Likelihood Inference for Exponential-Trawl Processes

Abstract

Integer-valued trawl processes are a class of serially correlated, stationary and infinitely divisible processes that Ole E. Barndorff-Nielsen has been working on in recent years. In this chapter, we provide the first analysis of likelihood inference for trawl processes by focusing on the so-called exponential-trawl process, which is also a continuous time hidden Markov process with countable state space. The core ideas include prediction decomposition, filtering and smoothing, complete-data analysis and EM algorithm. These can be easily scaled up to adapt to more general trawl processes but with increasing computation efforts.

Appendix: Proofs and Derivations

1.1 6.1 Heuristic Proof of Theorem 1

Our heuristic derivation starts from the following prediction decomposition of the Radon-Nikodym derivative:

$$\begin{aligned} \log \left( \dfrac{\mathrm {d}\mathbb {P}}{\mathrm {d} \mathbb {Q} }\right) _{\mathscr {F}_{T}^{X}|X_{0}}=\int _{t\in (0,T]}\log \left( \dfrac{ \mathrm {d}\mathbb {\mathbb {P}}}{\mathrm {d} \mathbb {Q} }\right) _{X_{t}|\mathscr {F}_{t-}^{X}}, \end{aligned}$$

(20)

where the integral over $t\in (0,T]$ means a continuous sum of the integrand random variables. Thus,

$$\begin{aligned} \left( \dfrac{\mathrm {d}\mathbb {\mathbb {P}}}{\mathrm {d} \mathbb {Q} }\right) _{X_{t}|\mathscr {F}_{t-}^{X}}= & {} \left( \dfrac{\mathrm {d}\mathbb { \mathbb {P}}}{\mathrm {d} \mathbb {Q} }\right) _{\varDelta X_{t}|\mathscr {F}_{t-}^{X}} \\= & {} \sum _{y\in \mathbb {Z} \backslash \left\{ 0\right\} }\dfrac{\mathbb {P}\left( \varDelta X_{t}=y| \mathscr {F}_{t-}^{X}\right) }{ \mathbb {Q} \left( \varDelta X_{t}=y|\mathscr {F}_{t-}^{X}\right) }1_{\left\{ \varDelta X_{t}=y\right\} }+\dfrac{\mathbb {P}\left( \varDelta X_{t}=0|\mathscr {F} _{t-}^{X}\right) }{ \mathbb {Q} \left( \varDelta X_{t}=0|\mathscr {F}_{t-}^{X}\right) }1_{\left\{ \varDelta X_{t}=0\right\} } \\= & {} \sum _{y\in \mathbb {Z} \backslash \left\{ 0\right\} }\dfrac{\lambda _{t-}^{\left( y\right) ,\mathbb { P}}\mathrm {d}t}{\lambda _{t-}^{\left( y\right) ,\mathbb { \mathbb {Q} }}\mathrm {d}t}1_{\left\{ \varDelta X_{t}=y\right\} }+\dfrac{1-\sum _{y\in \mathbb {Z} \backslash \left\{ 0\right\} }\lambda _{t-}^{\left( y\right) ,\mathbb {P}} \mathrm {d}t}{1-\sum _{y\in \mathbb {Z} \backslash \left\{ 0\right\} }\lambda _{t-}^{\left( y\right) , \mathbb {Q} }\mathrm {d}t}1_{\left\{ \varDelta X_{t}=0\right\} }, \end{aligned}$$

where the first equality follows because $X_{t-}$ is known in $\mathscr {F} _{t-}^{X}$; the third equality follows from (5). Therefore, (20) can be rewritten as

