Question

The number of arrivals at a store is modeled as a non-homogeneous Poisson process with a time-varying rate function given by: λ(t) = 2t + 1, for 0 ≤ t ≤ 3 λ(t) = 10, for t > 3 What is the expected number of arrivals between t = 1 and t = 4? What is the probability that there are no arrivals between t = 2 and t = 3? If the store has already had 5 arrivals by time t = 2, what is the probability that there are no arrivals between t = 2 and t = 4?

Answer 1

The expected number of arrivals between t = 1 and t = 4 is 20. The probability of no arrivals between t = 2 and t = 3 is approximately 0.0067. With 5 arrivals by t = 2, the probability of no arrivals from t = 2 to t = 4 is approximately 3.06 × 10^-7.

Step-by-step explanation:

Expected Number of Arrivals and Probabilities in Poisson Processes

For the expected number of arrivals between t = 1 and t = 4, we need to integrate the rate function λ(t) over the interval. This involves two sections:

1. From t = 1 to t = 3:

∫(2t + 1) dt from t = 1 to t = 3 = [(t^2 + t)] from t = 1 to t = 3 = (9 + 3) - (1 + 1) = 10.

1. From t = 3 to t = 4:

∫10 dt from t = 3 to t = 4 = 10.

Total expected arrivals = 10 + 10 = 20 arrivals.

For the probability of no arrivals between t = 2 and t = 3, we use the Poisson distribution with parameter λ = ∫(2t + 1) dt = 5. The formula for the Poisson probability is:

P(X = k) = λ^k * e^(-λ) / k!, where k is the number of occurrences. So, P(X = 0) = 5^0 * e^(-5) / 0! = e^(-5), which is approximately 0.0067.

If there have already been 5 arrivals by t = 2, the probability of no arrivals between t = 2 and t = 4 is the product of probabilities of no arrivals from t=2 to t=3 and from t=3 to t=4. We have already calculated the probability from t=2 to t=3 to be e^(-5). From t=3 to t=4, λ = 10, so P(X = 0) = 10^0 * e^(-10) / 0! = e^(-10). The combined probability is e^(-5)*e^(-10) = e^(-15), approximately 3.06 × 10^-7.