Suppose $z_{t} = x_{t}y_{t}$ where $x_{t}$ and $y_{t}$ are 0 mean, independent stationary stochastic process. What is the autocovariance function of $z_{t}$? Show that the spectral density can be written as $f_{z}(\omega) = \int_{-0.5}^{0.5}f_{x}(\omega-v)f_{y}(v)dv$.


If the series are independent, wouldn't the autocovariance function be $E[z_{t}z_{t+h}]$? This would then be $E[x_{t}y_{t}x_{t+h}y_{t+h}]$. I am unsure how to proceed from here. I am unsure how to prove how the spectral density can be written as the above integral. I know the autocorrelation function $\gamma_{z}(h)$ can be written in terms of the integral but I did not know there was an integral representation for $f_{z}(\omega)$.

So how does one represent the spectral density of z. From my textbook, this is $\sum_{h=-\infty}^{h=\infty}\gamma_{z}(h)e^{-2\pi\omega h}$


For the first question and because $\left( x_t \right)_t$ and $\left( y_t \right)_t$ are independent with zero expectations \begin{eqnarray*} \gamma_z \left( h \right) & = & E \left[ z_t z_{t - h} \right]\\ & = & E \left[ x_t y_t x_{t - h} y_{t - h} \right]\\ & = & E \left[ x_t x_{t - h} \right] E \left[ y_t y_{t - h} \right]\\ & = & \gamma_x \left( h \right) \gamma_y \left( h \right) \end{eqnarray*}

  • $\begingroup$ How does one find the spectral decomposition? $\endgroup$ – phil12 Apr 30 '13 at 19:50
  • $\begingroup$ @phil12 It is not a good idea to ask two questions in one post. Anyway as hint, write $f_z$ as a function of $\gamma_z$, replace $\gamma_z$ by the product found in the first question, apply the inverse transform to $\gamma_x$, interchange summation and integration (after justifying why you could do it), apply the inverse transform to $\gamma_y$ and you've got your answer. $\endgroup$ – Learner May 1 '13 at 3:57

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