# Lecture 01 (Jan 21, 2022)

### Autocorrelation

Let $(x_t)_{t=1}^T$ be a sequence of random variables. We calssify the sequence as:

• strict stationarity : for any fixed $k\geq 1$ and $t_1,\dots, t_k$ collection of positive integer, $(x_{t_1}, \dots, x_{t_k})$ has the same distributin of $(x_{t_1+l}, \dots, x_{t_k+l})$ for all $l\in \mathbb{Z}$.
• weak stationarity :
1. $\mathbb{E}(x_t) = \mu$ constant
2. $\text{Cov}(x_t, x_{t-l}) = \gamma_l$ for $l\in\mathbb{Z}$. (This is called lag-l auto-covariance of $x_t$)

properties of $\gamma_l$ for weak stationary $x_t$ :

1. $\gamma_0 = \text{Cov}(x_t, x_t) = \text{Var}(x_t)$
2. $\gamma_{-l} = \text{Cov}(x_t, x_{t+l}) = \text{Cov}(x_{t-l}, x_t) = \gamma_l$.

Def the correlation coefficient between $x_t$ and $x_{t-l}$ for a weakly stationary time series ${x_t}$ is called the lag-l autocorrelation of $x_t$ (ACF), denoted by $\rho_l$:

$\rho_l = \frac{\text{Cov}(x_t, x_{t-l})}{\sqrt{\text{Var}(x_t)}\sqrt{\text{Var}(x_{t-l})}} = \frac{\gamma_l}{\gamma_0}$

properties of $\rho_l$ :

1. $\rho_0 = 1$
2. $\rho_{-l} = \rho_l$.
3. $-1\leq \rho_l \leq 1$. (by correlation property which implied by Cauchy-Swartz)
4. $\rho_l = 0, \forall l\geq 1$ is equivalent to $x_t$ is not serielly correlated.

Def [lag-$l$ sample auto-correlation] :

$\hat{\rho}_l = \frac{\sum_{t=l+1}^T(x_t-\bar{x})(x_{t-l}-\bar{x})}{\sum_{t=1}^T(x_t-\bar{x})} \\\text{ where } \bar{x} = \frac{1}{T} \sum_{t=1}^T x_t \quad 0\leq l\leq T-1$

Remark: the numerator should be $1/(T-l)$ but when $l\ll T$, we will neglect the $-l$ and use $1/T$ directly and the $1/T$ cancles up and down and the $\hat{\rho}_l$ becomes what it is here.

Def [m-dependent sequence] A sequnce of random variables $y_1, y_2,\dots$ is m-dependent if for $\forall s \geq l$ and $s\in \mathbb{Z}$, ${y_1,\dots, y_s}$ is independent of ${y_{m+s+1}, y_{m+s+2},\dots}$ .

Example: $x_1,x_2,\dots$ independent sequence, define $y_i = x_i x_{i+m}$ $\frac{1}{n} \sum_{i=1}^n y_i = \frac{1}{n}\sum_{i=1}^n x_i x_{i+m}$ is the auto product moment at lag-m. Here $y_i$ is an m-dependent sequence.

Suppose $y_1,y_2,\dots$ is a staitonary sequence, i.e. $\mu=\mathbb{E}y_1$ and $\gamma_0 = \text{Var}(y_1)$ and $\gamma_l = \text{Cov}(y_1, y_{1+l})$ . If in addition, $y_t$ is m-dependent, then $\gamma_i = 0$ for $\forall i > m$. Let $S_n =\sum_{i=1}^n y_i$. Then $\mathbb{E}[S_n] = \sum_{i=1}^n \mathbb{E}[y_i] = n\mu$.

\begin{aligned} \text{Var}(S_n) &= \text{Cov} \left(\sum_{i=1}^n y_i, \sum_{j=1}^n y_j \right) \\ &= \sum_{i=1}^n \sum_{j=1}^n \text{Cov} (y_i, y_j) \\ &= \sum_{i=1}^n\sum_{j=1}^n \gamma_{i-j} \\ &= n \gamma_0 + 2(n-1) \gamma_1 + 2(n-2)\gamma_2+\dots + 2 (n-m) \gamma_m + \dots + 2\gamma_n \\ &= n \gamma_0 + 2(n-1)\gamma_1+ 2(n-2)\gamma_2 +\dots +2 (n-m) \gamma_m \\ &= n\gamma_0 + 2 \sum_{i=1}^m (n-i) \gamma_i \end{aligned}

where we used $\gamma_i = \gamma_{-i}$. Then

$\frac{1}{n} \text{Var}(S_n) = \gamma_0 + 2 \sum_{i=1}^m \left(1-\frac{i}{n}\right)\gamma_i$ If $m$ ix fixed , then as $n\rightarrow \infty$ , $\frac{\text{Var}(S_n)}{n} \rightarrow \gamma_0 + 2 \sum_{i=1}^m \gamma_i = \sum_{i=-m}^m \gamma_i$

Theorem [CLT for m-dependent sequence] Let $y_1,y_2,\dots$ be stationary $m$-dpendent sequence with finite variance and $S_n = \sum_{i=1}^n y_i$. Then

$\sqrt{n} \left(\frac{S_n}{n} - \mu\right) \rightarrow \mathcal{N}(0,\tau^2)$ as $n\rightarrow \infty$ where

$\tau^2 = \sum_{i=-m}^m \gamma_i = \gamma_0 + 2\sum_{i=1}^m \gamma_i \quad \mu = \mathbb{E}[y_1]$

Updated: