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.
[Math] Quadratic covariation of a jump process and a continuous process of finite variation always zero
martingalesquadratic-variationstochastic-processes
Related Solutions
Assume that $X$ solves $$ dX_t = \mu(t,X_t) dt + \sigma(t, X_t) dW_t. $$ In integral form, this means $$ X_t = X_0 + \int_0^t \mu(s,X_s) ds + \int_0^t \sigma(s,X_s) dW_s. $$ Now, the quadratic variation of a stochastic integral process $H\cdot Y$, where $(H\cdot Y)_t = \int_0^t H_s dY_s$, is $$ [H\cdot Y]_t = \int_0^t H_s^2 d[Y]_s. $$ You can find this in a special case as Proposition 3.2.17 of Karatzas and Shreve's "Brownian Motion and Stochastic Calculus", or in a general version for continuous semimartingales in Section IV.31 of Rogers and William's "Diffusions, Markov processes and Martingales". Using this, we obtain (with $\mu_t = \mu(t,X_t)$ and $\sigma_t = \sigma(t, X_t)$ as convenient shorthands, and $A_t = t$) $$ [X]_t = [\mu \cdot A]_t + [\sigma\cdot W]_t = \int_0^t \mu_s^2 d[A]_s + \int_0^t \sigma_s^2 d[W]_s \\ = \int_0^t \sigma(s,X_s)^2 ds, $$ where we have used $[A]_t = 0$, since all continuous processes of finite variation have zero quadratic variation, and $[W]_t = t$. As for the geometric Brownian motion, we have that $X$ is a GBM if it satisfies $$ dX_t = \mu X_t dt + \sigma X_t dW_t, $$ which yields $$ [X]_t = \int_0^t \sigma^2 X_s^2 ds. $$ In particular, $[X]$ isn't immediately seen to satisfy any SDE or ODE (as $[X]$ does not figure in the right-hand side of the above). The above is simply a representation of $[X]$ in terms of $X$, but I doubt that there in general exists and SDE or ODE satisfied by $[X]$.
An example is given in Appendix 5.2 of Coquet, Jakubowski, Mémin, and Słominski, Natural decomposition of processes and weak Dirichlet processes, pp. 81–116, Springer, Berlin, Heidelberg, 2006.
For completeness, I replicate the example below: we restrict ourselves to the time interval $[0,1]$. Let $f \in C[0,1]$ be defined by $f(t) = 0$ when $t = 1 - 2^{1-2p}$ and $f(t) = \frac{1}{p}$ when $t = 1 - 2^{-2p}$, where $p \in \mathbb{Z}^+$. We complete the construction of $f$ by linearly interpolating between these points. The graph of $f$ looks like a sequence of shrinking scalene triangles as you move forward in time.
By construction, is is clear that $f$ is of bounded variation on all intervals $[0,t]$ with $t < 1$; hence its quadratic variation on those intervals vanish.
By considering the quadratic variation on $[0,1]$ along the sample points $T = \{ 1 - 2^{-2k} \, : k \geq 1 \}$, you get that this quantity is infinite.
Finally, you may construct a sequence of refining partitions of $[0,1]$, say $(\pi_n)_{n=1}^\infty$, given by:
\begin{align*} \pi_n = \bigcup_{j \leq 2^{2n} - 1} \{j 2^{-2n}\} \cup \bigcup_{k \geq n} \{ 1 - 2^{-2k} \} \end{align*}
You can show that along this sequence, the pathwise quadratic variation takes up its entire mass at $t=1$.
Best Answer
There is the following general result:
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 $\langle f, g \rangle_t = 0$.
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 $\langle X, N \rangle = 0$.