## Some notes on Linear Filters

Let $\{ X_t \}$ and  $\{ Y_t \}$ be two stationary time series related by:

$X_{t} = M_{n}(L)Y_{t} =\sum_{j=- \infty}^{j=\infty} g_{j}Y_{t-j}$

where

$\sum_{j=- \infty}^{j=\infty}|g_{j}| < \infty$ and  $\sum_{j=- \infty}^{j=\infty}|g_{j}|^2 < \infty$

$\{ X_t \}$ is the filtered version of  $\{ Y_t \}$ and $M_{n}(L)$ is the filter. The effect of a linear filter is to change the importance of various cyclical components of the series and/or induce a shift with regard to the position in time.

Note that

${M_{n}(1) = 1}$: Trend Extraction
${M_{n}(1) = 0}$: Detrending

A Linear time-invariant filter demonstrates the following property:

$\displaystyle X_{t}=M(L)Y_{t} = M(L) \Psi (L) \epsilon_{t}$

We can show that the spectral density function f of the filtered series is equal to

$\displaystyle f_{y}(\lambda)=\underbrace{|M(e^{-i\lambda})|^{2}}_{| \Gamma ( \omega) |^{2}}f_{x}(\lambda)$

where $| \Gamma ( \omega) |^{2}|$ is the Transfer Function:

$\displaystyle | \Gamma ( \omega) |^{2} =\left| \sum_{j=-r}^{r} g_{j}e^{-i j \omega} \right| ^2$

$\displaystyle \Gamma ( \omega)$ is the Frequency Response Function:

$\displaystyle \Gamma ( \omega) = M_{n}(e^{-i \omega})=\sum_{j=-r}^{r} g_{j}e^{-i j \omega}$

and $\displaystyle M_{n}(\omega)$ is the Gain:

$\displaystyle M_{n}(\omega) = |M_{n}(e^{-i \omega}) |$

Τhe Gain indicates the factor by which the amplitude of a cyclical component changes as a result of applying the linear filter. On the other side the phase Ph(λ) measures the shift in time

$\displaystyle \boxed{Ph(\lambda)=tan^{-1}[-M^{\dotplus}(\lambda)/M^{*}(\lambda)], \; \; \; \; 0\leqslant\lambda\leqslant\pi }$

where

$\displaystyle M(e^{-i\lambda})=iM^{\dotplus}(\lambda)+M^{*}(\lambda)$

Note finally that if k multiple filters are applied consecutively

$\displaystyle X_{t}=M_{k}(L)...M_{2}(L)M_{1}(L)Y_{t}$

their combined effect is simply given by the product of the respective frequency response functions. Hence:

$\displaystyle M(e^{-i\lambda})=M_{k}(e^{-i\lambda})...M_{2}(e^{-i\lambda})M_{1}(e^{-i\lambda})$