Autocovariance

Definition

Let $\{Y_t\}$ set of random variable with mean ${\mu }_s$

$$\begin{array}{rl}{\mu }_t& =E[Y_t]\\ {\gamma }_{t,s}& =\mathrm{Cov}(Y_t,Y_s)=E[(Y_t-{\mu }_t)(Y_s-{\mu }_s)]\\ {\rho }_{t,s}& =\mathrm{Corr}(Y_t,Y_s)=\frac{{\gamma }_{t,s}}{\sqrt{{\gamma }_{t,t}{\gamma }_{s,s}}}\end{array}$$

Tip

autocov and autocorr is, respectively, cov and corr of 2 random variables from same sequence at two time points

Just memorize cov and corr formulas.

Recall

• $E[X]$ : Expectation of random variable $X$
• $\text{Cov}(X,Y)$ : Covariance
• $\text{Corr}(X,Y)$ : Correlation

Properties

$$\begin{array}{llc}{\gamma }_{t,t}=\mathrm{Var}(Y_t)& & {\rho }_{t,t}=1\\ {\gamma }_{t,s}={\gamma }_{s,t}& & {\rho }_{t,s}={\rho }_{s,t}\\ |{\gamma }_{t,s}|\le \sqrt{{\gamma }_{t,t},{\gamma }_{s,s}}& & |{\rho }_{t,s}|\le 1\end{array}$$

Exmaple: Autocovariance of an AR(1) model

Let $Y_t=0.7Y_{t-1}+e_t$, where $e_t\sim WN(0,{\sigma }_e^2)$

Notice that $0.7<1$. By Stationarity Condition, $Y_t$ is weakly stationary. This means:

1. $E[Y_t]=\mu$ is constant for all $t$
2. ${\gamma }_k$ only depends on lag $k$. It is independent of $t$

As a result of point 2, the variance is constant: ${\gamma }_0=\text{Var}(Y_t)={\sigma }^2$

Because $\{e_t\}$ is independent of its past values, $Y_t$ is also independent of its past values, since $Y_t$ is made up of values of $\{e_t\}$.

Notice that

$$\begin{array}{rl}\text{Var}(Y_t)& =E[Y_t^2]-E[Y_t{]}^2\\ ⟺E[Y_t^2]& =\text{Var}(Y_t)+{\mu }^2={\sigma }^2+{\mu }^2\end{array}$$

Its autocovariance at lag $1$ is

$$\begin{array}{rl}{\gamma }_1& =E[Y_t{Y}_{t-1}]-E[Y_t]E[Y_{t-1}]\\ & =E[(0.7Y_{t-1}+e_t)Y_{t-1}]-{\mu }^2\\ & =E[0.7Y_{t-1}^2+e_t{Y}_{t-1}]-{\mu }^2\\ & =0.7E[Y_{t-1}^2]+E[e_t{Y}_{t-1}]-{\mu }^2\\ & =0.7({\sigma }^2+{\mu }^2)+0-{\mu }^2\\ & =0.7{\sigma }^2\end{array}$$

Solving for ${\sigma }^2$:

$$\begin{array}{rl}\text{Var}(Y_t)& =\text{Var}(0.7Y_{t-1}+e_t)\\ & =0.49\text{Var}(Y_{t-1})+\text{Var}(e_t)+2\text{Cov}(Y_{t-1},e_t)\\ {\sigma }^2& =0.49{\sigma }^2+{\sigma }_e^2\\ 0.51{\sigma }^2& ={\sigma }_e^2\\ ⟺{\sigma }^2& =\frac{{\sigma }_e^2}{0.51}\end{array}$$

Continuing for autocovariance:

$$\begin{array}{rl}{\gamma }_1& =0.7\left(\frac{{\sigma }_e^2}{0.51}\right)\\ & =1.372{\sigma }_e^2\end{array}$$