# Almost Sure

## 27 December 11

### Compensators of Counting Processes

A counting process, X, is defined to be an adapted stochastic process starting from zero which is piecewise constant and right-continuous with jumps of size 1. That is, letting ${\tau_n}$ be the first time at which ${X_t=n}$, then

$\displaystyle X_t=\sum_{n=1}^\infty 1_{\{\tau_n\le t\}}.$

By the debut theorem, ${\tau_n}$ are stopping times. So, X is an increasing integer valued process counting the arrivals of the stopping times ${\tau_n}$. A basic example of a counting process is the Poisson process, for which ${X_t-X_s}$ has a Poisson distribution independently of ${\mathcal{F}_s}$, for all times ${t > s}$, and for which the gaps ${\tau_n-\tau_{n-1}}$ between the stopping times are independent exponentially distributed random variables. As we will see, although Poisson processes are just one specific example, every quasi-left-continuous counting process can actually be reduced to the case of a Poisson process by a time change. As always, we work with respect to a complete filtered probability space ${(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\ge0},{\mathbb P})}$.

Note that, as a counting process X has jumps bounded by 1, it is locally integrable and, hence, the compensator A of X exists. This is the unique right-continuous predictable and increasing process with ${A_0=0}$ such that ${X-A}$ is a local martingale. For example, if X is a Poisson process of rate ${\lambda}$, then the compensated Poisson process ${X_t-\lambda t}$ is a martingale. So, the compensator of X is the continuous process ${A_t=\lambda t}$. More generally, X is said to be quasi-left-continuous if ${{\mathbb P}(\Delta X_\tau=0)=1}$ for all predictable stopping times ${\tau}$, which is equivalent to the compensator of X being almost surely continuous. Another simple example of a counting process is ${X=1_{[\tau,\infty)}}$ for a stopping time ${\tau > 0}$, in which case the compensator of X is just the same thing as the compensator of ${\tau}$.

As I will show in this post, compensators of quasi-left-continuous counting processes have many parallels with the quadratic variation of continuous local martingales. For example, Lévy’s characterization states that a local martingale X starting from zero is standard Brownian motion if and only if its quadratic variation is ${[X]_t=t}$. Similarly, as we show below, a counting process is a homogeneous Poisson process of rate ${\lambda}$ if and only if its compensator is ${A_t=\lambda t}$. It was also shown previously in these notes that a continuous local martingale X has a finite limit ${X_\infty=\lim_{t\rightarrow\infty}X_t}$ if and only if ${[X]_\infty}$ is finite. Similarly, a counting process X has finite value ${X_\infty}$ at infinity if and only if the same is true of its compensator. Another property of a continuous local martingale X is that it is constant over all intervals on which its quadratic variation is constant. Similarly, a counting process X is constant over any interval on which its compensator is constant. Finally, it is known that every continuous local martingale is simply a continuous time change of standard Brownian motion. In the main result of this post (Theorem 5), we show that a similar statement holds for counting processes. That is, every quasi-left-continuous counting process is a continuous time change of a Poisson process of rate 1.

We start with the following result, which can be used to calculate the characteristic function of a quasi-left-continuous counting process in terms of its compensator. This result is similar to the fact that, for a continuous local martingale X, the process ${\exp(aX_t-\frac12a^2[X]_t)}$ is a local martingale. In fact, in both cases, this reduces to the statement that the Doléans exponential of a local martingale is itself a local martingale.

Lemma 1 Let X be a quasi-left-continuous counting process with compensator A. Then, A is a continuous increasing process such that

 $\displaystyle \exp\left(aX_t-(e^a-1)A_t\right)$ (1)

is a local martingale for all ${a\in{\mathbb C}}$.

Furthermore, for any ${a\in{\mathbb C}}$ such that ${e^a\not=1}$, A is the unique continuous increasing process such that ${A_0=0}$ and (1) is a local martingale.

Proof: As X is quasi-left-continuous and increasing, its compensator A is a continuous increasing process with ${A_0=0}$.

Set ${Y_t=\exp(aX_t)}$. As X is a counting process, we have ${\Delta X=1}$ at each jump time of X and, hence, ${Y_t=e^aY_{t-}}$ or, equivalently, ${\Delta Y_t=(e^a-1)Y_{t-}}$. As Y is a pure jump process,

$\displaystyle Y_t-Y_0=\sum_{s\le t}\Delta Y_s=(e^a-1)\int_0^tY_{s-}\,dX_s.$

Now, set ${b=e^a-1}$ and let ${M_t}$ be the process defined by (1). Then, ${M_t=e^{-bA_t}Y_t}$. Integration by parts gives

 $\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle M_t-M_0&\displaystyle=\int_0^te^{-bA_s}\,dY_s-b\int_0^tM_{s-}\,dA_s\smallskip\\ &\displaystyle=b\int_0^te^{-bA_s}Y_{s-}\,dX_s-b\int_0^tM_{s-}\,dA_s\smallskip\\ &\displaystyle=b\int_0^tM_{s-}\,d(X_s-A_s). \end{array}$ (2)

As the processes all have finite variation, this is just the standard integration by parts formula for Lebesgue-Stieltjes integration. Alternatively, using the stochastic integration by parts formula, the fact that the processes have finite variation means that the quadratic variation term is zero. Note that equation (2) just says that M is the Doléans exponential of ${b(X-a)}$. By preservation of the local martingale property, M is a local martingale as required.

