Let $N(t)$ be a Poisson process with rate $\lambda$. Then $(-1)^{N(t)}$ $N(t)$ wityh rarte $\lambda$ is a process which jumps between two states with between exponential waiting times between jumps.
a) I want to find $E[(-1)^N(t)]$
I get $\sum_{k>=0}(-1)^ke^{\lambda t}\frac{(\lambda t)^k}{k!}$= $e^{-\lambda t}e^{\lambda t}=1$ ??
Also, we can write $E[(-1)^N(t)] = P{(-1)^{N(t)=1}+(-1)*P{(-1)^{N(t)}=-1}=1$
Can we say that $P{(-1)^{N(t)=1}+P{(-1)^{N(t)}=-1}=1$ How do I now reconcile the two?
b) Next I want to show that $lim_{n \to \infty}P((-1)^{{N(t)}}=1)=1/2$ It makes intuitive sense but how to show it?
Finally I want to find $E(\int_0^{\tau}(-1)^{N(t)}dt)$ Also not so sure how to go about this.
Thanks for the consideration
Part (a) is: $$\operatorname{E}[(-1)^{N(t)}] = \sum_{k=0}^\infty (-1)^k e^{-\lambda t} \frac{(\lambda t)^k}{k!} = e^{\lambda t} \sum_{k=0}^\infty \frac{(-\lambda t)^k}{k!} = e^{-\lambda t} e^{-\lambda t} = e^{-2\lambda t}.$$
We can see the above is a special case of the probability-generating function of a Poisson distribution: $$\mathcal P_{N(t)}(z) = \operatorname{E}[z^{N(t)}] = e^{\lambda t (z - 1)},$$ where $z = -1$. Your error is in writing $e^{\lambda t}$ rather than $e^{-\lambda t}$ for the Poisson PMF.
Part (b) is unclear because the expression $$\lim_{n \to \infty} \Pr[(-1)^{N(t)} = 1]$$ has the limiting variable $n$, but you did not define how $n$ is related to $N(t)$. However, based on part (a), we can see that as $t \to \infty$, $\operatorname{E}[(-1)^{N(t)}] \to 0$, which suggests that $N(t)$ tends to be odd as often as it tends to be even; i.e. $\Pr[(-1)^{N(t)} = 1] = \Pr[(-1)^{N(t)} = -1] = \frac{1}{2}$. I leave it as an exercise to formally demonstrate this fact.
Part (c) is also unclear.