Let $X(t) = \int_{-\infty}^{\infty} h(u) Y(t-u)du $ be a process where $Y(t)$ is stationary with mean $0$. I want to calculate the mean and the autocovariance function of $X(t)$. So:
$$ E(X(t)) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} h(u) Y(t-u)du f_Y(t) dt $$
I know the answer must be zero. My question is: can I invert the integrals and compute it? Meaning:
$$ E(X(t)) = \int_{-\infty}^{\infty} h(u) \int_{-\infty}^{\infty} Y(t-u)f_Y(t) dt du =0 $$
Thanks!
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Edit.
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For the autocovariance function:
$$ E(X(t) X(t-h)) = E\left(\int_{\mathbb{R}} h(u) Y(t-u)du \int_{\mathbb{R}} h(v) Y(t-h-v)dv \right) $$
I know the result must be $$ \gamma_X(h) =\int_{\mathbb{R}} h(u) h(v) \gamma_Y(h+v-u) du dv $$ but I still can't figure why the expected value resulted in the above equation. For me it would be the same with:
$$ \gamma_Y(h+v+u) $$
Thanks!
Making this calculation is conceptually difficult to do unless we are clear on what is meant by \begin{equation} \int_{\mathbb{R}} h(u)Y(t-u)\,\mathrm{d}u. \end{equation} It is a random variable and so there is the additional variable that makes it random! Thus we have \begin{equation} X(t,\omega) = \int_{\mathbb{R}} h(u)Y(t-u,\omega)\,\mathrm{d}u. \end{equation}
Deriving the expectation is then just an application of Fubini's theorem and the argument would be \begin{equation} E_{\omega}\left[\int_{\mathbb{R}} h(u)Y(t-u,\omega)\,\mathrm{d}u\right] = \int_{\mathbb{R}} h(u)E_\omega[Y(t-u,\omega)]\,\mathrm{d}u =0. \end{equation} As a rule of thumb: "the expected value of the integral is the integral of the expected value".
For the autocovariance function we can proceed similarly (note that I have dropped the "random" argument). Again we use Fubini's theorem to interchange expectation and integral, \begin{align} E(X(t)X(t-k)) &= E\left(\int_{\mathbb{R}} h(u)Y(t -u) \, \mathrm{d}u\int_{\mathbb{R}} h(v)Y(t-k-v) \, \mathrm{d}v\right) \\ &= E\left(\int_{\mathbb{R}}\int_{\mathbb{R}} h(u)h(v)Y(t -u)Y(t-k-v) \, \,\mathrm{d}u\,\mathrm{d}v\right) \\ &= \int_{\mathbb{R}}\int_{\mathbb{R}} h(u)h(v)E[Y(t -u)Y(t-k-v)] \, \,\mathrm{d}u\,\mathrm{d}v. \end{align} Using the fact that $E[Y(t -u)Y(t-k-v)] = \text{Cov}_Y(t-u,t-k-v)$ and since $Y$ is stationary, $\text{Cov}_Y(t-u,t-k-v) = \gamma_Y(t-u -(t-k-v)) = \gamma_Y(k+v-u)$. Thus, \begin{equation} \gamma_X(k) = \int_{\mathbb{R}}\int_{\mathbb{R}} h(u)h(v)\gamma_Y(k+v-u) \, \,\mathrm{d}u\,\mathrm{d}v. \end{equation}
More detail on why this is OK to do:
We have, \begin{align} E_{\omega}\left[\int_{\mathbb{R}} h(u)Y(t-u,\omega)\,\mathrm{d}u\right] &= \int_{\mathbb{R}}\int_{\mathbb{R}} h(u)Y(t-u,\omega)\,\mathrm{d}u\, f(\omega)\,\mathrm{d}\omega \\ &= \int_{\mathbb{R}} h(u)\int_{\mathbb{R}} Y(t-u,\omega)f(\omega)\,\mathrm{d}\omega\,\mathrm{d}u \\ &= \int_{\mathbb{R}} h(u) E_\omega[Y(t-u,\omega)]\,\mathrm{d}u. \end{align} Fubini's theorem is used to interchange the integral signs in the second step (given of course that the conditions of Fubini's theorem are fulfilled).