Class 6

Counting Events in Continuous Time

Defaults, insurance claims, order arrivals and price jumps happen at unpredictable moments. To model them, split [0, t] into n slots of length \Delta t = t / n and let an event occur in each slot with probability \lambda \Delta t, independently across slots. The count up to time t is binomial with mean \lambda t, and its PGF converges to a Poisson PGF, \left(1 + \frac{\lambda t (z - 1)}{n}\right)^{n} \xrightarrow[n \to \infty]{} e^{\lambda t (z - 1)}. Events in disjoint intervals come from disjoint slots, so they are independent. A Poisson process with rate \lambda > 0 is a process \{N_{t}\} with N_{0} = 0, independent increments, and N_{t} - N_{s} \sim \text{Poisson}(\lambda (t - s)) for s < t. Its paths are step functions that jump up by one at each event, and \operatorname{E}(N_{t}) = \operatorname{V}(N_{t}) = \lambda t, \qquad \operatorname{P}(\mathop{}\!\mathrm{d}N_{t} = 1) = \lambda \mathop{}\!\mathrm{d}t+ o(\mathop{}\!\mathrm{d}t), \qquad \operatorname{P}(\mathop{}\!\mathrm{d}N_{t} \geq 2) = o(\mathop{}\!\mathrm{d}t). \tag{1} Over a short interval there is either no event or a single one, with probability proportional to the length of the interval.

Waiting Times

The first event comes after t exactly when no event occurs in [0, t], so \operatorname{P}(T_{1} > t) = \operatorname{P}(N_{t} = 0) = e^{-\lambda t}. The times T_{1}, T_{2}, \ldots between consecutive events are iid \text{Exp}(\lambda), and conversely adding up iid exponential waiting times builds a Poisson process. The exponential distribution is memoryless, \operatorname{P}(T > s + t \mid T > s) = e^{-\lambda t} = \operatorname{P}(T > t), so time already spent waiting tells us nothing about the remaining wait. In credit risk, \lambda is the hazard rate: given survival to t, a firm defaults in (t, t + \mathop{}\!\mathrm{d}t] with probability \lambda \mathop{}\!\mathrm{d}t+ o(\mathop{}\!\mathrm{d}t).

The time of the n-th event, W_{n} = T_{1} + \cdots + T_{n}, has a gamma distribution with mean n / \lambda and variance n / \lambda^{2}. Counts and arrival times describe the same paths, as Figure 1 shows, and \operatorname{P}(W_{n} \leq t) = \operatorname{P}(N_{t} \geq n) = 1 - \sum_{k = 0}^{n - 1} e^{-\lambda t} \frac{(\lambda t)^{k}}{k!}. \tag{2}

Figure 1: One simulated path of a Poisson process with rate \lambda = 1. The arrival times W_{n} are marked on the time axis, and the arrows below the path show the exponential interarrival times T_{n} = W_{n} - W_{n - 1}.

Martingales and Quadratic Variation

Let \mathcal{F}_{t} be the information generated by the path up to time t. Increments are independent of \mathcal{F}_{s}, so conditional expectations of future increments are ordinary expectations. N_{t} only moves up, but removing its expected growth gives three martingales: M_{t} = N_{t} - \lambda t, \qquad M_{t}^{2} - \lambda t, \qquad Z_{t} = \exp\left(\theta N_{t} - \lambda t (e^{\theta} - 1)\right). \tag{3} They mirror S_{n}, S_{n}^{2} - n and e^{\theta S_{n} - n K(\theta)} for the random walk, with K(\theta) = \lambda (e^{\theta} - 1) the CGF of N_{1}.

On a fine partition each subinterval holds at most one event, so each squared increment is 0 or 1 and equals the increment itself. The quadratic variation counts the jumps, [N, N]_{t} = [M, M]_{t} = N_{t}, or (\mathop{}\!\mathrm{d}N)^{2} = \mathop{}\!\mathrm{d}N. It is random, and its expectation \lambda t equals \operatorname{V}(M_{t}).