$$\begin{aligned} \int _{t\in \left( 0,T\right] }\log \left( \dfrac{\mathrm {d}\mathbb {\mathbb {P} }}{\mathrm {d} \mathbb {Q} }\right) _{X_{t}|\mathscr {F}_{t-}^{X}}= & {} \sum _{0<t\le T}\sum _{y\in \mathbb {Z} \backslash \left\{ 0\right\} }\log \left( \dfrac{\lambda _{t-}^{\left( y\right) ,\mathbb {P}}\mathrm {d}t}{\lambda _{t-}^{\left( y\right) ,\mathbb { \mathbb {Q} }}\mathrm {d}t}\right) 1_{\left\{ \varDelta X_{t}=y\right\} } \\&+\int _{\left\{ t\in \left( 0,T\right] :\varDelta X_{t}=0\right\} }\log \left( \dfrac{1-\sum _{y\in \mathbb {Z} \backslash \left\{ 0\right\} }\lambda _{t-}^{\left( y\right) ,\mathbb {P}} \mathrm {d}t}{1-\sum _{y\in \mathbb {Z} \backslash \left\{ 0\right\} }\lambda _{t-}^{\left( y\right) , \mathbb {Q} }\mathrm {d}t}\right) \\= & {} \sum _{0<t\le T}\sum _{y\in \mathbb {Z} \backslash \left\{ 0\right\} }\log \left( \dfrac{\lambda _{t-}^{\left( y\right) ,\mathbb {P}}}{\lambda _{t-}^{\left( y\right) ,\mathbb { \mathbb {Q} }}}\right) 1_{\left\{ \varDelta X_{t}=y\right\} } \\&-\int _{t\in \left( 0,T\right] }\sum _{y\in \mathbb {Z} \backslash \left\{ 0\right\} }\left( \lambda _{t-}^{\left( y\right) ,\mathbb { P}}-\lambda _{t-}^{\left( y\right) , \mathbb {Q} }\right) \mathrm {d}t, \end{aligned}$$

where the second equality follows from $\log \left( 1-x\right) \approx -x$ for small x and $\{ t\in \left( 0,T\right] :\varDelta X_{t}\ne 0\}$ has Lebesgue measure 0.

1.2 6.2 Heuristic Proof of Theorem 2

1.2.1 6.2.1 Update by Inactivity

We want to update $p_{\tau ,\tau }\left( \mathbf {j}\right)$ by incorporating the information $\mathscr {F}_{\left( \tau ,t\right) }\triangleq \sigma \left( \left\{ \varDelta Y_{s}=0, \tau <s<t\right\} \right)$ using Bayes’ Theorem:

$$\begin{aligned} \mathbb {P}\left( \left. \mathbf {C}_{t-}=\mathbf {j}\right| \mathscr {F} _{t-}\right)= & {} \mathbb {P}\left( \left. \mathbf {C}_{\tau }=\mathbf {j} \right| \mathscr {F}_{t-}\right) =\mathbb {P}\left( \left. \mathbf {C} _{\tau }=\mathbf {j}\right| \mathscr {F}_{\tau },\mathscr {F}_{\left( \tau ,t\right) }\right) \\\propto & {} \mathbb {P}\left( \mathscr {F}_{\left( \tau ,t\right) }\left| \mathscr {F}_{\tau },\mathbf {C}_{\tau }=\mathbf {j}\right. \right) \mathbb {P} \left( \left. \mathbf {C}_{\tau }=\mathbf {j}\right| \mathscr {F}_{\tau }\right) , \end{aligned}$$

where the first equality holds because there is no activity of $Y_{s}$ for $s\in \left( \tau ,t\right)$ and hence the hidden state $\mathbf {C}$ must stay the same.

Using the prediction decomposition, we have

$$\begin{aligned} \log \mathbb {P}\left( \mathscr {F}_{\left( \tau ,t\right) }\left| \mathscr {F}_{\tau },\mathbf {C}_{\tau }=\mathbf {j}\right. \right)= & {} \int _{s\in \left( \tau ,t\right) }\log \mathbb {P}\left( \varDelta Y_{s}=0| \mathscr {F}_{\tau },\mathscr {F}_{\left( \tau ,s\right) },\mathbf {C}_{\tau }= \mathbf {j}\right) \\= & {} \int _{s\in \left( \tau ,t\right) }\log \left( 1-\sum _{y\in \mathbb {Z} \backslash \left\{ 0\right\} }\nu \left( y\right) \mathrm {d}s-\sum _{y\in \mathbb {Z} \backslash \left\{ 0\right\} }\phi j_{y}\mathrm {d}s\right) \\= & {} -\sum _{y\in \mathbb {Z} \backslash \left\{ 0\right\} }\nu \left( y\right) \left( t-\tau \right) -\phi \left\| \mathbf {j}\right\| _{1}\left( t-\tau \right) , \end{aligned}$$