Conversely, consider any fixed ${a\in{\mathbb C}}$ such that ${e^a\not=1}$. Suppose that A is a continuous increasing process with ${A_0=0}$ and such that the process M defined by (1) is a local martingale. By (2), we have

$\displaystyle X_t-A_t=b^{-1}\int_0^tM_{s-}^{-1}\,dM_s.$

By preservation of the local martingale property, this means that ${X-A}$ is a local martingale and, hence, A is the compensator of X. $\Box$

The following result characterizes Poisson processes among the class of counting processes, and can be compared with Lévy’s characterization of Brownian motion which characterizes Brownian motion among the class of continuous local martingales.

Lemma 2 Let X be a counting process and ${\lambda}$ be a positive real number. Then, X is a Poisson process with rate ${\lambda}$ if and only if its compensator is ${A_t\equiv\lambda t}$.

Proof: It was previously shown, in the post on Poisson processes, that a counting process X is a Poisson process of rate ${\lambda}$ if and only if ${X_t-\lambda t}$ is a local martingale. Alternatively, putting ${A_t=\lambda t}$ and ${a=ib}$ in Lemma 1, then X has compensator ${\lambda t}$ if and only if

$\displaystyle M_t\equiv\exp(ibX_t-(e^{ib}-1)\lambda t)$

is a local martingale for all ${b\in{\mathbb R}}$. As this is uniformly bounded over finite intervals, it is equivalent to M being a martingale, or that ${X_t-X_s}$ has the characteristic function ${\exp((e^{ib}-1)\lambda(t-s))}$ independently of ${\mathcal{F}_s}$, for all ${t > s}$. Noting that this is the characteristic function of the Poisson distribution with parameter ${\lambda(t-s)}$ gives the result. $\Box$

Next, we show that a counting process tends to a finite limit at infinity if and only if the same is true of its compensator. This can be compared to the result that a continuous local martingale converges to a finite limit at infinity if and only if the same is true of its quadratic variation. In fact, it can be seen that the proof below applies to any increasing process X with uniformly bounded jumps.

Lemma 3 Let X be a counting process with compensator A. Then, with probability one, we have ${X_\infty < \infty}$ if and only if ${A_\infty < \infty}$.

Proof: Letting ${\tau}$ be the first time at which X hits the positive integer n, then ${X_\tau\le n}$, so we have

$\displaystyle {\mathbb E}[A_\tau]={\mathbb E}[X_\tau]\le n.$

Therefore, ${A_\tau < \infty}$ almost surely. On the event ${X_\infty < n}$, this gives ${\tau=\infty}$ and, hence, ${A_\infty < \infty}$ outside of a set of zero probability. Letting n increase to infinity shows that, almost surely, ${A_\infty < \infty}$ whenever ${X_\infty < \infty}$.

Conversely, for positive integer n, let ${\tau}$ be the first time at which ${A_t\ge n}$. Then, ${A_t < n}$ for all ${t < \tau}$. Furthermore, as X has jumps bounded by 1, the same is true of A. Therefore, ${A_\tau < n+1}$ and, as above, we get the inequality

$\displaystyle {\mathbb E}[X_\tau]={\mathbb E}[A_\tau] < n+1.$

So, ${X_\tau < \infty}$ almost surely. On the event ${A_\infty < \infty}$, this gives ${\tau=\infty}$ and, hence, ${X_\infty < \infty}$ outside of a set of zero probability. Letting n increase to infinity shows that, almost surely, ${X_\infty < \infty}$ whenever ${A_\infty < \infty}$. $\Box$

Recall that a continuous local martingale is constant on the same intervals on which its quadratic variation is constant. This result can also be modified to apply to counting processes and, more generally, to locally integrable increasing processes. That is, an increasing process X is constant on any interval for which its compensator is constant. Note that the converse does not hold here. For example, a Poisson process of rate ${\lambda > 0}$ is piecewise constant, and is constant over the intervals between its jump times. However, its compensator ${A_t=\lambda t}$ is not constant on any nontrivial intervals.