Superposition and Thinning

Independent Poisson processes with rates \lambda_{1} and \lambda_{2} add up to a Poisson process with rate \lambda_{1} + \lambda_{2}. Conversely, if each event is independently of type A with probability p, the type A and type B counts are independent Poisson processes with rates \lambda p and \lambda (1 - p). Many type A events signal a large total, which offsets the smaller share left for type B.

Compound Poisson Processes

When each event carries an iid size Y_{i}, independent of N_{t}, the process X_{t} = \sum_{i = 1}^{N_{t}} Y_{i} is a compound Poisson process. Conditioning on N_{t} gives \phi_{X_{t}}(u) = e^{\lambda t (\phi_{Y}(u) - 1)}, \qquad \operatorname{E}(X_{t}) = \lambda t \operatorname{E}(Y), \qquad \operatorname{V}(X_{t}) = \lambda t \operatorname{E}(Y^{2}), \tag{4} and M_{X_{t}}(\theta) = e^{\lambda t (M_{Y}(\theta) - 1)} when the MGF exists. The variance involves \operatorname{E}(Y^{2}) because both the number and the size of the jumps are random.

Credit Spreads and Insurance Ruin

If a firm defaults at the first event of a Poisson process, the default time \tau is \text{Exp}(\lambda). A zero-coupon bond paying one dollar at T if the firm survives, valued at its expected payoff discounted at the risk-free rate r, is worth P^{d}(T) = e^{-r T} \operatorname{P}(\tau > T) = e^{-(r + \lambda) T}, \tag{5} so the credit spread equals the default intensity.

An insurer with capital u collects premiums at rate c and pays compound Poisson claims, so its surplus is U_{t} = u + c t - X_{t}, with c > \lambda \operatorname{E}(Y). If the adjustment coefficient R > 0 solves \lambda (M_{Y}(R) - 1) = c R, \tag{6} then e^{-R U_{t}} is a martingale, and stopping it at the ruin time gives Lundberg’s inequality \psi(u) \leq e^{-R u}. The surplus jumps below zero by an unknown amount, so the argument yields a bound rather than an exact probability. For exponential claims with mean 1 / \eta, R = \eta - \lambda / c and the exact ruin probability is \psi(u) = \frac{\lambda}{c \eta} e^{-R u}.

Practice Problems

1. Trades in a thinly traded corporate bond arrive as a Poisson process with rate \lambda = 4 per hour.

   1. Compute the probability that no trade occurs during a 30-minute interval.
   2. A dealer has waited 20 minutes since the last trade. What is the expected additional waiting time until the next trade, and the probability that no trade occurs in the next 15 minutes?
   3. Compute the expected value and the standard deviation of the time until the third trade, and the probability that the third trade occurs within 30 minutes.
   4. Each trade is independently a customer buy with probability 0.55 and a customer sell otherwise. Compute the probability that an hour passes without a sell. Given that five buys occur during an hour, what is the expected number of sells during that hour?

2. A firm’s three-year zero-coupon bond has a continuously compounded yield of 5.5%, and the risk-free rate is 4%. Default arrives at the first event of a Poisson process, and bondholders receive nothing in default.

   1. Use (5) to find the default intensity \lambda implied by the bond price.
   2. Compute the probability that the firm survives three years, and the probability that it defaults during the third year given that it survives the first two.
   3. A second firm has a default intensity of 0.025 per year, independent of the first. A contract pays one dollar in three years if neither firm has defaulted. Use superposition to price it, and find the spread of its yield over the risk-free rate.

3. An insurer receives claims as a Poisson process with rate \lambda = 20 per year. Claim sizes are exponential with a mean of $0.5 million (\eta = 2), independent of each other and of the arrival times. Premiums arrive at c = \$12 million per year.

   1. Use (4) to compute the expected value and the standard deviation of total claims over one year. By what percentage do premiums exceed expected claims?
   2. Solve (6) for the adjustment coefficient R.
   3. How much initial capital u makes the Lundberg bound on the ruin probability equal to 1%? What is the exact ruin probability at that capital?