Question

A Poisson process \(N(t)\) with intensity \(\lambda(t)\) is a counting process with \- independent increments and such that \- \(N(t)-N(s)\) follows a Poisson distribution with parameter \(\int_{s}^{t} \lambda(u) d u\) for all \(0 \leq s \leq t\). (a) Find the compensator \(\Lambda\) of \(N\) and put \(M=N-\Lambda\). Show by a direct calculation that \(E\left(M(t) \mid \mathcal{F}_{s}\right)=M(s)\), where \(\mathcal{F}_{t}\) is the internal history \(N\). Is \(M\) a local square integrable martingale? (b) Find the compensator of \(M^{2}\).

Short answer

Short Answer: The compensator Λ of the Poisson process N(t) with intensity λ(t) is given by $$\Lambda(t) = \int_{0}^{t} \lambda(u) du$$. We define the process M as $$M(t) = N(t) - \Lambda(t)$$. It has been shown that M is a local square integrable martingale, as E(M(t) | 𝔽_s) = M(s), it has independent increments, and $$E(M(t)^2 | \mathcal{F}_s) = M(s)^2 + O(t-s)$$. The compensator of M^2 is given by $$\Lambda_{M^2}(t) = \operatorname{E}\left(M(t)^2 \right) = M(s)^2 + \int_{s}^{t} \lambda(u) du$$.

Step-by-step solution

Finding the compensator Λ

We know that the compensator Λ is given by: $$ \Lambda(t) = \int_{0}^{t} \lambda(u) du $$

Defining M

Now, the process M is defined as $$ M(t) = N(t) - \Lambda(t) $$

Showing E(M(t) | 𝔽_s) = M(s)

We need to calculate the conditional expectation of M(t) given 𝔽_s. By definition, the conditional expectation E(M(t) | 𝔽_s) is another random variable that satisfies: $$ E(M(t) - M(s) | 𝔽_{s}) = 0 $$ since M(t) - M(s) has mean zero. Now, let's substitute M(t) and M(s) with their expressions: $$ E((N(t) - \Lambda(t)) - (N(s) - \Lambda(s)) | 𝔽_{s}) = E((N(t) - N(s)) - (\Lambda(t) - \Lambda(s)) | 𝔽_{s}) $$ We know that N(t) - N(s) follows a Poisson distribution with parameter $$\int_{s}^{t} \lambda(u) du$$ and is independent of 𝔽_s. So, the conditional expectation becomes: $$ E(N(t) - N(s) | 𝔽_{s}) - E(\Lambda(t) - \Lambda(s) | 𝔽_{s}) = \int_{s}^{t} \lambda(u) du - \int_{s}^{t} \lambda(u) du = 0 $$ This proves that E(M(t) | 𝔽_s) = M(s).

Checking if M is a local square integrable martingale

It suffices to check if M has independent increments and if $$ E(M(t)^2 | \mathcal{F}_s) = M(s)^2 + O(t-s) $$ By the definition of M and the properties of Poisson processes, we see that M has independent increments. To verify the second condition, we’ll calculate $$ \operatorname{E}\left((M(t)-M(s))^{2} \mid \mathcal{F}_{s}\right) $$ Since this is a martingale, the expectation should be equivalent to $$ \operatorname{E}\left((M(t)-M(s))^{2}\right) $$ Now, substituting M(t) - M(s) with the expression we found earlier, we get: $$ \operatorname{E}\left((M(t)-M(s))^{2}\right) = \operatorname{E}\left((N(t)- N(s) - \Lambda(t) + \Lambda(s))^{2}\right) $$ By the property of the Poisson process, this expectation can be simplified as $$ \operatorname{E}\left((N(t)- N(s) - \Lambda(t) + \Lambda(s))^{2}\right) = \int_{s}^{t} \lambda(u) du $$ Moreover, M(s) = 0, hence $$ \operatorname{E}\left(M(t)^{2} \mid \mathcal{F}_s\right) = \int_{s}^{t} \lambda(u) du = M(s)^2 + \left( \int_{s}^{t} \lambda(u) du\right) $$ Thus, M is a local square integrable martingale. #Phase 2: Working on part (b)#

Finding the compensator of M^2

To find the compensator of M^2, we can use the Doob-Meyer decomposition, which states that a local square integrable martingale M can be decomposed into a martingale component and a compensator (predictable process) Λ_M. Moreover, $$ \Lambda_{M^2}(t) = \operatorname{E}\left(M(t)^2 \right) = M(s)^2 + \int_{s}^{t} \lambda(u) du $$ Since M is a local square integrable martingale, we found the compensator of M^2.