where the second equality intuitively holds because we know the instantaneous departure probability of a size y event at time s is $\phi C_{s-}^{\left( y\right) }\mathrm {d}s$ but $C_{s-}^{\left( y\right) }=C_{\tau }^{\left( y\right) }=j_{y}$ under $\mathscr {F}_{\left( \tau ,s\right) }$; the third equality follows from $\log \left( 1-x\right) \approx -x$ for small x. Therefore,

$$\begin{aligned} \mathbb {P}\left( \left. \mathbf {C}_{t-}=\mathbf {j}\right| \mathscr {F} _{t-}\right) \propto e^{-\phi \left\| \mathbf {j}\right\| _{1}\left( t-\tau \right) }\mathbb {P}\left( \left. \mathbf {C}_{\tau }=\mathbf {j} \right| \mathscr {F}_{\tau }\right) , \end{aligned}$$

where we throw out the term $\exp \left( -\sum _{y\in \mathbb {Z} \backslash \left\{ 0\right\} }\nu \left( y\right) \left( t-\tau \right) \right)$ because it doesn’t depend on $\mathbf {j}$. Normalizing the equation above leads to the desired result.

1.2.2 6.2.2 Update by Jump

We want to update $p_{\tau -,\tau -}\left( \mathbf {j}\right)$ by incorporating the piece of information, $\varDelta Y_{\tau }=y$. First note that

$$\begin{aligned} \mathbb {P}\left( \mathbf {C}_{\tau }=\mathbf {j}|\mathscr {F}_{\tau }\right)= & {} \mathbb {P}\left( \mathbf {C}_{\tau }=\mathbf {j}|\mathscr {F}_{\tau -},\varDelta Y_{\tau }=y\right) \\= & {} \mathbb {P}\left( \left. \mathbf {C}_{\tau }=\mathbf {j},\mathbf {C}_{\tau -}= \mathbf {j}-\mathbf {1}^{\left( y\right) }\right| \mathscr {F}_{\tau -},\varDelta Y_{\tau }=y\right) \\&+\,\mathbb {P}\left( \left. \mathbf {C}_{\tau }=\mathbf {j},\mathbf {C}_{\tau -}= \mathbf {j}+\mathbf {1}^{\left( -y\right) }\right| \mathscr {F}_{\tau -},\varDelta Y_{\tau }=y\right) , \end{aligned}$$

which corresponds to the arrival of a new size y event and the departure of an old size $-y$ event.

For the first term,

$$\begin{aligned}&\mathbb {P}\left( \left. \mathbf {C}_{\tau }=\mathbf {j},\mathbf {C}_{\tau -}= \mathbf {j}-\mathbf {1}^{\left( y\right) }\right| \mathscr {F}_{\tau -},\varDelta Y_{\tau }=y\right) \\&\quad =\dfrac{\mathbb {P}\left( \left. \mathbf {C}_{\tau }=\mathbf {j},\mathbf {C} _{\tau -}=\mathbf {j}-\mathbf {1}^{\left( y\right) },\varDelta Y_{\tau }=y\right| \mathscr {F}_{\tau -}\right) }{\mathbb {P}\left( \left. \varDelta Y_{\tau }=y\right| \mathscr {F}_{\tau -}\right) } \\&\quad =\dfrac{\mathbb {P}\left( \mathbf {C}_{\tau }=\mathbf {j},\varDelta Y_{\tau }=y\left| \mathbf {C}_{\tau -}=\mathbf {j}-\mathbf {1}^{\left( y\right) }, \mathscr {F}_{\tau -}\right. \right) \mathbb {P}\left( \left. \mathbf {C}_{\tau -}=\mathbf {j}-\mathbf {1}^{\left( y\right) }\right| \mathscr {F}_{\tau -}\right) }{\mathbb {P}\left( \left. \varDelta Y_{\tau }=y\right| \mathscr {F} _{\tau -}\right) } \\&\quad =\dfrac{\mathbb {P}\left( \varDelta \mathbf {C}_{\tau }=\mathbf {1}^{\left( y\right) }\left| \mathbf {C}_{\tau -}=\mathbf {j}-\mathbf {1}^{\left( y\right) },\mathscr {F}_{\tau -}\right. \right) \mathbb {P}\left( \left. \mathbf {C}_{\tau -}=\mathbf {j}-\mathbf {1}^{\left( y\right) }\right| \mathscr {F}_{\tau -}\right) }{\mathbb {P}\left( \left. \varDelta Y_{\tau }=y\right| \mathscr {F}_{\tau -}\right) } \\&\quad =\dfrac{\nu \left( y\right) }{\lambda _{\tau -}^{\left( y\right) }}\mathbb { P}\left( \left. \mathbf {C}_{\tau -}=\mathbf {j}-\mathbf {1}^{\left( y\right) }\right| \mathscr {F}_{\tau -}\right) {,} \end{aligned}$$

