Can someone please help me. Why is the positive/negative variation of the compensated Poisson process $A_t = \lambda t - N_t $:
$$ V^+_A([0,t]) = \lambda t\quad\text{ and }\quad V^-_A([0,t]) = N_t\quad? $$ I understand that \begin{align} V_A([0,t]) &= \sup_{\Pi} \sum_{i=1}^{k} A_{t_i}-A_{t_{i-1}} = \sup_{\Pi} \sum_{i=1}^{k} \lambda\,(t_{i}-t_{i-1}) - N_{t_{i}} + N_{t_{i-1}}\\[3mm] &= \lambda \, (t-0) - N_t + N_0 = \lambda \, t - N_t. \end{align} But I do not know how to calculate \begin{align} V^+_A([0,t]) &= \sup_{\Pi} \sum_{i=1}^{k} \max \{A_{t_{i}} - A_{t_{i-1}}, 0\}\\[3mm] &= \sup_{\Pi} \sum_{i=1}^{k} \max\Big\{\lambda \, (t_{i}-t_{i-1}) - N_{t_{i}} + N_{t_{i-1}}, 0\Big\}. \end{align} I don't understand how to get $\lambda t$ out.
The definition of $V^-_A([0,t])$ is
\begin{align} V^-_A([0,t]) &= \sup_{\Pi} \sum_{i=1}^{k}\max\Big\{-(A_{t_{i}} - A_{t_{i-1}}), 0\Big\}\\[3mm] &= \sup_{\Pi} \sum_{i=1}^{k} \max\Big\{\lambda\,(-t_{i}+t_{i-1}) + N_{t_{i}} - N_{t_{i-1}}, 0 \Big\}\,. \end{align}
It is trivial that for any partition $0=t_1<...<t_k=t$ \begin{align} \sum_{i=1}^kA_{t_i}-A_{t_{i-1}}&=\sum_{i=1}^k\lambda (t_i-t_{i-1})-N_{t_i}+N_{t_{i-1}}=\lambda\,t-N_t \end{align} holds. Therefore, the supremum over all partitions will be $\lambda t-N_t\,.$
To handle the other cases it is enough to consider a sequence of partitions $\Pi_n$ that satisfies the following two properties: $$ 0=t^n_1<...<t^n_{k_n}=t\quad \forall n\in\mathbb N\,, $$ $$ \{t^n_1,...,t^n_{k_n}\}\subset \{t^{n+1}_1,...,t^{n+1}_{k_{n+1}}\}\quad \forall n\in\mathbb N\,. $$ Since $N_{t^n_i}\ge N_{t^n_{i-1}}$ and $t^n_i\ge t^n_{i-1}\,,$ we have always $$ \lambda (t^n_i-t^n_{i-1})\ge 0\quad\text{ and }\quad N_{t^n_i}-N_{t^n_{i-1}}\le 0\,. $$ Case 1. If the path $t\mapsto N_t$ has no jump in $\bigcap\limits_{n\in\mathbb N}[t^n_i,t^n_{i+1}]$ then, for sufficiently large $n$, $$ N_{t^n_i}-N_{t^n_{i-1}}=0 $$ which means $$ \Big(\lambda (t^n_i-t^n_{i-1})-N_{t^n_i}+N_{t^n_{i-1}}\Big)^+= \lambda (t^n_i-t^n_{i-1})\,. $$ Case 2. When $t\mapsto N_t$ has a jump in $\bigcap\limits_{n\in\mathbb N}[t^n_i,t^n_{i+1}]$ then, for all $n$, $$ N_{t^n_i}-N_{t^n_{i-1}}>1 $$ which means that for sufficiently large $n$ $$ \Big(\lambda (t^n_i-t^n_{i-1})-N_{t^n_i}+N_{t^n_{i-1}}\Big)^+=0 $$ because $\lambda(t^n_i-t^n_{i-1})$ will be smaller than $1\,.$ This shows that \begin{align} \sup_{\Pi_n}\sum_{i=1}^{k_n}(A_{t^n_i}-A_{t^n_{i-1}})^+= \sup_{\Pi_n}\sum_{i=1}^{k_n}\lambda(t^n_i-t^n_{i-1})=\lambda\,t\,. \end{align} The proof of \begin{align} \sup_{\Pi_n}\sum_{i=1}^{k_n}(A_{t^n_{i}}-A_{t^n_{i-1}})^-=N_t \end{align} is similar.