Poisson Process

Study guide: Exam MAS-I

A Poisson Process \{N(t), t \geq 0\} is a counting process that models the number of events occurring in a time interval when events arrive independently at a constant average rate \lambda > 0.

P(N(t) = k) = \frac{e^{-\lambda t}(\lambda t)^k}{k!}

k = 0, 1, 2, \ldots

E[N(t)] = \text{Var}(N(t)) = \lambda t

• The number of events in non-overlapping intervals are independent
• The number of events in any interval of length t has distribution \text{Poi}(\lambda t)
• Interarrival times T_1, T_2, \ldots are i.i.d. \text{Exponential}(\lambda) with mean 1/\lambda
• The waiting time S_n = T_1 + \cdots + T_n to the n-th event follows a Gamma distribution with shape n and rate \lambda
• A Poisson process with rate \lambda thinned by probability p produces a Poisson process with rate \lambda p
• Superposition: two independent Poisson processes with rates \lambda_1 and \lambda_2 combine into a Poisson process with rate \lambda_1 + \lambda_2

Expected Number of Claims and Probability of More Than 5 Claims arrive at rate \lambda = 2 per month. Find the expected number of claims and the probability of more than 5 claims in 3 months.

Answer Over t = 3 months, N(3) \sim \text{Poi}(2 \times 3) = \text{Poi}(6). E[N(3)] = 6, \quad \text{Var}(N(3)) = 6 P(N(3) > 5) = 1 - P(N(3) \leq 5) = 1 - \sum_{k=0}^{5} \frac{e^{-6}6^k}{k!} \approx 1 - 0.4457 = 0.5543

Thinning a Poisson Process Large claims arrive at rate \lambda = 10 per year. Each claim independently has a 30% chance of exceeding the deductible. Find the rate of claims that exceed the deductible.

Answer By the thinning property, claims exceeding the deductible form a Poisson process with rate \lambda p = 10 \times 0.30 = 3 per year.