where the fourth equality follows from (3) (using $\mathscr {C}_{\tau -}\supseteq \mathscr {F}_{\tau -}$) and (5).

Using similar arguments, the second term is

$$\begin{aligned} \mathbb {P}\left( \left. \mathbf {C}_{\tau }=\mathbf {j},\mathbf {C}_{\tau -}= \mathbf {j}+\mathbf {1}^{\left( -y\right) }\right| \mathscr {F}_{\tau -},\varDelta Y_{\tau }=y\right) =\dfrac{\phi \left( j_{-y}+1\right) }{\lambda _{\tau -}^{\left( y\right) }}\mathbb {P}\left( \left. \mathbf {C}_{\tau -}= \mathbf {j}+\mathbf {1}^{\left( -y\right) }\right| \mathscr {F}_{\tau -}\right) . \end{aligned}$$

Combining all of these gives us the required result.

1.3 6.3 Heuristic Proof of Theorem 3

The case of updating smoothing distribution $p_{\tau -,T}\left( \mathbf {j} \right)$ due to inactivity is trivial because the hidden configuration $\mathbf {C}$ must stay unchanged because of the inactivity during the time period $[t,\tau )$.

1.3.1 6.3.1 Update by Jump

We now consider the case of (backward) updating the smoothing distribution $p_{\tau ,T}\left( \mathbf {j}\right)$ due to the jump $\varDelta Y_{\tau }=y$. Then

$$\begin{aligned} \mathbb {P}\left( \mathbf {C}_{\tau -}=\mathbf {j}|\mathscr {F}_{T}\right)= & {} \mathbb {P}\left( \left. \mathbf {C}_{\tau -}=\mathbf {j},\mathbf {C}_{\tau }= \mathbf {j}+\mathbf {1}^{\left( y\right) }\right| \mathscr {F}_{T}\right) + \mathbb {P}\left( \left. \mathbf {C}_{\tau -}=\mathbf {j},\mathbf {C}_{\tau }= \mathbf {j}-\mathbf {1}^{\left( -y\right) }\right| \mathscr {F}_{T}\right) \\= & {} \mathbb {P}\left( \mathbf {C}_{\tau -}=\mathbf {j}\left| \mathscr {F}_{T}, \mathbf {C}_{\tau }=\mathbf {j}+\mathbf {1}^{\left( y\right) }\right. \right) \mathbb {P}\left( \left. \mathbf {C}_{\tau }=\mathbf {j}+\mathbf {1}^{\left( y\right) }\right| \mathscr {F}_{T}\right) \\&+\,\mathbb {P}\left( \mathbf {C}_{\tau -}=\mathbf {j}\left| \mathscr {F}_{T}, \mathbf {C}_{\tau }=\mathbf {j}-\mathbf {1}^{\left( -y\right) }\right. \right) \mathbb {P}\left( \left. \mathbf {C}_{\tau }=\mathbf {j}-\mathbf {1}^{\left( -y\right) }\right| \mathscr {F}_{T}\right) . \end{aligned}$$

