Solution: Doob-Meyer decomposition for a counting process (Exam 2022)

Since \( Z_i\geq 0 \) for all \( i \) we must have that \( X(t)-X(s)\geq 0 \) for all \( t\geq s \). Thereby, for \( t>s \),

\[ \mbox{E}[X(t)|{\cal F}_s] = \mbox{E}[(X(t)-X(s))+X(s)|{\cal F}_s] = \mbox{E}[X(t)-X(s)|{\cal F}_s] + X(s) \geq X(s), \]

and thus \(X(t)\) is a sub-martingale.

In a counting process at most one event can happen at the same time, so \( dN(t)\in\{0,1\} \). If \( dN(t)=0 \) we clearly have \( dX(t)=0 \), whereas if \( dN(t)=1 \) we have \( dX(t)=Z_{N(t)} \). We can write this more compactly as

\[ dX(t) = dN(t)\cdot Z_{N(t)}. \]

Since \( Z_{N(t)}\in\{0,1\} \) we have \( dN(t)\cdot Z_{N(t)}\in\{0,1\} \), which gives that

\( \begin{align*} dX^\star(t) &= \mbox{E}[dX(t)|{\mathscr F}_{t-}] = \mbox{E}[dN(t)\cdot Z_{N(t)}|{\mathscr F}_{t-}]\\ &= P(dN(t)\cdot Z_{N(t)} = 1|{\mathscr F}_{t-})\\ &= P(dN(t)=1,Z_{N(t)}=1|{\mathscr F}_{t-})\\ &= P(dN(t)=1|{\mathscr F}_{t-})\cdot P(Z_{N(t)}=1|{\mathscr F}_{t-},dN(t)=1). \end{align*} \)

The definition of the intensity process \( \lambda (t) \) gives that \( P(dN(t)=1|{\mathscr F}_{t-})=\lambda(t)dt \). When \( {\mathscr F}_{t-} \) is given and we know that \( dN(t)=1 \), we know the value of \( Z_{N(t-)} \) and that \( N(t)=N(t-)+1 \). Since the counting process \( N(t) \) is independent of the \( Z_i \) chain, the Markov structure of \( \{Z_i\}_{i=1}^\infty \) thereby implies that

\( \begin{align*} P(Z_{N(t)}=1|{\mathscr F}_{t-},dN(t)=1) &= P(Z_{N(t)}=1|Z_{N(t-)})\\ &= \alpha \mathbb{I}(Z_{N(t-)}=0) + (1-\beta) \mathbb{I}(Z_{N(t-)}=1)\\ &= \alpha (1-Z_{N(t-)}) + (1-\beta) Z_{N(t-)}. \end{align*} \)

Thereby we have

\[ dX^\star(t) = (\alpha (1-Z_{N(t-)}) +(1-\beta) Z_{N(t-)})\lambda(t)dt. \]

To get the compensator we integrate the incremental process \( dX^\star(t) \) from zero to \( t \),

\( \begin{align*} X^\star(t) &= \int_0^t dX^\star(u)du = \int_0^t (\alpha (1-Z_{N(u-)}) +(1-\beta) Z_{N(u-)})\lambda(u)du. \\ \end{align*} \)

Using that \( Z_{N(u-)} \) is constant between two subsequent event times \( T_{i-1} \) and \( T_i \) we get

\( \begin{align*} X^\star(t) &= \left[\sum_{i=1}^{N(t-)} \int_{T_{i-1}}^{T_i} (\alpha (1-Z_{N(u-)}) +(1-\beta) Z_{N(u-)})\lambda(u)du \right] \\ &~~~+ \int_{T_{N(t-)}}^t (\alpha (1-Z_{N(u-)}) +(1-\beta) Z_{N(u-)})\lambda(u)du \\ &=\left[\sum_{i=1}^{N(t-)} \int_{T_{i-1}}^{T_i} (\alpha (1-Z_{i-1}) +(1-\beta) Z_{i-1})\lambda(u)du \right] \\ &~~~+ \int_{T_{N(t-)}}^t (\alpha (1-Z_{N(u-)}) +(1-\beta) Z_{N(u-)})\lambda(u)du \\ &=\left[\sum_{i=1}^{N(t-)} (\alpha (1-Z_{i-1}) +(1-\beta) Z_{i-1}) \int_{T_{i-1}}^{T_i} \lambda(u)du \right] \\ &~~~+ (\alpha (1-Z_{N(u-)}) +(1-\beta) Z_{N(u-)}) \int_{T_{N(t-)}}^t \lambda(u)du \\ &= \left[\sum_{i=1}^{N(t-)} (\alpha (1-Z_{i-1}) +(1-\beta) Z_{i-1}) (\Lambda(T_i)-\Lambda(T_{i-1}))\right] \\ &~~~+ (\alpha (1-Z_{N(t-)}) +(1-\beta) Z_{N(t-)}) (\Lambda(t)-\Lambda(T_{N(t-)})),\\ \end{align*} \)

where \( \Lambda(t)=\int_0^t\lambda(u)du \) is the integrated intensity process of the counting process \( N(t) \).