Suppose that $X_t$ is a weakly stationary process; then, its autocovariance function can be represented as: $$\gamma_X(h) = \int_{(-\pi , \pi ]} e^{i h v} dF(v)$$ The function $F$ is called the spectral distribution of $X_t$. If $F$ admits a density with respect to the Lebesgue measure, say $f$, we say that $f$ is the spectral density of $X_t$. I am interested in the following problem (taken from Brockwell and Davis, Time Series: Theory and Methods, Exercise 4.15):

Suppose $x_t = w_t - 2w_{t-1}$, where $w_t$ is a mean zero white noise sequence with variance $\sigma_w^2$. Given $\epsilon > 0$, find an integer $k$ and constants $a_0, \ldots , a_k$, with $a_0=1$, such that if $f_y$ is the spectral density of the process \begin{align*} y_t = \sum_{j=0}^k a_j x_{t-j} \end{align*} then, \begin{align*} \sup_{-0.5 \leq \omega \leq 0.5} \left| f_y(\omega ) - \frac{\mathrm{Var}({y_t})}{2\pi} \right| < \epsilon \end{align*}

Motivation: The key to this problem should be that for every symmetric spectral density (e.g. the variance of $y_t$), there exists an invertible $MA(q)$ process $X_t$ such that the spectral density of $X_t$ is within an epsilon of the initial symmetric spectral density. However, the process in this question takes away some freedom from which MA processes I can choose from to approximate the variance of $y_t$, which makes it non-obvious how to calculate the desired coefficients.


1 Answer 1


The solution of problem (taken from the book of Brockwell and Davis) will be given in the following.

The process $ Z_t, X_t$ and $Y_t$ are stationary processes and their spectral density $ f_Z(\lambda), f_X(\lambda)$ and $f_Y(\lambda)$ are, respectively, \begin{align*} f_Z(\lambda)&=\frac{\sigma^2}{2\pi},\\ f_X(\lambda)&=|1-2\mathrm{e}^{-i\lambda}|^2\frac{\sigma^2}{2\pi} =|1-\tfrac12\mathrm{e}^{-i\lambda}|^2\frac{2\sigma^2}{\pi},\\ f_Y(\lambda)& =\Big|\sum_{j=0}^{k}a_j\mathrm{e}^{-ij\lambda} \Big|^2 f_X(\lambda)\\ &=\Big|\sum_{j=0}^{k}a_j\mathrm{e}^{-ij\lambda} \Big|^2 |1-\tfrac12\mathrm{e}^{-i\lambda}|^2\frac{2\sigma^2}{\pi} . \end{align*} Now if $ a_j=2^{-j}, j\ge1 $, then \begin{align*} &\bigg|1- \Big|\sum_{j=0}^{k}a_j\mathrm{e}^{-ij\lambda} \Big|^2 |1-\tfrac12\mathrm{e}^{-i\lambda}|^2\bigg| \\ &\qquad=\bigg|1- \Big|\sum_{j=0}^{k}(\tfrac12\mathrm{e}^{-i\lambda})^j \Big|^2 |1-\tfrac12\mathrm{e}^{-i\lambda}|^2 \bigg| \\ &\qquad =|1- |1-(2\mathrm{e}^{i\lambda})^{-(k+1)}|^2 |\\ &\qquad \le 2^{-2(k+1)}<\frac{\varepsilon}{2\sigma^2},\qquad \text{if } k\ge \dfrac{-\log(\varepsilon/\sigma^2)}{\log4} . \end{align*} and \begin{gather*} \sup_{-\pi\le\lambda\le\pi}\Big|f_Y(\lambda)-\frac{2\sigma^2}{\pi}\Big| <\frac{\varepsilon}{2},\\ \Big|\frac{\mathsf{Var[Y_t]}}{2\pi}-\frac{2\sigma^2}{\pi}\Big| =\Bigg|\frac{1}{2\pi}\int_{-\pi}^{\pi}\Big[f_Y(\lambda)-\frac{2\sigma^2}{\pi}\Big]\, \mathrm{d}\lambda \Bigg|\le \frac{\varepsilon}{2}. \end{gather*} Hence, \begin{equation*} \sup_{-\pi\le\lambda\le\pi}\Big|f_Y(\lambda)-\frac{\mathsf{Var}[Y_t]}{2\pi}\Big| <\varepsilon. \end{equation*}

  • $\begingroup$ Looks good! Is there a typo in your factorisation of $f_X$ (second equation from the top)? $\endgroup$ Commented Jun 11, 2021 at 13:25
  • $\begingroup$ Thank you for your reply. I use the following equation: $|1-2e^{-i\lambda}|=2|1-\frac{1}{2}e^{i\lambda}|=2|1-\frac12e^{-i\lambda}|$. $\endgroup$
    – JGWang
    Commented Jun 12, 2021 at 3:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .