Poisson Processes and Poisson Distributions. Poisson Process - Deals with the number of occurrences per interval.

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1 Poisson Processes and Poisson Distributions Poisson Process - Deals with the number of occurrences per interval. Eamples Number of phone calls per minute Number of cars arriving at a toll both per hour Number of failures per 1000 hours Number of bumps along a road per mile Number of flaws in fabric per square yard Number of pieces of debris per cubic meter of sea water Conditions For any interval over the real numbers: 1. Assume the occurrences happen at random throughout the interval. 2. Partition the interval into subintervals such that, a. The probability of more than one occurrence per subinterval is zero. b. The probability of one occurrence in a subinterval is the same for all the other subintervals, and is proportional to the length of the subinterval. c. An occurrence in any one subinterval is independent of an occurrence in any of the other subintervals. 3. Then the random occurrences are said to be a Poisson Process, such that; If the mean number of occurrences in the interval is λ > 0, then the random variable X, (where X equals the number of occurrences in the interval) has a Poisson Distribution with parameter λ such that; Probability Mass Function f () e! for = 0, 1, 2,... Mean: μ = E(X) = λ Variance: σ 2 = V(X) = λ

2 Eponential Distributions Eponential Distribution - Deals with the distance or time between successive occurrences within an interval of a Poisson Process. Eamples Time between phone calls Time between cars arriving at a toll both per hour Time between failures of electronic components Distance between successive bumps along a highway Number of square yards of fabric per flaw Number of cubic meters of sea water per piece of debris Conditions Given a Poisson Process with mean λ, then the random variable X, (which equals the distance or time between successive occurrences within a Poisson Process), has an Eponential Distribution with parameter λ such that; Probability Density Function f() =λ e λ for 0 < Mean: μ = E(X) = 1 / λ Variance: σ 2 = V(X) = 1 / λ 2 a b P(a X b) = λ e d = e e λ λa λb P(X ) e λ λ = λ d = 1 e 0 λ λ P(X ) = λ e d = e Solving problems by the eponential approach Probability of no events in interval, use P(X > ) = e -λ. Probability of at least one event in interval, use P(X < ) = 1 - e -λ. Probability of the first event in interval, use P(X < ) = 1 - e -λ. Probability of operating at least hours (or more than hours or no less than hours), use P( X > ) = e -λ. Probability of operating less than hours (or no more than hours or at most hours), use P( X < ) = 1 - e -λ. Probability of failing within hours (or failing in less than hours or failing in the net hours) is the same as operating for less than hours, use P( X < ) = 1 - e -λ. ability of failing after hours (or failing in more than hours or of not failing within the net hours) is the same as operating for more than hours, use P( X > ) = e -λ. Note: Problems dealing with two or more events in a time interval, need to be worked using the Poisson Distribution probability formula.

3 Hints for Solving Poisson Processes and Eponential Distribution Problems 1. Suppose the number of phone calls arriving at switchboard follow a Poisson Process with an average of 3 calls per 60 minutes ( λ = 3 calls per 60 minutes). a. What is the probability that there are no calls within a 15 minute period? Poisson Approach λ = 3 calls per 60 minutes λ' / λ = New Interval / Old Interval λ' = λ (New Interval / Old Interval) λ' = 3 calls (15 minutes / 60 minutes) = 0.75 calls per 15 minutes Probability of no calls = P(X = 0) X e λ λ e (0.75) P(X = ) = P(X = 0) = = e = ! 0! Eponential Approach Although the problem is stated in terms of a Poisson Process (calls per hour), we can also address the question as a time interval problem using the eponential distribution. "What is the probability of no calls within 15 minutes" is same as asking "what is the probability that the net phone call will not occur until at least 15 minutes has elapsed"; that is to ask - what is P(X > 15 minutes)? P(X > ) = e λ P(X > 15) = e -(3/60)15 = e = b. What is the probability that at least one call arrives within a 25 minute period? Poisson Approach λ = 3 calls per 60 minutes λ' / λ = New Interval / Old Interval λ' = λ (New Interval / Old Interval) λ' = 3 calls (25 minutes / 60 minutes) = 1.25 calls per 25 minutes Probability of at least one = P(X 1) = 1 - P(X < 1) = 1 - P(X = 0) 1-0 e λ λ P(X = 0) = 1 = 0! e λ 1 = 1 - e = = ! Eponential Approach "What is probability of at least one call within 25 minutes: is the same as asking "what is the probability of a call arriving in less than 25 minutes"; that is to ask, what is P(X < 25 minutes)? P(X < ) = 1 e λ = 1 - e -(3/60)25 = 1 - e = =

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5 Additional Hints for Solving Poisson and Eponential Distribution Problems Poisson Processes deal with the number of events per interval. Eponential Distributions deal with the length of the interval between events. Intervals can be any dimension: time, length, area, volume. The following eamples use time. Solving problems by the eponential approach Probability of no events in interval, use P(X > ) = e -λ. Probability of at least one event in interval, use P(X < ) = 1 - e -λ. Probability of the first event in interval, use P(X < ) = 1 - e -λ. Probability of operating at least hours (or more than hours or no less than hours), use P( X > ) = e -λ. Probability of operating less than hours (or no more than hours or at most hours), use P( X < ) = 1 - e -λ. Probability of failing within hours (or failing in less than hours or failing in the net hours) is the same as operating for less than hours, use P( X < ) = 1 - e -λ. Probability of failing after hours (or failing in more than hours or of not failing within the net hours) is the same as operating for more than hours, use P( X > ) = e -λ. Note: Problems dealing with two or more events in hours, need to be worked using the Poisson Distribution probability formula.

6 Even More Additional Hints for Solving Eponential Distribution Problems Eponential Distributions deal with the length of the interval between events. Intervals can be any dimension: time, length, area, volume. The following eamples use time. Probability of no events in interval use P(X > ) Probability of the first event in interval use P(X < ) Probability of at least one event in interval use P(X < ) Probability of operating at least hours use P(X > ) Probability of operating no less than hours use P(X > ) Probability of operating for more than hours use P(X > ) Probability of operating less than hours use P(X < ) Probability of operating at most hours use P(X < ) Probability of operating no more than hours use P(X < ) Probability of failing in less than hours use P(X < ) Probability of failing within hours use P(X < ) Probability of failing in the net hours use P(X < ) Probability of failing before hours use P(X < ) Probability of failing after hours use P(X > ) Probability of failing in more than hours use P(X > ) Probability of not failing in the net hours use P(X > ) Probability of not failing within the net hours use P(X > ) Probability of not failing in less than hours use P(X > ) P(X > ) = e -λ P(X < ) = 1 - e -λ P(a < X < b) = e -λa - e -λb P(X < (t 1 + t 2 ) X > t 1 ) = P(X < t 2 ) P(X > (t 1 + t 2 ) X > t 1 ) = P(X > t 2 ) To find t, given P( < t) = y To find t, given P( > t) = y ln(1 y) t = λ ln(y) t = λ