Example: 1. You have observed that the number of hits to your web site follow a Poisson distribution at a rate of 2 per day.

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1 16 The Exponential Distribution Example: 1. You have observed that the number of hits to your web site follow a Poisson distribution at a rate of 2 per day. Let T be the time (in days) between hits. 2. You observe the number of calls that arrive each day over a period of a year, and note that the arrivals follow a Poisson distribution with an average of 3 per day. Let T be the waiting time between calls. 3. Records show that job submissions to a busy computer centre have a Poisson distribution with an average of 4 per minute. Let T be the time in minutes between submissions. 4. Records indicate that messages arrive to a computer server following a Poisson distribution at the rate of 6 per hour. Let T be the time in hours that elapses between messages.

You observe the number of calls that arrive each day over a period of a year, and note that the arrivals follow a Poisson distribution with an average of 3 per day.

2 Probability Density Function Of Waiting Times Generally the exponential distribution describes waiting time between Poisson occurrences Proof: Let T = time that elapses after a Poisson event. P(T > t) = probability that no event occurred in the time interval of length t. The probability that no Poisson event occurred in the time interval [, t]: P(, t) = e λt. where λ is the average Poisson occurrence rate in a unit time interval. So: P(T > t) = e λt, Hence the CDF is: F(t) = P(T t) = 1 e λt and, working backwards, the PDF is f(t) = F (t) = λe λt

The probability that no Poisson event occurred in the time interval [, t]: P(, t) = e λt.

3 The PDF: f(t) = λe λt, t > = otherwise The CDF: P(T t) = F(t) = t λe λt dt = [ e λt ] t T= = e λt + 1 i.e. F(t) = 1 e λt, t > = otherwise.

4 Example: If jobs arrive every 15 seconds on average, λ = 4 per minute, what is the probability of waiting less than or equal to 3 seconds, i.e.5 min? P(T.5). dexp(x, 4) P (X <=.5) x P(T.5) =.5 4e 4t dt = [ e 4t].5 t= = 1 e 2 =.86 From R pexp(.5,4) [1]

seconds, i.e.5 min? P(T.5). dexp(x, 4) 1 2 3 4 P (X <=.5)..5 1. 1.5 2.

5 What is the maximum waiting time between two job submissions with 95% confidence? We need to find k so that This is the quantile function. qexp(.95, 4) [1] P(T k) =.95 The probability that there will be.74 min, about 45 seconds, between two job submissions is.95. Applications of the Exponential Distribution: 1. Time between telephone calls 2. Time between machine breakdowns 3. Time between successive job arrivals at a computing centre

95 The probability that there will be.74 min, about 45 seconds, between two job submissions is.95. Applications of the Exponential Distribution: 1.

6 Example Accidents occur with a Poisson distribution at an average of 4 per week. i.e. λ = 4 1. Calculate the probability of more than 5 accidents in any one week 2. What is the probability that at least two weeks will elapse between accident? Solution 1. Poisson: In R 1-ppois(5, 4) [1] Exponential: P(X > 5) = 1 P(X 5) P(Time between occurrences > 2) = 2 λe λt dt In R = [ e λt ] T=2 = e 8 =.34 pexp(2, 4) [1] pexp(2, 4) [1]

What is the probability that at least two weeks will elapse between accident? Solution 1.

7 Density of the Exponential Distribution with λ = 2, 3, 4 and 6 lambda = 2 lambda = 3 dexp(x, 2) dexp(x, 3) x x lambda = 4 lambda = 6 dexp(x, 4) 2 4 dexp(x, 6) x x par(mfrow = c(2,2)) curve(dexp(x, 2),, 3, main ="lambda = 2") curve(dexp(x, 3),, 3, main ="lambda = 3") curve(dexp(x, 4),, 3, main ="lambda = 4") curve(dexp(x, 6),, 3, main ="lambda = 6")

8 The Markov Property of Exponential Examples: 1. The distribution of the remaining life does not depend on how long the component has been operating. i.e. the component does not age - its breakdown is a result of some sudden failure not a gradual deterioration 2. Time between telephone calls Waiting time for a call is independent of how long we have been waiting

operating. i.e. the component does not age - its breakdown is a result of some sudden

9 The Markov property Show: P(T x + t T > t) = P(T x) Proof: E 1 = T x + t, and E 2 = T > t Then: Now P(E 1 E 2 ) = P(E 1 E 2 ) P(E 2 ) P(E 1 E 2 ) = P(t < T x + t) = x+t t λe λt dt = e λt [1 e λx ] and thus now P(E 2 ) = t λe λt dt = e λt P(E 1 E 2 ) = e λt [1 e λx ] e λt = 1 e λx 1 e λx = F(x) = P(T x)

10 Examples 1. Calls arrive at an average rate of 12 per hour. Find the probability that a call will occur in the next 5 minutes given that you have already waited 1 minutes for a call i.e. Find P(T 15 T > 1) From the Markov property P(T 15 T > 1) = P(T 5) So: P(T 5) = 1 e 5λ The average rate of telephone calls is λ = 2 in a minute, then P(T 5) = 1 e (5)( 2) = 1 e 1 = 1.37 =.63 In R > pexp(5,.2) [1]

11 Examples 2. The average rate of job submissions in a busy computer centre is 4 per minute. If it can be assumed that the number of submissions per minute interval is Poisson distributed, calculate the probability that: (a) at least 15 seconds will elapse between any two jobs. (b) less than 1 minutes will elapse between jobs. (c) If no jobs have arrived in the last 3 seconds, what is the probability that a job will arrive in the next 15 seconds? Solution λ = 4 per minute (a) P(t > 15 sec.) = P(T >.25 min) =.25 λe λt dt = [ e λt ] T=.25 = e 1 =.37 In R > 1- pexp(.25, 4) [1]

seconds will elapse between any two jobs. (b) less than 1 minutes will elapse between jobs.

12 Mean of the Exponential Distribution Recall that when X is continuous: E(X) = x xf(x)dx For the exponential distribution: E(T) = tλe λt dt Integration by parts: udv = (uv) tλe λt dt = udv Trick is to spot the u and v: vdu Take and u = t dv = λe λt dt which gives v = e λt Then tλe λt dt = ( te λt) + e λt dt = ( te λt λ 1 e λt) = 1 λ

udv = (uv) tλe λt dt = udv Trick is to spot the u and v: vdu Take and u = t dv = λe

13 R Functions for the Exponential Distribution Density Function: dexp dexp(1, 4) Cumulative Distribution Function pexp pexp(5, 4) P(T 5) with λ = 4 Quantile Function qexp qexp(.95, 4) Choose k so that > qexp(.95, 4) 1] P(T k).95.

pexp pexp(5, 4) P(T 5) with λ = 4 Quantile Function qexp

Homework set 4 - Solutions

Homework set 4 - Solutions Math 495 Renato Feres Problems R for continuous time Markov chains The sequence of random variables of a Markov chain may represent the states of a random system recorded at