1. Busy people arrive at the park according to a Poisson process with rate λ1=3/hour and stay in the park for exactly 1/6 of an hour. Relaxed people arrive at the park according to a Poisson process with rate λ2=2/hour and stay in the park for exactly half an hour. The arrivals of busy and relaxed people are independent processes. Assume that no other people arrive at the park. Is the process of total arrivals at the park a Poisson process? If yes, enter the rate of that process in the answer box below. If it is not, enter 0.
2. Whenever a relaxed person exits the park, he/she enters a nearby coffee shop. (Assume, for simplicity, that going from the park to the coffee shop takes zero time.) Is the process of arrivals of relaxed persons at the coffee shop a Poisson process? If yes, enter the rate of that process in the answer box below. If it is not, enter 0.
use this. surely this is correct
a) e^(-0.5)
b) 3/2*e^(-3/2)
For part 2.
Rate of total arrival is 5
Rate of relaxed persons to coffee shop is 2
a) 5
b) 2
1.
a) e^(-0.5)
b) 3/2*e^(-3/2)
2.
5
2
1. The process of total arrivals at the park is not a Poisson process. The rate of arrivals depends on two independent Poisson processes with different rates and durations of stay. Therefore, the rate of the total arrivals cannot be determined by a single value and is not constant over time.
2. The process of arrivals of relaxed persons at the coffee shop can be considered a Poisson process. Since relaxed people enter the coffee shop whenever they exit the park, their arrivals at the coffee shop are independent and follow a Poisson distribution. The rate of this process can be calculated based on the rate of arrivals of relaxed people at the park and the duration of their stay. However, without knowing the specific rate, it is not possible to provide an exact answer.
To determine whether the processes described are Poisson processes, we need to check if they satisfy the properties of a Poisson process.
1. For the total arrivals at the park to be a Poisson process, the inter-arrival times must be exponentially distributed and independent. Let's consider the arrivals of busy people and relaxed people separately.
- The arrivals of busy people follow a Poisson process with rate λ1 = 3/hour. This means that the inter-arrival times for busy people are exponentially distributed with parameter λ1.
- The arrivals of relaxed people follow a Poisson process with rate λ2 = 2/hour. This means that the inter-arrival times for relaxed people are exponentially distributed with parameter λ2.
Since the arrivals of busy and relaxed people are independent, the inter-arrival times for each group are independent as well.
To determine whether the total arrivals at the park form a Poisson process, we need to consider the combined inter-arrival times of busy and relaxed people. Since the inter-arrival times for the two groups are independent, the total inter-arrival times will be the sum of the inter-arrival times for each group.
- The inter-arrival times for busy people are 1/6 hour.
- The inter-arrival times for relaxed people are 1/2 hour.
To determine if the total arrivals form a Poisson process, we need to check if the sum of these inter-arrival times is exponentially distributed with some rate parameter.
1/6 + 1/2 = 5/6
The sum of the inter-arrival times, 5/6 hour, is not a constant rate but a fixed duration. Therefore, the process of total arrivals at the park is not a Poisson process. The answer is 0.
2. For the arrivals of relaxed persons at the coffee shop to be a Poisson process, the same properties of a Poisson process need to be satisfied: exponentially distributed inter-arrival times and independence.
Since the relaxed people leave the park and immediately enter the coffee shop, the inter-arrival times at the coffee shop will be the same as the departure times from the park for relaxed people, which are fixed at 1/2 hour.
Since the inter-arrival times are not exponentially distributed, the process of arrivals of relaxed persons at the coffee shop is not a Poisson process. The answer is 0.
a/ e^(-3)
b/ 3/2*e^(-3/2)