually you can better imagine this result by using whole numbers. So you scale it up to the whole number 1. Or thus about, by scaling it also, in this case to whole number, that is 1, to 5 . 0.18 in 5 . 1 hour interval, or thus to about 1 time in a 5 hour interval. So you use the about proportion ( 0.180447044 times ) 5 . ( 0.180447044 times )  =  ( 1 hour interval ) 5 . ( 1 hour interval ) in general, you convert the calculated fraction to a whole number 1, by multiplying by (1 divided by this fraction). 1 () (new times) fraction ( your just calculated times it will occur)  =  (new interval) 1 ( your used interval ) () fraction   As this scaling might occur all the time, you might as well fill it in directly in the Poisson formula: from [1] k L L . e  k! you get thus, using the above terminology [2] (new total occurrences) (old total occurrences)  =  (new time interval) (old time interval) which can be rewritten as: [3] (new time interval) . (old total occurrences) (new total occurrences) =  (old time interval) filling in the right side of [3] in [1], you get [4] k (new interval).(old occurrences) (new interval).(old occurrences) () () . (old time interval) (old interval) e  k!   You can ask yourself then questions like:  If given that 24 calls per day are coming in on average per agent, what is the probability that you get 5 calls per 1 hour? Thus = ( 24 times )  ( 1 working day) = 1/8 . ( 24 times )  1/8 . ( 8 hours ) = ( 3 times )  ( 1 hour ) Thus in this 1 hour interval, you have L = totally 3 occurrences per time interval on average or thus here L = totally 3 calls per hour on average or thus L = 3 and k = 5 occurrences per time interval or thus here k = 5 calls per hour or thus k = 5  Using the Poisson distribution, you get that this probability equals k L L . e =  k! 5 3 = (3) . e  5! = 0.100818813... Thus a probability of 1 in 10, or thus this should happen once every 10 hours, or thus about once in one to two days.   Another example, using the cumulative Poisson distribution  If given that 24 calls per day are coming in on average per agent, what is the probability that you get 5 or fewer calls per hour? Thus = ( 24 times )  ( 1 working day) = 1/8 . ( 24 times )  1/8 . ( 8 hours ) = ( 3 times )  ( 1 hour ) Thus in this 1 hour interval, you have: L = totally 3 occurrences per time interval on average or thus L = totally 3 calls per hour on average or thus L = 3 and because is asked for 'fewer' than 5, you get: k = 0, 1, 2, 3, 4 or 5 occurrences per time interval or thus here k = 0, 1, 2, 3, 4 or 5 calls per hour  Using the Poisson distribution, you get that this probability equals (similar to the Binomial distribution, where you also have to add together all possibilities smaller or equal to in this case) k L = SUM( k from 0 to 5 ) of ( L . e )  k! 0 3 1 3 2 3 (3) . e (3) . e (3) . e =  +  +  + 0! 1! 2! 3 3 4 3 5 3 (3) . e (3) . e (3) . e +  +  +  3! 4! 5! = 0.0497870684 + 0.149361205 + 0.224041808 + 0.224041808 + 0.168031356 + 0.100818813 = 0.9160820584 Thus a probability of about 9 in 10, thus about in 90% of the hours, you should have 5 or less calls.   Note: History: Sim<e'>on Denis Poisson (17811840) (a student of Simon de Laplace and JosephLouis Lagrange) introduced this Poisson equation for the first time in 1837 in his work 'Recherches sur la probabilit<e'> des jugements'   Book: see also:  [book: Bronshtein / Semendyayev  a guide book to mathematics  ISBN 0 387911065  p. 605 'Poisson distribution' (short formula, 1 numeric example / 1 numeric example of Poisson distribution approximation to binomial distribution)]  [book: Childers, Donald G.  probability and random processes (using MatLab with applications to continuous and discrete time systems)  publisher: McGrawHill  ISBN 0256133611  p. 90 'Expectation (Averages)' (introducting the mean as the statistical average, or in a physical way as the 'center of gravity' (also called 'first moment' here) of the Poisson graph in vertical direction, so putting it balancing on a cord will leave it in equilibrium) / 'Poisson random processes' (with good example relation telephone calls arriving according to a Poisson distribution, and with the telephone call time length following an exponential distribution, with a formula workout and numeric example, in a network, M/M/1 queue)]  [book: Daintith, John / Nelson, R. David  the penguin dictionary of mathematics  ISBN 0140511199  p. 254 'Poisson distribution' (only a very short definition)]  [book: Gellert, W. / K<u..>stner, H. / Hellwich, M.  kleine Enzyklop<a..>die Mathematik  ISBN 3871443239  p. 640 'Poissonverteilung' (nice description, good example with e.g. probability of taking out 1 black ball out of a vase (with putting it back afterwards) with a very large amount of white balls)]  [book: Green, David R. / Lewis, John  science with pocket calculators  ISBN O0851095607  p. 152 'Poisson distribution']  [book: Hastings, Kevin J.  Probability and statistics  ISBN 0201 592789  p. 80 'Poisson distribution' with example, p. 82 'Poisson probability mass function' (definition), p. 83 (comparison values Binomial and Poisson distribution), p. 83 (example cumulative Poisson distribution, for users wanting to access a server), p. 84 'Poisson random processes' (rate parameter, solution of differential equation gives Poisson formula), p. 215 'Poisson interarrival times', p. 237 'Generating functions' (superposition of Poisson arrival rates, distribution of arrival times, about telephone calls arriving at a switch board)]  [book: Kramer, Edna Erestine  the nature and growth of modern mathematics  p. 333 'Realm of random variables' (Shows example of how to approximate a binomial distribution with a Poisson distribution)]  [book: Lipschutz, Seymour  Probability  McGrawHill (Schaum)  p. 108 'Poisson distribution' (4 graphs, approximation to binomial distribution via Taylor expansion)]  [book: Montgomery, Douglas, C. / Runger, George C. / Hubele, Norma, F.  engineering statistics  publisher: John Wiley  ISBN 047117026 7  p. 92 'Poisson process' (example with transmission of n bits over a communication channel, flaws in a copper cable and contamination of optical storage disks]  [book: Spiegel, Murray R.  Probability and statistics  McGrawHill (Schaum series)  p. 123 'Mean equals N . p', p. 129 'The Poisson distribution', p. 239 'fitting of numerical data (automobile accidents')]  [book: Weisstein, Eric W.  CRC concise encyclopedia of mathematics  ISBN 0849396409  p. 1388 'Poisson distribution' (1 graph, lot of formulas about this curve)]  [book: Zwillinger, Daniel  CRC standard mathematical tables and formula  p. 582 'Poisson distribution' (only a very short definition) / 'Poisson arrivals' (only a formula with no further explanation)]   Internet: see also:  Poisson Distribution http://home.ched.coventry.ac.uk/Volume/Vol4/listof.htm  Poisson Distribution http://mathworld.wolfram.com/PoissonDistribution.html  Poisson Distribution http://en.wikipedia.org/wiki/Poisson_distribution  Sim<e'>on Denis Poisson http://wwwgroups.dcs.stand.ac.uk/~history/Mathematicians/Poisson.html  Math: Probability: Distribution: Link: Overview: Can you give an overview of links? http://www.faqts.com/knowledge_base/view.phtml/aid/32917/fid/815 