# Lecture 16 (Feb 28, 2022)

Model checking:

for the residual, we could standardize it $\tilde{r}_t = r_t/\sigma_t$ and we do Ljung Box test on $\tilde{r}_t$ and $\tilde{r}_t^2$.

 Since $\sigma_t^2 = \alpha_0 + \alpha_1 r_{t-1}^2 + \dots + \alpha_p r_{t-p}^2$ and since $\mathbb{E}[r_t \mathcal{F}_{t-1}] = \sigma_t^2$ we could approximately write
$r_t^2 = \alpha_0 + \alpha_1 r_{t-1}^2+\dots \alpha_pr_{t-p}^2$

and we could use PACF to determine the order $p$.

### Generalized ARCH

 Input : $x_t$ and model $r_t = x_t - \mathbb{E}[x_t \mathcal{F}_{t-1}]$ and $r_t = \sigma_t \epsilon_t$ with $\epsilon_t$ iid.

The generalized ARCH models the conditional variance as

$\sigma_t^2 = \alpha_0 + \sum_{i=1}^p \alpha_i r_{t-i}^2 + \sum_{j=1}^q \beta_j \sigma_{t-j}^2$

The variance of $r_t$ is

$\text{Var}[r_t^2] = \mathbb{E}[\sigma_t^2] = \alpha_0 + \sum_{i=1}^p \alpha_i \mathbb{E}[r_{t-i}^2] + \sum_{j=1}^q \beta_j \mathbb{E}[\sigma_{t-j}^2]$

Thus, we have

$\text{Var}[r_t] = \frac{\alpha_0}{1-\sum_{i=1}^p \alpha_i -\sum_{j=1}^q \beta_j}$

Therefore we need $\sum_{i=1}^p \alpha_i + \sum_{j=1}^q \beta_j < 1$

if we define $\eta_t = r_t^2-\sigma_t^2$. Then, we could write

\begin{aligned} r_t^2 &= \eta_t + \sigma_t^2 = \eta_t + \alpha_0 + \sum_{i=1}^p \alpha_i r_{t-i}^2 + \sum_{j=1}^q \beta_j (\sigma_{t-j}^2 - r_{t-j}^2 + r_{t-j}^2) \\ &= \eta_t + \alpha_0 + \sum_{i=1}^{p\vee q} (\alpha_i + \beta_i) r_{t-i}^2 - \sum_{j=1}^q \beta_j\eta_{t-j} \end{aligned}

which is in the ARMA form but the error $\eta$ is not independent.

The error sequence has mean zero

$\mathbb{E}[\eta_t] = \mathbb{E}[r_t^2] - \mathbb{E}[\eta_t^2] = 0$

Then

\begin{aligned} \mathbb{E}[\eta_t\eta_{t-j}] &= \mathbb{E}[\mathbb{E}[(r_t^2-\sigma_t^2)(r_{t-j}^2-\sigma_{t-j}^2)|\mathcal{F}_{t-1}]] \\ &= \mathbb{E}[\sigma_t^2\sigma_{t-j}^2(\epsilon_{t-j}^2-1)\mathbb{E}[\epsilon_t^2-1]] = 0 \end{aligned}

Example: GARCH(1,1) model.

$\sigma_t^2 = \alpha_0 + \alpha_1r_{t-1}^2 + \beta_1\sigma_{t-1}^2$

The unconditional Kurtosis is then K

$K = \frac{\mathbb{E}r^4_t}{\mathbb{E}[r_{t}^2]^2} = \frac{3[1-(\alpha_1+\beta_1)^2]}{1 - (\alpha_1+\beta_1)^2 -2 \alpha_1^2} > 3$

provided that $(\alpha_1+\beta_1)^2+2\alpha_1^2 < 1$. Thus, the tail distribution of GARCH(1,1) residuals is heavier than that of the $\mathcal{N}(0,1)$.

### Forecast:

$1$ step ahead prediction of $\sigma_t^2$ :

$\widehat{\sigma_{t+1}^2} = \mathbb{E}[\sigma_{t+1}^2| \mathcal{F}_t] = \alpha_0 + \alpha_1 r_{t}^2 + \beta_1 \sigma_{t}^2$

Note:

$\sigma_{t+1}^2 = \alpha_0 + (\alpha_1+\beta_1) \sigma_t^2 + \alpha_1(r_t^2-\sigma_t^2)$

Thus,

$\sigma_{t+2}^2 = \alpha_0 + (\alpha_1 + \beta_1)\sigma_{t+1}^2 + \alpha_1(\epsilon_{t+1}^2-1)\sigma_{t+1}^2$

and we could recurse to get

$\widehat{\sigma_{t+2}^2} = \alpha_0 + (\alpha_1+\beta_1)\mathbb{E}[\sigma_{t+1}^2| \mathcal{F}_t] = \alpha_0 + (\alpha_1+\beta_1)\widehat{\sigma_{t+1}^2}$

which forms a recursion for forecast. This exponential structure shows the mean reverting behaviour and the variance converge to the uncoditional variance $\mathbb{E}[r_t^2]$.

Updated: