Quadratic covariation of a jump process and a continuous process of finite variation always zero?

Question

Suppose $N_t$ is a Poisson process (so $\langle N \rangle_t=N_t$, its quadratic variation is itself), and $X_t$ is a continuous process of finite variation (so $\langle X \rangle_t=0$). Then does it follow that $\langle X,N \rangle_t=0$? I have proof which is similar to the proof of $\langle X \rangle_t=0$ as in the "Finite variation processes" section here: https://en.wikipedia.org/wiki/Quadratic_variation , but I want to confirm this is indeed the case.

Answer 1

There is the following general result:

Let $f: [0,\infty) \to \mathbb{R}$ be a continuous function and $g: [0,\infty) \to \mathbb{R}$ be a function of bounded variation (on compact intervals). Then $$ \langle f, g \rangle_t = 0$$ for all $t \geq 0$, i.e. $$\langle f,g \rangle_t = \lim_{|\Pi| \to 0} \sum_{t_j \in \Pi} (f(t_{j+1})-f(t_j)) (g(t_{j+1}-g(t_j))=0$$ where $\Pi = \{0=t_0< \ldots < t_n = t\}$ denotes a partition of $[0,t]$ with mesh size $|\Pi|$.

Proof: For fixed $t >0$ set

$$\|g\|_{\text{BV}} := \sup_{\Pi} \sum_{t_j \in \Pi} |g(t_{j+1})-g(t_j)|$$

where the supremum is taken over all partitions of the interval $[0,t]$. By assumption, $\|g\|_{\text{BV}}<\infty$. Moreover, because $f$ is uniformly continuous on the interval $[0,t]$, we have

$$\lim_{|\Pi| \to 0} \sup_{|s-r| \leq |\Pi|, s,r \in [0,t]} |f(r)-f(s)| = 0. \tag{1}$$

Consequently,

$$\begin{align*} \left| \sum_{t_j \in \Pi} (f(t_{j+1})-f(t_j)) (g(t_{j+1}-g(t_j)) \right| &\leq \sup_{|r-s| \leq |\Pi|, r,s \in [0,t]} |f(r)-f(s)| \cdot \sum_{t_j \in \Pi} |g(t_{j+1}-g(t_j)| \\ &\leq \|g\|_{\text{BV}} \cdot \sup_{|r-s| \leq |\Pi|, r,s \in [0,t]} |f(r)-f(s)| \\ &\xrightarrow[(1)]{|\Pi| \to 0} 0. \end{align*}$$

By definition, this shows $\langle f, g \rangle_t = 0$.

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Back to your original question: Since $X$ has, by assumption, continuous sample paths and $(N_t)_{t \geq 0}$ has sample paths of bounded variation (since the sample paths are non-decreasing), an application of the above statement proves $\langle X, N \rangle = 0$.