1 Discrete Fourier Transformation

Denote ${\displaystyle c^f=(\cos (2\pi ft),t=0,1,\cdots ,n-1)}$, and ${\displaystyle s^f=(\sin (2\pi ft),t=0,1,\cdots ,n-1)}$.

${\displaystyle \cos (2\pi ft)=\frac{{\mathrm{e}}^{2\pi \mathrm{i}ft}+{\mathrm{e}}^{-2\pi \mathrm{i}ft}}{2},\sin (2\pi ft)=\frac{{\mathrm{e}}^{2\pi \mathrm{i}ft}-{\mathrm{e}}^{-2\pi \mathrm{i}ft}}{2\mathrm{i}}}$.

Denote ${\displaystyle u^f=(\exp (2\pi \mathrm{i}ft),t=0,1,\cdots ,n-1)}$. ${\displaystyle f=0,\frac{1}{n},\frac{2}{n},\cdots ,\frac{n-1}{n}}$.

Now, given time series data ${\displaystyle y_0,y_1,\cdots ,y_{n-1}}$. Because ${\displaystyle u^0,u^{\frac{1}{n}},\cdots ,u^{\frac{n-1}{n}}}$ form a basis, $$y=a_0{u}^0+a_1{u}^{\frac{1}{n}}+\cdots +a_{n-1}{u}^{\frac{n-1}{n}},$$ for some ${\displaystyle a_0,a_1,\cdots ,a_{n-1}\in \mathbb{C}}$. Then $$\begin{array}{rl}⟨y,u^{\frac{j}{n}}⟩& =⟨a_0{u}^0+a_1{u}^{\frac{1}{n}}+\cdots +a_{n-1}{u}^{\frac{n-1}{n}},u^{\frac{j}{n}}⟩\\ & =\end{array}$$
Here ${\displaystyle a_j=\frac{1}{n}⟨y,u^{\frac{1}{n}}⟩}$. So define discrete Fourier Transform (DFT) of ${\displaystyle y=(y_0,\cdots ,y_{n-1})}$: (${\displaystyle y_i\in \mathbb{R}}$) $$\begin{array}{rl}{b}_j& =⟨y,u^{\frac{j}{n}}⟩=\sum _{t=0}^{n-1}{y}_t\exp (-2\pi \mathrm{i}\frac{j}{n}t)\\ & =\sum _{t=0}^{n-1}{y}_t\cos \left(2\pi \frac{j}{n}t\right)-\mathrm{i}\sum _{t=0}^{n-1}{y}_t\sin \left(2\pi \frac{j}{n}t\right),j=0,1,\cdots ,n-1.\end{array}$$
We can use np.fft.fft() to compute it.

1.1 Properties of DFT

1. $$b_0=\sum _{t=0}^{n-1}{y}_t=y_0+\cdots +y_{n-1}.$$
2. $$\text{begin{align*}}$$

&= \overline{\sum_{t=0}^{n-1}y_{t}\exp \left(-2\pi \mathrm{i} \frac{j}{n}\right)t}=\overline{b_{j}}.
\end{align*}$$

Inverse DFT

The formula is given by $$y_t=\frac{1}{n}\sum _{j=0}^{n-1}{b}_j\exp \left(\frac{2\pi \mathrm{i}jt}{n}\right).$$

2 Periodogram

Define periodogram $$I\left(\frac{j}{n}\right)=\frac{|b_j{|}^2}{n},0<\frac{j}{n}<\frac{1}{2}.$$

Since $$b_j=\sum _{t=0}^{n-1}{y}_t\exp (-\frac{2\pi \mathrm{i}jt}{n})=\sum _{t=0}^{n-1}{y}_t(\cos \frac{2\pi jt}{n}-\mathrm{i}\sin \frac{2\pi jt}{n}),$$ periodogram can be written as $$I\left(\frac{j}{n}\right)=\frac{1}{n}[{(\sum _{t=0}^{n-1}{y}_t\cos \frac{2\pi jt}{n})}^2+{(\sum _{t=0}^{n-1}{y}_t\sin \frac{2\pi jt}{n})}^2],0<\frac{j}{n}\le \frac{1}{2}.$$

Use of periodogram:

1. If you are trying to fit ${\displaystyle y_t={\beta }_0+{\beta }_1\cos 2\pi ft+{\beta }_2\sin 2\pi ft+{\epsilon }_t}$, we know by here $$\mathrm{posterior}(f)\propto {\left(\frac{1}{\mathrm{RSS}(f)}\right)}^{\frac{n-3}{2}}|X_f^{\mathrm{T}}{X}_f{|}^{-\frac{1}{2}}I\{0<f<\frac{1}{2}\}.$$ If you use Fourier frequencies, ${\displaystyle \mathcal{F}=\{\frac{1}{n},\cdots ,\frac{[n/2]}{2}\}}$, then $$\mathrm{RSS}(f)=\sum _{t=1}^n(y_t-\bar{y}{)}^2-2I(f).$$

2. Periodogram can suggest other models for the data. For $$y_t={\beta }_0+{\beta }_1\cos 2\pi f_{1}t+{\beta }_2\sin 2\pi f_{1}t+{\beta }_3\cos 2\pi f_{2}t+{\beta }_4\sin 2\pi f_{2}t+{\epsilon }_t.$$