How to calculate the jump of $e^{{\rm i}N_ty-\operatorname E[N_t](e^{{\rm i}y}-1)}$ for a Poisson process $N$?

Let $(N_t)_{t\ge0}$ be a (nonhomogeneous) càdlàg Poisson process on a probability space $(\Omega,\mathcal A,\operatorname P)$. Assume $\alpha(t):=\operatorname E\left[N_t\right]$ for $t\ge0$ is continuous. As usual, let $x(t-):=\lim_{s\to t-}x(s)$ and $\Delta x(t):=x(t)-x(t-)$ for $t\ge0$. Now let $$Z_t:=e^{{\rm i}N_ty-\alpha(t)(e^{{\rm i}y}-1)}\;\;\;\text{for }t\ge0$$ for some $y\in\mathbb R$.

In the proof of Lemma 4.3.3 in this lecture notes, it seems like it is claimed that $$\Delta Z_t=e^{{\rm i}\Delta N_ty\color{red}{-1}}e^{-\alpha(t)(e^{{\rm i}y}-1)}\tag1\;\;\;\text{for all }t\ge0.$$ I don't get where the "$\color{red}{-1}$" comes from. Since $\Delta(f\circ x)(t)=f(x(t))-f(x(t-))$ for every continuous $f$, shouldn't we have $$\Delta Z_t=\underbrace{\left(e^{{\rm i}N_ty}-e^{{\rm i}N_ty}\right)}_{=\:e^{{\rm i}\Delta N_ty}}e^{-\alpha(t)(e^{{\rm i}y}-1)}\tag2$$ instead?

Since the "$-1$" is crucial in the remainder of the lecture notes, I doubt that it is a typo. So, there seems to be something I'm missing.

Solution 1:

Why is $\Delta(f\circ x)(t)=f\bigl(\Delta x(t)\bigr)$? (The latter is $f\bigl(x(t)-x(t-)\bigr)$ and the former is $f\bigl(x(t)\bigr)-f\bigl(x(t-)\bigr)$.)

We have \begin{align*} \Delta Z_t:\!&={\rm e}^{{\rm i}uN_t-\alpha(t)({\rm e}^{{\rm i}u}-1)}-{\rm e}^{{\rm i}uN_{t-}-\alpha(t-)({\rm e}^{{\rm i}u}-1)}\\[.4em] &={\rm e}^{{\rm i}uN_t-\alpha(t)({\rm e}^{{\rm i}u}-1)}\left(1-{\rm e}^{{\rm i}uN_{t-}-\alpha(t-)({\rm e}^{{\rm i}u}-1)-\bigl[{\rm i}uN_t-\alpha(t)({\rm e}^{{\rm i}u}-1)\bigr]}\right)\\[.4em] &={\rm e}^{{\rm i}uN_t-\alpha(t)({\rm e}^{{\rm i}u}-1)}\left(1-{\rm e}^{{\rm i}u\bigl(N_{t-}-N_t\bigr)-\bigl(\alpha(t-)-\alpha(t)\bigr)({\rm e}^{{\rm i}u}-1)}\right)\\[.4em] &={\rm e}^{{\rm i}uN_t-\alpha(t)({\rm e}^{{\rm i}u}-1)}\left(1-{\rm e}^{-{\rm i}u\Delta N_t}\right)\!,\\[.4em] \end{align*} because $\alpha$ is continuous. Then $$ \Bigl|\Delta Z^j_t\Delta Z^k_t\Bigr| =\exp\!\left(-\alpha^j(t)({\rm e}^{\rm i u^j}-1)\right)\exp\!\left(-\alpha^k(t)({\rm e}^{\rm i u^k}-1)\right)\Bigl|1-{\rm e}^{-{\rm i}u^j\Delta N^j_t}\Bigr|\,\Bigl|1-{\rm e}^{-{\rm i}u^k\Delta N^k_t}\Bigr|. $$