Derive PSD of an MA(1) process

Question

Suppose there is a moving average process of order 1 that also includes a zeroth term thus:

\begin{equation} Y_t=\theta\left(B\right)X_t = \left(\theta_0+\theta_1B\right)X_t \end{equation}

where $X_t$ is white noise (normally distributed). Now I know that the PSD can be determined from:

\begin{equation} f\left(\omega\right) = \sigma_w^2\left|\theta\left(\mathrm{e}^{-2\pi\mathrm{i}\omega}\right)\right|^2 = \sigma_w^2\left|\theta_0+\theta_1\mathrm{e}^{-2\pi\mathrm{i}\omega}\right|^2 \end{equation}

I know the standard result is:

\begin{equation} \sigma_w^2\left(\theta_0^2+\theta_1^2+2\theta_0\theta_1 \cos \left(2\pi\omega\right)\right) \end{equation}

But when I multiply out the terms I cannot see how to simplify:

\begin{equation} \theta_0^2+\theta_0\theta_1\cos\left(-2\pi\omega\right) +\mathrm{i}\theta_0\theta_1\sin\left(-2\pi\omega\right)+ \end{equation} \begin{equation} \theta_0\theta_1\cos\left(-2\pi\omega\right) +\theta_1^2\cos^2\left(-2\pi\omega\right) + \mathrm{i}\theta_1^2\cos\left(-2\pi\omega\right)\sin\left(-2\pi\omega\right)+ \end{equation} \begin{equation} \mathrm{i}\theta_0\theta_1\sin\left(-2\pi\omega\right)+\mathrm{i}\theta_1^2\cos\left(-2\pi\omega\right)\sin\left(-2\pi\omega\right)-\theta_1^2\sin^2\left(-2\pi\omega\right) \end{equation}

I can see how to obtain:

\begin{equation} \theta_0^2+\theta_1^2\left(\cos^2\left(2\pi\omega\right)+\sin^2\left(2\pi\omega\right)\right) \end{equation}

but I can't simplify the rest. I need to do this for more complicated processes, so any guidance on how to deal with he extra terms would be much appreciated.

Answer 1

Keep in mind, if you have $\lvert A(\omega)\rvert^{2}$, this can be written as $A(\omega)A^{*}(\omega)$. So, doing this for $f(\omega)$, we can write

\begin{equation} f(\omega) = \sigma_{\omega}^{2}\lvert\theta_{0}+\theta_{1}e^{-2\pi i\omega}\rvert^{2} = \sigma_{\omega}^{2}(\theta_{0}+\theta_{1}e^{-2\pi i\omega})(\theta_{0}+\theta_{1}e^{2\pi i\omega})\end{equation}

Expanding this out, we get

\begin{equation}\sigma_{\omega}^{2}[\theta_{0}^{2} + \theta_{1}^{2} + \theta_{0}\theta_{1}e^{-2\pi i\omega} + \theta_{0}\theta_{1}e^{2\pi i\omega}] = \sigma_{\omega}^{2}[\theta_{0}^{2} + \theta_{1}^{2} + \theta_{0}\theta_{1}(e^{2\pi i\omega} + e^{-2\pi i\omega})]\end{equation}

Using Euler's formula for a cosine, we get

\begin{equation}f(\omega) = \sigma_{\omega}^{2}[\theta_{0}^{2} + \theta_{1}^{2} + 2\theta_{0}\theta_{1}cos(2\pi\omega)]\end{equation}

Let me know if you need anything further!