Lemma 4 Let X be an adapted, right-continuous and increasing process with compensator A. Then, X is almost-surely constant on all intervals on which A is constant. That is, with probability 1, we have ${X_t=X_s}$ for all times ${s < t}$ for which ${A_s=A_t}$.

Proof: For each pair of positive real numbers ${s,\epsilon > 0}$, define the stopping time

$\displaystyle \tau_{s,\epsilon}=\inf\left\{t\ge s\colon A_t\ge A_s+\epsilon\right\}.$

As this is predictable, there exists stopping times ${\tau_n < \tau_{u,\epsilon}}$ increasing to ${\tau_{u,\epsilon}}$. Letting n increase to infinity,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle{\mathbb E}\left[X_{\tau_{s,\epsilon}-}-X_s\right]&\displaystyle=\lim_{n\rightarrow\infty}{\mathbb E}\left[X_{\tau_n}-X_s\right]\smallskip\\ &\displaystyle=\lim_{n\rightarrow\infty}{\mathbb E}[A_{\tau_n}-A_s]\le\epsilon. \end{array}$

Now, for any time ${t > s}$ at which ${A_t=A_s}$, right-continuity of the compensator implies that ${\tau_{s,\epsilon}}$ is strictly greater than t,

$\displaystyle {\mathbb E}\left[\sup_{t > s}1_{\{A_t=A_s\}}(X_t-X_s)\right]\le{\mathbb E}\left[X_{\tau_{s,\epsilon-}}-X_s\right]\le\epsilon.$

However, this inequality holds for all positive ${\epsilon}$, so the expectation of ${\sup_{t > s}1_{\{A_t=A_s\}}(X_t-X_s)}$ must be zero, showing that this term is almost-surely zero. By countable additivity, with probability 1, we have that ${X_t=X_s}$ for all ${t > s}$ for which ${A_t=A_s}$ and such that s is rational. Finally, right-continuity of X and A extends this statement to all real s. $\Box$

Finally, we prove the main result of this post, that every quasi-left-continuous process is a continuous time-change of a homogeneous Poisson process. For example, consider a totally inaccessible stopping time ${\tau}$ such that ${\tau < \infty}$, and let A be its compensator. Then, ${X=1_{[\tau,\infty)}}$ is a quasi-left-continuous counting process, so we can write ${X_t=Y_{A_t}}$ for a Poisson process Y. Then, ${A_\tau}$ is the first jump time of Y, so it must have the exponential distribution with parameter 1. Therefore, Theorem 5 gives a significant generalization of Corollary 6 of the previous post.

Theorem 5 can also be compared with the statement that every continuous local martingale is a continuous time change of standard Brownian motion. Similarly, here we must also consider a possible enlargement of the underlying probability space. That is, we consider a probability space ${(\Omega^\prime,\mathcal{F}^\prime,{\mathbb P}^\prime)}$ and a measurable onto map ${f\colon\Omega^\prime\rightarrow\Omega}$ preserving probabilities, so ${{\mathbb P}(A)={\mathbb P}^\prime(f^{-1}(A))}$ for all ${A\in\mathcal{F}}$. Then, any random variable ${X\colon\Omega\rightarrow{\mathbb R}}$ can be lifted to the random variable ${X^\prime=X\circ f}$ on the space ${\Omega^\prime}$. Similarly, any stochastic process can be lifted to a process defined on ${\Omega^\prime}$. We do not need to consider enlargements of the filtration, so that we just set

$\displaystyle \mathcal{F}^\prime_t=\left\{f^{-1}(A)\colon A\in\mathcal{F}_t\right\}$

(and, if necessary, add in any zero probability sets to keep the filtration complete). Then, we enlarge the probability space by working under the filtered probability space ${(\Omega^\prime,\mathcal{F}^\prime,\{\mathcal{F}^\prime_t\}_{t\ge0},{\mathbb P}^\prime)}$. By doing this, it is always possible to assume that there exists a Poisson process independent of the original filtration ${\mathcal{F}_\cdot}$. That is, if ${(E,\mathcal{E},\mu)}$ is any probability space which has a Poisson process defined on it, we can take ${\Omega^\prime=\Omega\times E}$, ${\mathcal{F}^\prime=\mathcal{F}\otimes\mathcal{E}}$, ${{\mathbb P}^\prime={\mathbb P}\otimes\mu}$, and let ${f\colon\Omega\times E\rightarrow\Omega}$ be the projection onto ${\Omega}$.

It can be seen from the proof of Theorem 5 that enlarging the probability space is only necessary if ${A_\infty}$ is finite (equivalently, by Lemma 3, ${X_\infty}$ is finite) with positive probability. This is because, in this case, the process X only has finitely many jumps, so that the Poisson process Y cannot be expressed purely as a function of X, and it may be necessary to add extra randomness to the probability space in order to construct Y.

