Let $(X(t))_{t \geq 0}$ be a Poisson process, $Z$ a Bernoulli random variable, independent of $X(t)$.
We define $$Y(t)=Z(-1)^{X(t)}$$
It is clear that $Y(t)$ is sationary.
Now we define $$S(t)=\int_{0}^{t}{Y(v)dv}$$ Q : Calculate $E(S(t))$ and $Var(S(t))$
My attempt : $$\begin{align*} E[S(t)]&=E[\int_{0}^{t}{Y(v)dv}]\\ &=E\Bigg[\int_{0}^{t}{Z(-1)^{X(v)}dv}\Bigg]\\ &=E\Bigg[Z\int_{0}^{t}{(-1)^{X(v)}dv}\Bigg]\\ \end{align*}$$ Here, I defined an operator : $$T : (\Omega,\mathcal F, \mathbb P) \longrightarrow \mathbb R$$ $$\hspace{5.5cm} W(t) \longrightarrow T(W(t)) = \int_{0}^{t}{W(v)dv}$$
I would then define $$f : (\Omega,\mathcal F, \mathbb P) \longrightarrow (\Omega,\mathcal F, \mathbb P) $$ $$ X(t) \longrightarrow f(X(t))=(-1)^{X(t)}$$
Then I would have $$E[S(t)]= E\Bigg[Z\int_{0}^{t}{(-1)^{X(v)}dv}\Bigg]=E\Big[Z[T(f(X))]\Big]$$
Now I know that if we have two independent r.v's $X$ and $Y$ then for any measurable functions $\varphi$ and $\psi$, the new random variables $\varphi(X)$,$\psi(Y)$ are independent.
The same is applied on $Z$ and $f(X)$. Now the problem lies with the operator $T$. In case there is an independance, it will result that $$ E\Big[Z[T(f(X))]\Big]=E[Z]\times E[T(f(X))]=0$$
I am not sure how to proceed. This was just an intuitive idea.
The whole calculation: \begin{align} E\left[\int_0^t Z (-1)^{X(v)}\,dv\right] &= \underbrace{E[Z]}_{=\frac{1}{2}}\cdot \int_0^t E\left[(-1)^{X(v)}\right]\,dv \\ &= \frac{1}{2} \int_0^t \sum_{k=0}^\infty (-1)^k\frac{(v\lambda)^ke^{-v\lambda}}{k!}\,dv \\ &= \frac{1}{2} \int_0^t \left(\sum_{k=0}^\infty (-1)^k\frac{(v\lambda)^k}{k!}\right)e^{-v\lambda}\,dv \\ &= \frac{1}{2} \int_0^t e^{-2v\lambda}\,dv \\ &= \frac{1}{4\lambda}\left(1-e^{-2\lambda t}\right) \end{align}
please check it before using^^