If I have a Poisson process $N\left(t\right)$ with intensity rate parameter $\lambda$, I understand that $E\left(N\left(t\right)\right)=\lambda t$ and $E\left[dN\left(t\right)\right]=\lambda dt$.
Also I understand if the compound Poisson process is defined as $Y\left(t\right)=\sum_{n=1}^{N\left(t\right)}J_{n}$, where $J_{n}$ is n random variables following some specified i.i.d, I can have
\begin{align} E\left(dY\left(t\right)\right) & = E\left(J_{n}\right)\lambda dt \end{align}
My question is if I have a jump process as $X\left(t\right)=\sum_{n=1}^{N\left(t\right)}\left(-1\right)^n$. What will $E\left(dX\left(t\right)\right)$ be? Intuitively I think the we should have $E\left(X\left(t\right)\right)=(-1)^{\lambda t}$ or $E\left(X\left(t\right)\right)=(-1)^{\lfloor\lambda t \rfloor}$, I have no idea how to derive the result. Especially, I have no idea what will $E\left( dX\left(t\right)\right)$ be.
I think this question is equivalent to considering the corresponding continuous time Markov chain $\varepsilon_{t}$ with transition rate matrix
\begin{align} \begin{pmatrix}-\lambda & \lambda\\ \lambda & -\lambda \end{pmatrix} \end{align}
then find what is $E\left(\varepsilon_{t}\right)$ and $E\left(d\varepsilon_{t}\right)$.
Since
$$X_t(\omega) = \begin{cases} -1, & N_t(\omega) \, \text{is odd} \\ 0, & N_t(\omega) \, \text{is even} \end{cases}$$
we have
$$\begin{align*} \mathbb{E}(X_t) &= \sum_{n=0}^{\infty} \mathbb{E}(X_t 1_{\{N_t = 2n+1\}}) + \sum_{n=0}^{\infty} \mathbb{E}(X_t 1_{\{N_t = 2n\}}) \\ &= -\sum_{n=0}^{\infty} \mathbb{P}(N_t = 2n+1) + 0 \\ &= - \mathbb{P}(N_t \, \text{is odd}) = \mathbb{P}(N_t \, \text{is even})-1, \end{align*}$$
so everything boils down to calculating the probability that a Poisson distributed random variable takes an even value. Using that $\mathbb{P}(N_t=k) = e^{-\lambda t} (\lambda t)^k/k!$, we find
$$\begin{align*} \mathbb{P}(N_t \, \text{is even}) &= e^{-\lambda t} \sum_{k=0} \frac{(\lambda t)^{2k}}{(2k)!} \\ &= e^{-\lambda t} \frac{1}{2} \left( \sum_{k=0}^{\infty} \frac{(\lambda t)^k}{k!} + \sum_{k=0}^{\infty} \frac{(-\lambda t)^k}{k!} \right) \\ &= \frac{1}{2} e^{-\lambda t} (e^{\lambda t} + e^{-\lambda t}) = \frac{1}{2}(1+e^{-2\lambda t}). \end{align*}$$
Hence,
$$\mathbb{E}(X_t) = \frac{1}{2} (e^{-2\lambda t}-1).$$