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Input Modeling. Modeling and Simulation CS 313. Poisson distribution.

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Input modeling

Input Modeling

Modeling and Simulation

CS 313


Poisson distribution
Poisson distribution

  • It is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time and/or space if these events occur with a known average rate and independently of the time since the last event.

  • Example:

    • Suppose you typically get 4 pieces of mail per day. That becomes your expectation, but there will be a certain spread: sometimes a little more, sometimes a little less, and once in a while nothing at all.

    • Given only the average rate, for a certain period of observation (pieces of mail per day), the Poisson distribution will tell you how likely it is that you will get 3, or 5, or 11, or any other number, during one period of observation.


Poisson distribution1
Poisson distribution

  • The distribution equation:

  • If the expected number of occurrences in a given interval is λ, then the probability that there are exactly k occurrences (k being a non-negative integer, k = 0, 1, 2, ...) is equal to:

  • Where:

    • e is the base of the natural logarithm (e = 2.71828...)

    • k is the number of occurrences of an event— the probability of which is given by the function (The random number)

    • k! is the factorial of k

    • λ is a positive real number, equal to the expected number of occurrences during the given interval (the average rate)

  • For instance, if the events occur on average 4 times per minute, and one is interested in the probability of an event occurring k times in a 10 minute interval, one would use a Poisson distribution as the model with λ = 10×4 = 40.


Poisson distribution example
Poisson distribution example

  • Consider in an office 2 customers arrived today. Calculate the possibilities for exactly 3 customers to arrive tomorrow.

  • Step1:

    • Find e-λ where, λ=2 and e=2.718

    • e-λ = (2.718)-2 = 0.135

  • Step2:

    • Find λx where, λ=2 and x=3

    • λx = 23 = 8

  • Step3:

    • Find f(x)

    • f(x) = e-λλx / x!

    • f(3) = (0.135)(8) / 3! = 0.18

  • Hence there is18% possibility for 3 customers to arrive tomorrow.


Exponential distribution
Exponential distribution

  • It is a family of continuous probability distributions. It describes the time between events in a Poisson process, i.e. a process in which events occur continuously and independently at a constant average rate.

  • The probability density function (pdf) of an exponential distribution is

    • λ is the parameter of the distribution

    • x is the random variable


Binomial distribution
Binomial distribution

  • Definition:

    • The Binomial distribution is one of the discrete probability distributions. It is used when there are exactly two mutually exclusive outcomes of a trial. These outcomes are appropriately labeled Success and Failure.

    • The Binomial distribution is used to obtain the probability of observing r successes in n trials, with the probability of success on a single trial denoted by p.

  • Function:

    • P(X = r) = nCr pr (1-p)n-r

    • Where:

      • n = number of events

      • r = number of successful events

      • p = probability of success on a single trial

      • nCr= ( n! / (n-r)! ) / r!

      • 1-p = probability of failure


Binomial distribution example
Binomial distribution example

  • Toss a coin for 12 times. What is the probability of getting exactly 7 heads?

  • Step 1:

    • Number of trials n = 12

    • Number of successes r = 7 since we define getting a head as success

    • Probability of success on any single trial p = 0.5

  • Step 2: calculate nCr

    • nCr = ( n! / (n-r)! ) / r!

    • = ( 12! / (12-7)! ) / 7!

    • = ( 12! / 5! ) / 7!

    • = ( 479001600 / 120 ) / 5040

    • = ( 3991680 / 5040 )

    • = 792

  • Step 3: find pr

    • pr = 0.57 = 0.0078125

  • Step 4: to Find (1-p)n-rcalculate 1-p and n-r

    • 1-p = 1-0.5 = 0.5

    • n-r = 12-7 = 5

  • Step 5: find (1-p)n-r

    • = 0.55 = 0.03125

  • Step 6: solve P(X = r) = nCrpr(1-p)n-r

    • = 792 * 0.0078125 * 0.03125

    • = 0.193359375

  • The probability of getting exactly 7 heads is 0.19


Goodness of fit tests
Goodness-of-Fit Tests

  • Conduct hypothesis testing on input data distribution using:

    • Kolmogorov-Smirnov test

    • Chi-square test

  • Goodness-of-fit tests provide helpful guidance for evaluating the suitability of a potential input model.

  • No single correct distribution in a real application exists.

  • If very little data is available, it is unlikely to reject any candidate distribution.

  • If a lot of data is available, it is likely to reject all candidate distributions.


Chi square test
Chi-Square test

  • Intuition: comparing the histogram of the data to the shape of the candidate density or mass function.

  • Valid for large sample sizes when parameters are estimated by maximum likelihood

  • By arranging the n observations into a set of k class intervals or cells, the test statistics is:

    • Where:

      • Oiis the observed frequency (number of occurrences per interval/value)

      • Ei is the expected frequency

      • Ei = n * piwhere pi is the theoretical probability of the ith interval and n is the total number of data values

      • Suggested minimum for Ei= 5


Chi square test1
Chi-Square test

  • The hypothesis of a chi-square test is:

    • H0: the random variable, X, conforms to the distributional assumption with the parameter(s) given by the estimate(s).

    • H1: the random variable X does not conform.

  • If the distribution tested is discrete and combining adjacent cell is not required (so that Ei > minimum requirement), each value of the random variable should be a class interval, unless combining is necessary, and:

    pi = p(xi) = P(X = xi)


Chi square test example
Chi-Square test example

  • Vehicle Arrival Example (continued) :

  • The histogram appears to be Poisson

  • We find the estimated mean (λ)to be 3.64

  • Using Poisson pmf:

  • For λ = 3.64, the probabilities are:

    • p(0) = 0.026 p(6) = 0.085

    • p(1) = 0.096 p(7) = 0.044

    • p(2) = 0.174 p(8) = 0.020

    • p(3) = 0.211 p(9) = 0.008

    • p(4) = 0.192 p(10) = 0.003

    • p(5) = 0.140 p(≥11) = 0.001


Chi square test example cont
Chi-Square test example cont.

  • H0: the random variable is Poisson distributed.

  • H1: the random variable is not Poisson distributed.

  • Because the hypothesis (H0) is rejected


Chi square test2
Chi-Square test

  • Chi-square test can accommodate estimation of parameters.

  • Chi-square test requires data to be placed in intervals.

  • Changing the number of classes and the interval width affects the value of the calculated and tabulated chi-square.

  • A hypothesis can be accepted if the data is grouped in one way and rejected in another way.


Kolmogorov smirnov test
Kolmogorov-Smirnov test

  • Intuition: formalize the idea behind examining a q-q plot

  • A more powerful test, particularly useful when:

    • Sample sizes are small

    • No parameters have been estimated from the data