Let N(t) be a Poisson process with rate l and let Y1, Y2, ..., Yn be independent and identically distributed random variables with mean m and standard deviation s . Assume that the Yi's are independent of N(t). A process X(t) is said to be a compound Poisson process if
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The process X(t) is a random sum of random variables since the number of terms in the sum, N(t), is itself random.
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X(t) is a continuous-time stochastic process
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If the Yi's are continuous random variables, then X(t) has continuous states
Refer to Example 8.4-3 on page 343. Note that the values given in the question below are different.
The number of students using a check-cashing service at the Student Union is a Poisson process with a a rate of l =40 per hour. The amount of each check is a random variable with a mean of $25 and a standard deviation of $6. Complete the following table.
Round all your answers to the nearest dollar.
E[ X(t) ] |
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VAR[ X(t) ] |
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STD [ X(t) ] |
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Solution
Refer to example 8.4-2 on page 344. Note that the values given in the question below are different.
The number of cars comming to a parking garage is a Poisson process with a rate of 20 per hour. The amount charged for parking depends on the time the car spends in the garage. The mean amount is $5.75 and the standard deviation is $1.35. Find the expected value, variance, and standard deviation for the toatal amount of money collected over a 12 hour period.
E[ X(t) ] |
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VAR[ X(t) ] |
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STD [ X(t) ] |
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Solution
Consider Exercise 8.4-4 on page 344. The number of boats for which a certain drawbridge must be raised is a Poisson process with a rate of 0.6 per hour. The average time that the bridge is up is 10 minutes with a standard deviation of 3 minutes. The bridge is raised and lowered individually for each boat. What is the probability that the bridge is up more than 60 minutes in an 12 hour period?
Solution:
Begin by converting all values so they are in terms of hours.
10 minutes = 10/60 hours. So m =1/6 hours
3 minutes = 3/60 hours. So s =1/20 hours
The time that the bridge is up in a 12-hour period is approximately normally distributed with a mean of
E[ X(t) ] = ltm = (0.6)(12)(1/6) =0.12
VAR[ X(t) ] = lt (m^2 + s ^2) = (0.6)(12)[ (1/6)^2 + (1/20)^2] =0.218
STD[ X(t) ] = sqrt( 0.218)= 0.47
P [ X (12) >1 ) ] = P( Z > (1-1.2)/ 0.47)
= 1- P( Z < (1-1.2)/ 0.47)
=1-P( Z < -0.43 )
=0.6664
Consider the following variation to the problem.The number of boats for which the drawbridge must be raised is a Poisson process with a rate of 0.9 per hour. The average time that the bridge is up is 12 minutes with a standard deviation of 4 minutes. The bridge is raised and lowered individually for each boat. What is the probability that the bridge is up more than 60 minutes in an 8 hour period?
E[ X(t) ] |
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VAR[ X(t) ] |
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STD [ X(t) ] |
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P(X (8) >1) |
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Solution
Solutions
Solution to Example 8.4-3
The number of students using a check-cashing service at the Student Union is a Poisson process with a a rate of l =40 per hour. The amount of each check is a random variable with a mean of $25 and a standard deviation of $6. Complete the following table.
The total dollar amount X(t) of checks cashed in an 8-hour day has a mean of
E[ X(t) ] = 40* 8 * 25= $8000
VAR[ X(t) ] = 40* 8 [ 25^2 + 6^2] =211,520
STD [ X(t) ] =sqrt ( 211,520)=459.9= 460
Solution to Example 8.4-2
The total dollar amount X(t) collected over a 12-hour day has a mean of
E[ X(t) ] = 20* 12 * 5.75= $1380
VAR[ X(t) ] = 20* 12[ 5.75^2 + 1.35^2] = $8372.40
STD [ X(t) ] =sqrt ( 8372.40)=459.9= $91.50
Solution to Exercise 8.4-4:
Begin by converting all values so they are in terms of hours.
12minutes = 12/60 hours. So m =1/5 hours
4 minutes = 4/60 hours. So s =1/15 hours
The time that the bridge is up in an 8-hour period is approximately normally distributed with a mean of
E[ X(t) ] = ltm = (0.9)(8)(1/5) =1.44
VAR[ X(t) ] = lt (m^2 + s ^2) = (0.9)(8)[ (1/5)^2 + (1/15)^2] =0.32
STD[ X(t) ] = sqrt( 0.32)= 0.57
P [ X (12) >1 ) ] = P( Z > (1-1.44)/ 0.57)
= 1- P( Z < (1-1.44)/ 0.57)
=1-P( Z < -0.477 )
=0.779350
=0.7794
