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Postscript version of these notes

STAT 804

Lecture 22 Notes

Properties of the Periodogram

The discrete Fourier transform

\begin{displaymath}{\hat X}(\omega) = \frac{1}{\sqrt{T}}\sum_{t=0}^{T-1} X_t \exp(2\pi \omega t i)

is periodic with period 1 because all the exponentials have period 1. Moreover,

\begin{displaymath}{\hat X}(1-\omega) = \frac{1}{\sqrt{T}}\sum_{t=0}^{T-1} X_t \...
\omega t i) \exp(2\pi t i) = {\overline{{\hat X}(\omega)}}

so that the periodogram satisfies

\begin{displaymath}\vert{\hat X}(1-\omega)\vert^2 = \vert{\hat X}(\omega)\vert^2 \,

Thus the periodogram is symmetric around $\omega=1/2$ which is called the Nyquist or folding frequency. (The value is always 1/2 in cycles per point but usually it would be converted to cycles per time unit like year or day or whatever.)

Similarly the power spectral density fX given by

\begin{displaymath}f_X(\omega) = \sum_{-\infty}^\infty C_X(h) \exp(2 \pi h \omega i)

is periodic with period 1 and satisfies

\begin{displaymath}f_X(-\omega) = f_X(\omega)

which is equivalent to

\begin{displaymath}f_X(1-\omega) = f_X(\omega) \, .

Spectra of Some Basic Processes

Here I compute the spectra of a few basic processes directly and then indirectly by a more powerful technique.

Direct from the definition

White Noise:
Since $C_\epsilon(k)=0$ for all non-zero k we have

\begin{displaymath}f_\epsilon(\omega) \equiv C_\epsilon(0) = \sigma^2_\epsilon \, .

For $X_t = \epsilon_t - b \epsilon_{t-1}$ we have $C_X(0)= \sigma^2(1+b^2)$ and $C_X(\pm 1) = -b\sigma^2$ so that

\begin{eqnarray*}f_X(\omega)& =& \sigma^2(1+b^2)-b\sigma^2 (\exp(2\pi\omega i)
...\omega i))\\
& = & \sigma^2(1+b^2)-2b\sigma^2 \cos(2\pi\omega)

We have $C_X(k) = \rho^{\vert k\vert} C_X(0)$ and

\begin{eqnarray*}f_X(\omega)& =& C_X(0)( 1+\sum_{k>0} \rho^k(\exp(2\pi\omega ki)...
& = & \frac{\sigma_\epsilon^2}{1+\rho^2 -2\rho\cos(2\pi\omega)}

Using filters

Any mean 0 ARMA(p,q) process can be rewritten in MA form as

\begin{displaymath}X_t = \sum_{s=0}^\infty a_s \epsilon_{t-s}

and then the covariance of X is

\begin{displaymath}C_X(h) = \sum_{r=0}^\infty \sum_{s=0}^\infty a_r a_s Cov(\epsilon_{t+h-r},

Although the covariance simplifies for white noise, let us simply write C(t+h-r -(t-s)) = C(h+s-r) for the covariance in this double sum so that the calculation will apply to any stationary $\epsilon$. Then plug this double sum into the definition of fX to get

\begin{displaymath}f_X(\omega) = \sum_{h=-\infty}^\infty \exp(2\pi\omega h i)
\sum_{r=0}^\infty \sum_{s=0}^\infty a_r a_s C(h+s-r) \\

Now write the h in the complex exponential in the form (h+s-r) +r-sand bring the sum over h to the inside to get

\begin{displaymath}f_X(\omega) = \sum_{r=0}^\infty a_r e^{2\pi\omega ri} \sum_{s...
... si} \sum_{h=-\infty}^\infty \exp(2\pi\omega(h+s-r)i)

Finally make the substitution k=h+s-r in the inside sum and define

\begin{displaymath}A(\omega) = \sum_{r=0}^\infty a_r e^{2\pi\omega ri}

to see that

\begin{displaymath}f_X(\omega) = A(\omega) {\overline{A(\omega)}} f_\epsilon(\omega)


\begin{displaymath}f_X(\omega) = \vert A(\omega)\vert^2 f_\epsilon(\omega) \, .

The function A (or $\bar A$) is called the frequency response function and |A|2 the power transfer function. The jargon gain is sometimes used for |A|.

The Spectrum of an ARMA(p,q)

An ARMA(p,q) process X satisfies

\begin{displaymath}\sum_{s=0}^p a_s X_{t-s} =
\sum_{r=0}^q b_r \epsilon_{t-r}

so that if Y is the process $\sum_{s=0}^p a_s X_{t-s}$ then

\begin{displaymath}f_Y(\omega) = \vert A(\omega)\vert^2 f_X(\omega)

where $A(\omega) = \sum_{s=0}^p a_s \exp(2\pi\omega si)$. At the same time $Y_t = \sum_{r=0}^q b_r \epsilon_{t-r}$ so

\begin{displaymath}f_Y(\omega) = \vert B(\omega)\vert^2 f_\epsilon(\omega)

where $B(\omega) = \sum_{s=0}^q b_s \exp(2\pi\omega si)$. Hence

\begin{displaymath}f_X(\omega) = \frac{\vert B(\omega)\vert^2}{\vert A(\omega)\vert^2}\, .

For example, in the ARMA(1,1) case $X_t-aX_{t-1} = \epsilon_t-b\epsilon_{t-1}$we find (referring to our MA(1) calculation above that $\vert B(\omega)\vert^2 = 1+b^2-2b \cos(2\pi\omega)$ and $\vert A(\omega)\vert^2 = 1+a^2-2a \cos(2\pi\omega)$ so that

\begin{displaymath}f_X(\omega) = \frac{1+b^2-2b \cos(2\pi\omega)}{1+a^2-2a \cos(2\pi\omega} \, .

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Richard Lockhart