Approximating the spectral density of white noise by a moving average process 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.
 A: 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*}