Note that

$$\begin{aligned} \mathbb {P}\left( \mathbf {C}_{\tau -}=\mathbf {j}\left| \mathscr {F}_{T}, \mathbf {C}_{\tau }=\mathbf {k}\right. \right)= & {} \mathbb {P}\left( \mathbf {C} _{\tau -}=\mathbf {j}\left| \mathscr {F}_{\tau },\mathbf {C}_{\tau }= \mathbf {k}\right. \right) \\= & {} \dfrac{\mathbb {P}\left( \mathbf {C}_{\tau }=\mathbf {k}|\mathbf {C}_{\tau -}= \mathbf {j},\mathscr {F}_{\tau }\right) \mathbb {P}\left( \mathbf {C}_{\tau -}= \mathbf {j}|\mathscr {F}_{\tau }\right) }{\mathbb {P}\left( \mathbf {C}_{\tau }= \mathbf {k}|\mathscr {F}_{\tau }\right) } \nonumber \\= & {} \dfrac{ \begin{array}{c} \mathbb {P}\left( \mathbf {C}_{\tau }=\mathbf {k}|\mathbf {C}_{\tau -}=\mathbf {j} ,\mathscr {F}_{\tau }\right) \mathbb {P}\left( \varDelta Y_{\tau }=y|\mathbf {C} _{\tau -}=\mathbf {j},\mathscr {F}_{\tau -}\right) \\ \times \mathbb {P}\left( \mathbf {C}_{\tau -}=\mathbf {j}|\mathscr {F}_{\tau -}\right) \end{array} }{\mathbb {P}\left( \mathbf {C}_{\tau }=\mathbf {k}|\mathscr {F}_{\tau }\right) \mathbb {P}\left( \varDelta Y_{\tau }=y|\mathscr {F}_{\tau -}\right) } \nonumber \\= & {} \dfrac{\mathbb {P}\left( \mathbf {C}_{\tau }=\mathbf {k},\varDelta Y_{\tau }=y| \mathbf {C}_{\tau -}=\mathbf {j},\mathscr {F}_{\tau -}\right) }{\lambda _{\tau -}^{\left( y\right) }\mathrm {d}t}\dfrac{\mathbb {P}\left( \mathbf {C}_{\tau -}= \mathbf {j}|\mathscr {F}_{\tau -}\right) }{\mathbb {P}\left( \mathbf {C}_{\tau }= \mathbf {k}|\mathscr {F}_{\tau }\right) }, \nonumber \end{aligned}$$

(21)

where the first equality holds due to the Markov property of $\mathbf {C}_{t}$, a heuristic derivation is given later; the second and third equalities follow from the Bayes’ Theorem. Since

$$\begin{aligned} \mathbb {P}\left( \left. \mathbf {C}_{\tau }=\mathbf {j}+\mathbf {1}^{\left( y\right) },\varDelta Y_{\tau }=y\right| \mathbf {C}_{\tau -}=\mathbf {j}, \mathscr {F}_{\tau -}\right)= & {} \mathbb {P}\left( \left. \varDelta \mathbf {C} _{\tau }=\mathbf {1}^{\left( y\right) }\right| \mathbf {C}_{\tau -}= \mathbf {j},\mathscr {F}_{\tau -}\right) \\= & {} \nu \left( y\right) \mathrm {d}t, \\ \mathbb {P}\left( \left. \mathbf {C}_{\tau }=\mathbf {j}-\mathbf {1}^{\left( -y\right) },\varDelta Y_{\tau }=y\right| \mathbf {C}_{\tau -}=\mathbf {j}, \mathscr {F}_{\tau -}\right)= & {} \mathbb {P}\left( \left. \varDelta \mathbf {C} _{\tau }=-\mathbf {1}^{\left( -y\right) }\right| \mathbf {C}_{\tau -}= \mathbf {j},\mathscr {F}_{\tau -}\right) \\= & {} \phi j_{-y}\mathrm {d}t, \end{aligned}$$

combining all of these gives us the required result.

