Quadratic variation of a Poisson process where time depends on a function F

Question

Let $(N_t)_{t \geq 0}$ be a Poisson process of rate $1$. Then the quadratic variation of $N$ is itself (so $\langle N\rangle_t=N_t$). For a suitable function $f$, what would be the quadratic variation of the jump process $\left(N_{\int_0^t f(s)\ \mathsf ds}\right)_{t \geq 0}$? Would it be true that $\langle N_{\int_0^\cdot f(s)\ \mathsf ds}\rangle_t=\langle N\rangle _{\int_0^t f(s)\ \mathsf ds}$? If so, why?

Answer 1

The jump process $\left(N_{\int_0^t f(s)\ \mathsf d s}\right)_{t\geqslant 0}$ is a non-homogeneous Poisson process with rate $f(s)$. If we define the time transformation $$u = F(t) = \int_0^t f(s)\ \mathsf ds,$$ then the process $\left(M_u\right)_{u\geqslant 0}$ is homogeneous with rate $1$. Therefore it follows that $$\left\langle N_{\int_0^\cdot f(s)\ \mathsf ds } \right\rangle_t = \langle M\rangle_u = M_u = N_{\int_0^t f(s)\ \mathsf d s} = \langle N\rangle_{\int_0^t f(s)\ \mathsf ds}. $$