Assume $N_1(t)$ and $N_2(t)$ are two independent Poisson processes with parameters $\lambda_1, \lambda_2$ respectively. We define $Z(t)=YN_1(t)(-1)^{N_2(t)}, t>0$ in which $P_Y(y)=\frac{1}{2}\delta(y+1)+\frac{1}{2}\delta(y-1)$ is independent of $N_1, N_2$. I want to calculate the mean and autocorrelation of $Z(t)$.
I calculated the mean as $$\mathbb{E}\left[Z(t)\right]=\mathbb{E}\left[YN_1(t)(-1)^{N_2(t)}\right] = \mathbb{E}\left[Y\right]\mathbb{E}\left[N_1(t)\right]\mathbb{E}\left[(-1)^{N_2(t)}\right] = 0$$
Cause $\mathbb{E}\left[Y\right] = 0$.
I also tried to calculate the autocorrelation as follow
$$\mathbb{E}\left[Z(t_1)Z(t_2)\right] = \mathbb{E}\left[YN_1(t_1)(-1)^{N_2(t_1)}YN_1(t_2)(-1)^{N_2(t_2)}\right]=\mathbb{E}\left[Y^2\right]\mathbb{E}\left[N_1(t_1)N_1(t_2)\right]\mathbb{E}\left[(-1)^{N_2(t_1)+N_2(t_2)}\right]$$
where we know $\mathbb{E}\left[Y^2\right]=\frac{1}{2}(-1)^2+\frac{1}{2}(+1)^2=1$. Also $\mathbb{E}\left[N_1(t_1)N_1(t_2)\right]=\lambda_1^2t_1t_2-\min\{t_1,t_2\}$ is known since $N_1$ is a Poisson counting process.
My question
I tried to calculate the third part as follow:
$$\mathbb{E}\left[(-1)^{N_2(t_1)+N_2(t_2)}\right] = \sum _{n_1 = 0}^{\infty}\sum_{n_2 = 0}^{\infty}(-1)^{n_1+n_2}\frac{e^{-\lambda_2t_1}(\lambda_2t_1)^{n_1}}{n_1!}\frac{e^{-\lambda_2t_2}(\lambda_2t_2)^{n_2}}{n_2!}$$
I don't know how to calculate the last summation! Any help would be appreciated!
P.S. for the last summation I actually don't know if the expression for the expectation is correct because $N_2(t_1)$ and $N_2(t_2)$ are not independent. Can someone write the correct expression?
Observe that $(-1)^{N_2(t_1) + N_2(t_2)}$ is either $-1$ or $+1$, depending on whether $N_2(t_1) + N_2(t_2)$ is odd or even.
Assume, $0< t_1 < t_2$. Then $$N_2(t_1) + N_2(t_2) = 2N_2(t_1) + N_2(t_1, t_2),$$ where $N_2(t_1,t_2)$ denotes the number of arrivals between $t_1$ and $t_2$.
Since $(-1)^{2N_2(t_1)} = 1$, we have \begin{align} (-1)^{N_2(t_1) + N_2(t_2)} & = (-1)^{2N_2(t_1) + N_2(t_1, t_2)}\\ & = (-1)^{N_2(t_1, t_2)} \end{align} We know that $N_2(t_1, t_2)$ has Poisson distribution with parameter $\mu = \lambda_2(t_2 - t_1)$. That is, $\mathbb{P}[N_2(t_1, t_2) = k] = e^{-\mu} \frac{\mu^k}{k!}$. So \begin{align*} \mathbb{E}[(-1)^{N_2(t_1, t_2)}] & = \sum_{k = 0}^\infty (-1)^k e^{-\mu} \frac{\mu^k}{k!}\\ & = e^{-\mu} \sum_{k = 0}^\infty \frac{(-\mu)^k}{k!}\\ & = e^{-2\mu}. \end{align*}