Stochastic Processes

Study guide: Exam MAS-I

A Stochastic Process \{X(t) : t \in T\} is a collection of Random Variables indexed by a parameter t (usually time), describing how a random quantity — such as an insurer's cumulative claim count or surplus — evolves over an index set T.

\{X(t) : t \in T\}

• T is the index set: a discrete-time process has t = 0, 1, 2, \ldots, while a continuous-time process has t \geq 0. A Time Series — values of a variable recorded at equally spaced times — is a common discrete-time example
• The set of values X(t) can take is the state space, likewise discrete or continuous
• A counting process N(t) records the number of events by time t; the Poisson Process is the central continuous-time example
• A process has the Markov property when the future depends on the past only through the present state:

P\big(X(t_{n+1}) = j \mid X(t_n), \ldots, X(t_0)\big) = P\big(X(t_{n+1}) = j \mid X(t_n)\big)

• For a Poisson counting process the increments are independent and stationary, giving N(t) - N(s) \sim \text{Poi}(\lambda(t-s)) for s < t

Cumulative Claims as a Counting Process Claims hit a portfolio as a Poisson Process at rate \lambda = 4 per month. Let N(t) be the cumulative number of claims by month t. Classify the process and find the expected count and variance over the first half-year.

Answer N(t) is a continuous-time (t \geq 0), discrete-state (0, 1, 2, \ldots) counting process. Over t = 6 months: N(6) \sim \text{Poi}(4 \times 6) = \text{Poi}(24) E[N(6)] = \text{Var}(N(6)) = 24 On average the insurer expects 24 claims in six months.

A Two-State Risk-Class Markov Chain Each year an insured is classed Standard (S) or Preferred (P). The process is a discrete-time Markov Chain with one-step transition matrix \begin{array}{c|cc} & S & P \\ \hline S & 0.8 & 0.2 \\ P & 0.3 & 0.7 \end{array} An insured is Standard this year. Find the probability they are Preferred two years from now.

Answer Condition on the class next year (the Markov property lets us ignore earlier history):

\begin{align*}
P(X_2 = P \mid X_0 = S) &= P_{SS}P_{SP} + P_{SP}P_{PP} \\
  &= (0.8)(0.2) + (0.2)(0.7) \\
  &= 0.16 + 0.14 \\
  &= 0.30
\end{align*}

There is a 30% chance the insured is Preferred after two years.