Theorem 5 Any quasi-left-continuous counting process X is a continuous time-change of a homogeneous Poisson process of rate 1 (possibly under an enlargement of the probability space).

More precisely, let A be the compensator of X. Then, there exists a Poisson process Y of rate 1, with respect to a filtration ${\{\mathcal{G}_t\}_{t\ge 0}}$, such that, for each ${t\ge0}$, ${\omega\mapsto A_t(\omega)}$ is a ${\mathcal{G}_\cdot}$-stopping time and ${X_t=Y_{A_t}}$.

Proof: Without loss of generality, we can suppose that the underlying filtration ${\{\mathcal{F}_t\}_{t\ge0}}$ is right-continuous. If not, we can replace ${\mathcal{F}_t}$ by ${\mathcal{F}_{t+}=\bigcap_{s > t}\mathcal{F}_s}$, which does not affect the conditions of the theorem. Now, define the random times

$\displaystyle \tau_t=\inf\left\{s\ge 0\colon A_s > t\right\}.$

By continuity of the compensator, we have ${A_{\tau_t}=t\wedge A_\infty}$. Also, as ${\{\tau_t < s\}=\{A_s > t\}\in\mathcal{F}_s}$, ${\tau_t}$ is a stopping time. Then,

$\displaystyle {\mathbb E}[X_{\tau_t}]=E[A_{\tau_t}]\le t,$

so the process ${Z_t\equiv X_{\tau_t}}$ is integrable. Now, define the new filtration ${\mathcal{G}^0_t=\mathcal{F}_{\tau_t}}$ and the local martingale ${M=X-A}$. As the stopped process ${M^{\tau_t}}$ is dominated in ${L^1}$ (by ${t\wedge Z_t}$), it is a proper martingale. By optional sampling,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle{\mathbb E}[Z_t-A_{\tau_t}\;\vert\mathcal{F}_{\tau_s}]&\displaystyle=\lim_{u\rightarrow\infty}{\mathbb E}[M_{\tau_t\wedge u}\;\vert\mathcal{F}_{\tau_s}]\smallskip\\ &\displaystyle=\lim_{u\rightarrow\infty}M_{\tau_s\wedge u}=Z_s-A_{\tau_s}, \end{array}$

for all ${t\ge s}$, so ${Z_t-A_{\tau_t}}$ is a ${\mathcal{G}^0_\cdot}$-martingale. For any fixed time t, we have ${\tau_{A_t}\ge t}$ and A is constant on the interval ${[t,\tau_{A_t}]}$. By Lemma 4, X is also constant on this interval. So,

$\displaystyle X_t=X_{\tau_{A_t}}=Z_{A_t}.$

This shows that X is a continuous time-change of Z. We also need to show that Z is a counting process. That is, it needs to be shown that Z is right-continuous, increasing and piecewise-constant with ${Z_0=0}$, and jumps ${\Delta Z\in\{0,1\}}$. However, as ${Z_{A_t}}$ satisfies these properties for the continuous increasing process A, it follows that Z also satisfies these properties on the interval ${[0,A_\infty)}$. Also, by construction, Z is constant on ${[A_\infty,\infty)}$ and left-continuous at ${A_\infty}$, so it is indeed a counting process.

We have shown that Z is a counting process with compensator ${A_{\tau_t}=t\wedge A_\infty}$. If it is known that ${A_\infty=\infty}$ almost surely, then Lemma 2 implies that Z is a Poisson process and we are done. More generally, it is necessary to enlarge the underlying probability space. By doing so, we can assume the existence of a Poisson process, U (with respect to its natural filtration) of rate 1, which is independent of ${\mathcal{F}_\infty}$. Then, ${U_t-t}$ is a martingale under its natural filtration. By independence, letting ${\mathcal{G}_\cdot}$ be the filtration generated by ${\mathcal{G}^0_\cdot}$ and U, then ${Z_t-t\wedge A_\infty}$ and ${U_t-t}$ remain martingales under this enlarged filtration. As ${A_\infty}$ is a ${\mathcal{G}_\cdot}$-stopping time, if we set ${Y_t=Z_t+U_t-U_{t\wedge A_\infty}}$ then

$\displaystyle Y_t-t = (Z_t-t\wedge A_\infty)+(U_t-t)-(U_{t\wedge A_\infty}-t\wedge A_\infty)$

is a ${\mathcal{G}}$-martingale. Therefore, Y is a counting process with compensator t and, by Lemma 2, is a Poisson process of rate 1. Finally, as ${A_t\le A_\infty}$,

$\displaystyle Y_{A_t}=Z_{A_t}+U_{A_t}-U_{A_t\wedge A_\infty}=Z_{A_t}=X_t$

as required. $\Box$