1.3.2 6.3.2 Derivation of (21)

Let $\mathscr {F}_{(\tau ,T]}\triangleq \sigma \left( \left\{ Y_{t}\right\} _{\tau <t\le T}\right)$ and $\mathscr {C}_{(\tau ,T]}\triangleq \sigma \left( \left\{ \mathbf {C}_{t}\right\} _{\tau <t\le T}\right)$. Note that heuristically the Bayes’ Theorem implies

$$\begin{aligned} \mathbb {P}\left( \mathbf {C}_{\tau -}=\mathbf {j}\left| \mathscr {F}_{T}, \mathbf {C}_{\tau }=\mathbf {k}\right. \right)= & {} \mathbb {P}\left( \mathbf {C} _{\tau -}=\mathbf {j}\left| \mathscr {F}_{\tau },\mathscr {F}_{(\tau ,T]}, \mathbf {C}_{\tau }=\mathbf {k}\right. \right) \\= & {} \dfrac{\left( \dfrac{\mathrm {d}\mathbb {P}}{\mathrm {d} \mathbb {Q} }\right) _{\mathscr {F}_{(\tau ,T]}\left| \mathscr {F}_{\tau },\mathbf {C} _{\tau }=\mathbf {k},\mathbf {C}_{\tau -}=\mathbf {j}\right. }}{\left( \dfrac{ \mathrm {d}\mathbb {P}}{\mathrm {d} \mathbb {Q} }\right) _{\mathscr {F}_{(\tau ,T]}\left| \mathscr {F}_{\tau },\mathbf {C} _{\tau }=\mathbf {k}\right. }}\mathbb {P}\left( \mathbf {C}_{\tau -}=\mathbf {j} \left| \mathscr {F}_{\tau },\mathbf {C}_{\tau }=\mathbf {k}\right. \right) . \end{aligned}$$

Since $\mathscr {F}_{(\tau ,T]}\subseteq \mathscr {C}_{(\tau ,T]}$ (each $Y_{t}=\sum _{y\in \mathbb {Z} \backslash \left\{ 0\right\} }C_{t}^{\left( y\right) }$), the Markov property of $\mathbf {C}_{t}$ implies

$$\begin{aligned} \left( \dfrac{\mathrm {d}\mathbb {P}}{\mathrm {d} \mathbb {Q} }\right) _{\mathscr {F}_{(\tau ,T]}\left| \mathscr {F}_{\tau },\mathbf {C} _{\tau }=\mathbf {k},\mathbf {C}_{\tau -}=\mathbf {j}\right. }=\left( \dfrac{ \mathrm {d}\mathbb {P}}{\mathrm {d} \mathbb {Q} }\right) _{\mathscr {F}_{(\tau ,T]}\left| \mathscr {F}_{\tau },\mathbf {C} _{\tau }=\mathbf {k}\right. }, \end{aligned}$$

because given the current information $\mathbf {C}_{\tau }$ the information in the past $\mathbf {C}_{\tau -}$ is irrelevant. This then proves that

$$\begin{aligned} \mathbb {P}\left( \mathbf {C}_{\tau -}=\mathbf {j}\left| \mathscr {F}_{T}, \mathbf {C}_{\tau }=\mathbf {k}\right. \right) =\mathbb {P}\left( \mathbf {C} _{\tau -}=\mathbf {j}\left| \mathscr {F}_{\tau },\mathbf {C}_{\tau }= \mathbf {k}\right. \right) . \end{aligned}$$

1.4 6.4 Proof of Theorem 4

Since each $C_{t}^{\left( y\right) }$ is independent for different y, the complete-data log-likelihood can be written as

$$\begin{aligned} l_{\mathscr {C}_{T}}\left( {\theta }\right) =\sum _{y\in \mathbb {Z} \backslash \left\{ 0\right\} }l_{\mathscr {C}_{T}^{\left( y\right) }|C_{0}^{\left( y\right) }}\left( {\theta }\right) +\sum _{y\in \mathbb {Z} \backslash \left\{ 0\right\} }l_{C_{0}^{\left( y\right) }}\left( \mathbf { \theta }\right) , \end{aligned}$$

where we recall that $\mathscr {C}_{t}^{\left( y\right) }$ is the natural filtration generated by $C_{t}^{\left( y\right) }$,

