I encountered the following problem while studying stochastic process: Let $N = (N(t),t\geq0)$ be a Poisson process with parameter $\lambda$. $T_0$ is a r.v. independent of $N$, with $P(T_0 = 1) = P(T_0=-1)=1/2$. Define $$T(t) = T_0 (-1)^{N(t)}, t\geq0$$ and $$X(t) = \int_0^tT(s)ds$$ compute $EX(t)$ and $Cov(X(s),X(t))$.
I've tried to compute $E(T(t))$ and $Cov(T(s),T(t))$, which I found all to be zero. Because,
$E(T(t)) = E[E(T(t)|N(t)=k)] = E(T_0)E((-1)^{k}) = 0$,
And since that $T(s)T(t) = 1$, iff $N(t), N(s)$ are all evens or all odds, $T(s)T(t) = -1$, iff one of $N(t), N(s)$ is even, and one of them is odd. If we denote the first event by $A$, and the second by $B$, we have $P(A) = P(B) = 1/2$.
$$E(T(s)T(t)) = P(A) + (-1)P(B)=0 $$
Therefore, $Cov(T(s),T(t))=0$.
I think this is somehow related to the computation of the moments of $X(t)$, but I can't figure it out. I don't think total expectation is going to work here, because the integrand $T(s)$ has a different variable than $t$, and cannot be fixed under the condition $N(t) = k$. If we fix $T_0 = 1$ or $-1$, I'll have to compute $$\int_0^t (-1)^{N(s)}ds$$ which kind of doesn't even make sense to me. How to integrate such a function that has a r.v. as its power?
I guess the answer should be something like $0$, or $t$. Can anyone give me a hint?
Besides, $N(t)$ is a function, with $t$ taken over the positive real line, but its values are r.v.'s. How can it be integrated anyway?
When you computed the mean of the integral you changed the order of certain integrations. Formally, you did the following
$$E[X(t)] = E\left[\int_0^tT(s)ds\right]= \int_0^tE[T(s)]ds. $$
First
$$E[X(t)]=\int_{\Omega}X_{\omega}(t)P(d\omega)=\int_{\Omega}\int_0^tT_0(\omega)(-1)^{N_{\omega}(s)}\ dsP(d\omega)$$
At this point, it turns out that before we could talk about the expectation above we have to define the probability space $[\Omega,\mathcal A, P]$ over which $(-1)^{N_{\omega}(s)}$ is a stochastic process. $(-1)^{N_{\omega}(s)}$ is neither continuous in $s$ nor it is bounded or monotonous. So, defining such a stochastic process must be a hard task. Also, it is a question that $(-1)^{N_{\omega}(s)}$ can be integrated being an everywhere discontinuous function.
Assume for now that we are done with all this.
Second
Assuming that we have defined our stochastic process and our function can be integrated for any $\omega$, so we can talk about the mean.
$$E[X(t)]=\int_{\Omega}X_{\omega}P(d\omega)=\int_{\Omega}\int_0^tT_0(\omega)(-1)^{N_{\omega}(s)}\ dsP(d\omega).$$
according to Fubini's theorem these integrals can be switched if the following integral exists
$$\iint_{\Omega\times [0,t)}\mid T_0(\omega)(-1)^{N_{\omega}(s)}\mid d(P\times\lambda)=t.$$
It does. So, switching the integrals seems to be all right. But for me the definition of this stochastic process and the integrability of the function are serious problems.