Autocovariance function of a Poisson process transformation here is the problem formulation:
Let $\{N_t,t \ge 0\}$ follow a Poisson process with rate parameter $\lambda$ and let $A$ be a random variable with zero mean and unit variance, $A$ is independent of {N_t} for all $t \ge 0$. We define $X_t=A(-1)^{N_t}$, $t\ge0$. Derive the autocovariance function of $\{X_t, t \ge 0\}$.
I am having particular difficulties understanding the $(-1)^{N_t}$ transformation in formal terms. For instance, I believe we need to establish independence of $A$ and $(-1)^{N_t}$ first but I am not sure how to accomplish that... Thanks for your help!
 A: Let's start with the computation of the expectation. The independence of $A$ and $(N_t)_{t \geq 0}$ implies
$$\mathbb{E}X_t = \mathbb{E}(A \cdot (-1)^{N_t}) = \underbrace{\mathbb{E}(A)}_{0} \cdot \underbrace{\mathbb{E}((-1)^{N_t})}_{\leq 1} = 0$$
Similarly,
$$\begin{align*} \text{cov}(X_s,X_t) &= \mathbb{E}(X_s \cdot X_t) = \mathbb{E}\big(A^2 \cdot (-1)^{N_t} \cdot (-1)^{N_s}\big) \\ &= \underbrace{\mathbb{E}(A^2)}_{1} \cdot \mathbb{E}\big((-1)^{N_t+N_s}\big) \tag{1} \end{align*}$$
for any $0 \leq s \leq t$. Since $(N_t)_{t \geq 0}$ has independent and stationary increments, we have
$$\mathbb{E}\big((-1)^{N_t+N_s} \big) = \mathbb{E}\big((-1)^{(N_t-N_s)} \cdot (-1)^{2N_s} \big) = \mathbb{E}\big((-1)^{N_{t-s}}\big) \cdot \mathbb{E}\big((-1)^{2N_s}\big) = \mathbb{E}\big((-1)^{N_{t-s}}\big). \tag{2}$$
By assumption, $N_{t-s} \sim \text{Poi}(\lambda \cdot (t-s))$. Therefore,
$$\mathbb{E}((-1)^{N_{t-s}}) = \sum_{k=0}^{\infty} (-1)^k \cdot \mathbb{P}(N_{t-s}=k) = e^{-\lambda \cdot (t-s)} \cdot \sum_{k=0}^{\infty} (-1)^k \cdot \frac{(\lambda \cdot (t-s))^{k}}{k!} = e^{-2\lambda \cdot (t-s)} \tag{3}$$
Combining $(1)$, $(2)$, $(3)$, we find
$$\text{cov}(X_s,X_t) = e^{2\lambda \cdot (t-s)}$$