$$\begin{aligned} l_{\mathscr {C}_{T}^{\left( y\right) }|C_{0}^{\left( y\right) }}\left( {\theta }\right)= & {} \sum _{0<t\le T}\left( \log \left( \nu \left( y\right) \right) 1_{\left\{ \varDelta C_{t}^{\left( y\right) }=1\right\} }+\log \left( \phi C_{t-}^{\left( y\right) }\right) 1_{\left\{ \varDelta C_{t}^{\left( y\right) }=-1\right\} }\right) \\&-\int _{t\in (0,T]}\left( \nu \left( y\right) +\phi C_{t-}^{\left( y\right) }\right) \mathrm {d}t \\= & {} \log \left( \nu \left( y\right) \right) N_{T}^{\left( y\right) ,\mathrm {A} }-\nu \left( y\right) T+\log \left( \phi \right) N_{T}^{\left( y\right) , \mathrm {D}}-\phi \int _{t\in (0,T]}C_{t-}^{\left( y\right) }\mathrm {d}t, \end{aligned}$$

where the first equality follows directly from Theorem 1 (ignoring the constant), and

$$\begin{aligned} l_{C_{0}^{\left( y\right) }}\left( {\theta }\right) =C_{0}^{\left( y\right) }\left( \log \nu \left( y\right) -\log \phi \right) -\dfrac{\nu \left( y\right) }{\phi } \end{aligned}$$

because of $C_{0}^{\left( y\right) }\backsim \mathrm {Poisson}\left( \nu \left( y\right) /\phi \right)$. Thus, collecting terms will give us the required result (16). The derivations of the MCLE are elementary.

Let

$$\begin{aligned} \left\| \nu \right\| \triangleq \int \nu \left( \mathrm {d}y\right) =\sum _{y=1}^{\infty }\nu \left( y\right) . \end{aligned}$$

The ergodicity of $D_{t-}$ implies that as $T\rightarrow \infty$

$$\begin{aligned} \dfrac{1}{T}\int _{t\in (0,T]}D_{t-}\mathrm {d}t\rightarrow \mathbb {E}\left( D_{t-}\right) =\dfrac{\left\| \nu \right\| }{\phi }. \end{aligned}$$

Since $\dfrac{N_{T}^{\mathrm {D}}}{T}\approx \dfrac{N_{T}^{\mathrm {A}}}{T} \rightarrow \left\| \nu \right\|$, we have

$$\begin{aligned} \dfrac{\varXi _{T}}{T}=\dfrac{N_{T}^{\mathrm {D}}}{T}-\dfrac{D_{0}+T^{-1}\int _{t \in (0,T]}D_{t-}\mathrm {d}t}{T}\rightarrow \left\| \nu \right\| \text {, too.} \end{aligned}$$

Thus,

$$\begin{aligned} \hat{\phi }_{\mathrm {MCLE}}= & {} \frac{\dfrac{\varXi _{T}}{T}+\sqrt{\left( \dfrac{ \varXi _{T}}{T}\right) ^{2}+4T^{-1}\dfrac{N_{T}^{\mathrm {A}}+N_{T}^{\mathrm {D}} }{T}T^{-1}\int _{t\in (0,T]}D_{t-}\mathrm {d}t}}{2T^{-1}\int _{t\in (0,T]}D_{t-} \mathrm {d}t} \\\rightarrow & {} \frac{\left\| \nu \right\| +\sqrt{\left\| \nu \right\| ^{2}+0}}{2\dfrac{\left\| \nu \right\| }{\phi }}=\phi . \end{aligned}$$

Finally, for any $y\in \mathbb {Z} \backslash \left\{ 0\right\}$, $\dfrac{N_{T}^{\left( y\right) }}{T} \rightarrow \nu \left( y\right)$ and $\hat{\phi }_{\mathrm {MCLE} }^{-1}\rightarrow \phi ^{-1}<\infty$, so we easily have

$$\begin{aligned} \hat{\nu }_{\mathrm {MCLE}}\left( y\right) =\dfrac{\dfrac{N_{T}^{\left( y\right) }}{T}+\dfrac{C_{0}^{\left( y\right) }}{T}}{1+\dfrac{\hat{\phi }_{ \mathrm {MCLE}}^{-1}}{T}}\rightarrow \nu \left( y\right) \text { as well.} \end{aligned}$$

