# Stationarity of convolution filter of process

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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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).

• Excelent! Thank you so much! I will edit my comment as I try to evaluate the autocovariance function - and see if the answer matches. – YetAnotherUsr Sep 3 '19 at 6:09
• I edited my question – YetAnotherUsr Sep 4 '19 at 21:53
• @M.Gonzalez I edited my answer – Timothy Hedgeworth Sep 4 '19 at 23:10
• Thanks, Timothy! :) – YetAnotherUsr Sep 5 '19 at 13:41