Why is the positive variation of the compensated Poisson process $A_t = \lambda t - N_t $: $V^+_A([0,t]) = \lambda t$ 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}
 A: 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.