1.5 6.5 Proof of Proposition 1

As $C_{t}^{(y)}\ge 0$, (19) implies that

$$\begin{aligned} C_{0}^{\left( y\right) }\ge C_{0,T}^{\left( y\right) ,\mathrm {L} }=\sup _{t\in \left[ 0,T\right] }\left( N_{t}^{(-y)}-N_{t}^{(y)}\right) ,\ \ \ \ y=1,2,\ldots , \end{aligned}$$

where we set $N_{0}^{\left( y\right) }\triangleq 0$ conventionally. Now

$$\begin{aligned} C_{0}^{\left( y\right) }=\frac{Y_{0}-\sum _{y^{\prime }\ne y}y^{\prime }C_{0}^{(y^{\prime })}}{y}\le \left\lfloor \frac{Y_{0}-\sum _{y^{\prime }\ne y}y^{\prime }C_{0,T}^{\left( y^{\prime }\right) ,\mathrm {L}}}{y} \right\rfloor =C_{0,T}^{\left( y\right) ,\mathrm {U}}, \end{aligned}$$

so we have

$$\begin{aligned} C_{0,T}^{\left( y\right) ,\mathrm {U}}\ge C_{0}^{\left( y\right) }\ge C_{0,T}^{\left( y\right) ,\mathrm {L}}. \end{aligned}$$

Let $N_{t}^{\left( -y\right) ,*}$ be the counting processes of $-y$ jumps resulting from the departures of the initial events of size y that constitute $C_{0}^{\left( y\right) }$. Let $\tau$ be the time when $N^{\left( -y\right) ,*}$ achieve $C_{0}^{\left( y\right) }$. Then we have

$$\begin{aligned} C_{0,T}^{\left( y\right) ,\mathrm {L}}= & {} C_{0,\tau }^{\left( y\right) , \mathrm {L}}\vee \sup \limits _{t\in (\tau ,T]}\left( N_{t}^{\left( -y\right) ,*}-\left( N_{t}^{\left( y\right) }-\left( N_{t}^{\left( -y\right) }-N_{t}^{\left( -y\right) ,*}\right) \right) \right) \\= & {} C_{0,\tau }^{\left( y\right) ,\mathrm {L}}\vee \left( C_{0}^{\left( y\right) }-\inf \limits _{t\in (\tau ,T]}\left( N_{t}^{\left( y\right) }-\left( N_{t}^{\left( -y\right) }-N_{t}^{\left( -y\right) ,*}\right) \right) \right) . \end{aligned}$$

Observe that $N_{t}^{\left( y\right) }-\left( N_{t}^{\left( -y\right) }-N_{t}^{\left( -y\right) ,*}\right)$ is a M/G/$\infty$ queue initiated at state 0, so by the ergodicity we must have with probability 1

$$\begin{aligned} \lim \limits _{T\rightarrow \infty }\inf \limits _{t\in (\tau ,T]}\left( N_{t}^{\left( y\right) }-\left( N_{t}^{\left( -y\right) }-N_{t}^{\left( -y\right) ,*}\right) \right) =0. \end{aligned}$$

This then shows that actually

$$\begin{aligned} \lim \limits _{T\rightarrow \infty }C_{0,T}^{\left( y\right) ,\mathrm {L} }=C_{0,\tau }^{\left( y\right) ,\mathrm {L}}\vee C_{0}^{\left( y\right) }=C_{0}^{\left( y\right) }, \end{aligned}$$

where the last equality follows because $C_{0,\tau }^{\left( y\right) , \mathrm {L}}\le C_{0}^{\left( y\right) }$. Correspondingly,

$$\begin{aligned} \lim \limits _{T\rightarrow \infty }C_{0,T}^{\left( y\right) ,\mathrm {U} }=\left\lfloor \frac{Y_{0}-\sum _{y^{\prime }\ne y}y^{\prime }C_{0}^{\left( y\right) }}{y}\right\rfloor =C_{0}^{\left( y\right) }\text {.} \end{aligned}$$