Key concepts

Intensity

In the context of a Poisson process, the term intensity refers to the rate at which events occur. It's a function, usually denoted as \(lambda(t)\), that indicates the expected number of events occurring in a small interval of time. The idea is that if you look at a very short time span \(dt\), the probability of an event happening during that time is \(lambda(t)dt\).

For a Poisson process, the intensity function plays a crucial role because it essentially characterizes the process. The number of events occurring between time \(s\) and \(t\) follows a Poisson distribution with parameter \(\int_{s}^{t}lambda(u) du\), which is the cumulative intensity over that period. In simpler terms, the higher the intensity at a given time, the higher the chances that more events will take place around that time.

Compensator

The compensator, often represented as \(Lambda(t)\), acts as a correcting mechanism in a Poisson process. It transforms the process into a martingale, which is a type of stochastic process that models a fair game. The compensator is essentially the cumulative intensity function, obtained by integrating the intensity function over time.

To illustrate, for a Poisson process with an intensity function \(lambda(t)\), the compensator is given by \(Lambda(t) = \int_{0}^{t} lambda(u) du\). This cumulative function accounts for the expected number of events up to time \(t\), making the adjusted process \(M(t) = N(t) - Lambda(t)\) centered and without drift. This adjusted process, with its mean 'compensated' to zero, has properties similar to a martingale, which we'll cover next.

Martingale

A martingale is a type of stochastic process that has no predictable trends or patterns in its future evolution. One way to think about a martingale is as a fair betting game, where the expected winnings at any future time are always equal to the current fortune.

In technical terms, for a process \(X(t)\) to be a martingale with respect to some filtration \(mathcal{F}_t\), which is the history or information up to time \(t\), it must satisfy \(E(X(t) \mid mathcal{F}_s) = X(s)\) for all \(s < t\). This means that the conditional expectation of the process at a future time \(t\), given all current and past information, is simply the current value of the process at time \(s\).

In the context of the exercise problem, after compensating the Poisson process by subtracting its compensator, the resultant process \(M(t)\) becomes a martingale. The calculation shows that the expected value of \(M(t)\) at a future time, given the past history, is equal to its current value, reinforcing the martingale property.

Conditional Expectation

The concept of conditional expectation, denoted as \(E(X(t) \mid mathcal{F}_s)\), is a crucial notion in the theory of stochastic processes. It represents the expected value of a random variable \(X(t)\) at some future time \(t\), given the information up to an earlier time \(s\). Essentially, it is an average that takes into account the information we have up to the present moment.

Returning to our Poisson process example, the conditional expectation is used to demonstrate that \(M(t)\), the process obtained after subtracting the compensator from the original process, has properties of a martingale. This is shown by the fact that \(E(M(t) - M(s) \mid mathcal{F}_{s}) = 0\), meaning the expected change in the process, given the information up to time \(s\), is zero.

Doob-Meyer Decomposition

The Doob-Meyer decomposition is a theorem in probability theory that allows us to separate a submartingale into two distinct components: a martingale and an increasing predictable process known as the compensator. This decomposition is especially useful for demonstrating the properties of different types of stochastic processes.

In the given exercise, we're concerned with finding the compensator of \(M^2\), where \(M\) is a martingale constructed from a Poisson process. The Doob-Meyer decomposition is applied to express \(M^2\) as the sum of a martingale and its compensator. This compensator represents the cumulative predictable change in \(M^2\), accounting for the trend in the variance of the martingale over time, and is crucial in demonstrating that \(M\) itself is a local square integrable martingale.