Summary Sheet

ARMA model forecast

$$(1-0.5B)(Z_t - 3) = a_t$$

Let $X_t = Z_t - 3$, then $(1-0.5B)X_t = a_t\sim AR(1)$. And because the root $B=2$ lies outside of the unit circle, the series is stationary. Rewrite as a $MA(\infty)$

$$(1-0.5B)(1+\psi_1B + \psi_2B^2 + ...) = 1$$

$\psi_1 = 1/2$
$\psi_2 = 1/4$
$\psi_k = 2^{-k}$

Then, $\hat X_t(1) = \sum_{0}^\infty 2^{-(i+l)}a_{t-i}, \hat Z_t(l) = 3 + \hat X_t(l)$

$e_t(l)=\sum_{i=0}^{l-1} \psi_i a_{t+l-i}$,
then $var(e_t(l)) = \sum_{i=0}^{l-1} \psi_i^2 a_{t+l-i}^2 = \sigma^2 \sum_0^{l-1}\psi_i^2$

$$(1-B+0.25B^2)(Z_t-1)=a_t$$

Let $X_t = Z_t -1$, $(0.25B^2 - B + 1)=(0.5B-1)^2 \Rightarrow B=2$ is a stationary $AR(2)$ model.

$X_t = X_{t-1}+\frac{1}{4}X_{t-2} + a_t$, then
$\hat Z_t(1) = 1+X_t + \frac{1}{4} X_{t-1}$
$\hat Z_t(2) =1+ X_t + \frac{1}{4} X_{t-1}+ \frac{1}{4} X_t=\frac{5}{4}X_t + \frac{1}{4} X_{t-1}$
$\hat Z_t(3) = 1+\frac{5}{4} X_t + \frac{1}{4} X_{t-1}+ \frac{1}{4}(X_t+\frac{1}{4}X_{t-1})=\frac{3}{2}X_t + \frac{5}{16}X_{t-1}$ $\hat Z_t(4) = 1+\frac{3}{2}X_t + \frac{5}{16}X_{t-1}+\frac{5}{16}X_t + \frac{1}{16}X_{t-1} = \frac{29}{16}X_{t}+\frac{3}{8}X_{t-1}$

$(1-B+0.25B^2)(1+\psi_1B+\psi_2B^2+...) = 1$
$\psi_1 -1 = 0\Rightarrow \psi_1 = 1$
$\psi_2 -\psi_1 +0.25 = 0\Rightarrow \psi_2 = 0.75$
$\psi_3 -\psi_2 +0.25\psi_1 = 0\Rightarrow \psi_3 = 0.5$

Therefore,
$var(e_t(1)) = \sigma^2$
$var(e_t(2)) = 2\sigma^2$
$var(e_t(3)) = \frac{41}{16}\sigma^2$

$$(1-B+0.25B^2)(Z_t + 3) = (1+0.25B)a_t$$

All B lies out of the unit circle, it's a stationary and invertible ARMA(2,1) model.

$(1-B+0.25B^2)(1+\psi_1B+\psi_2B^2 + ...) = 1+0.25B$
$\psi_1 = 0.25+1=1.25$
$\psi_2 = \psi_1 = 1.25$
$\psi_3 = \psi_2 -0.25\psi_1 = \frac{15}{16}$
$\psi_4 = \psi_3 - 0.25\psi_2 = \frac{5}{8}$
$\psi_5 = \psi_4 - 0.25\psi_3 = \frac{25}{64}$

$var(e_t(1)) = \sigma^2$
$var(e_t(2)) = \frac{41}{16}\sigma^2$
$var(e_t(3)) = \frac{33}{8}\sigma^2$

TFN

Consider a dynamic regression model $y_t = \sum_0^k v_i x_{t-i}+n_t$ where both $x_t, n_t$ are stationary and invertible ARMA model given by $\phi_x(B)x_t = \theta(B)a_t, \phi_n(B)n_t = \theta_n(B)e_t$ and $cov(e_t,a_s)= 0$

state the prewhitening process for how to identify the value of k

Apply $\phi_x(B)/\theta(B)$ on the regression model, we get

$$\frac{\phi_x(B)}{\theta(B)}y_t = \sum_0^k v_i \frac{\phi_x(B)}{\theta(B)} x_{t-i} + \frac{\phi_x(B)}{\theta(B)} n_t$$

$$\hat y_t = v(B)a_t + \epsilon_t$$

By having $\phi_n(B)=\phi_x(B), \theta_n(B)=\theta(B)$, we have $\epsilon \sim e$ so that

$$\hat y_t = v(B)a_t + e_t$$

then, multiply both sides by $a_{t-k}$ and take expectations

$$E(\hat y_t a_{t-k}) = v(B) E(a_t a_{t-k}) + E(e_t a_{t-k})$$

$$cov(\hat y_t, a_{t-k}) = v_k\sigma_a^2$$

$$v_k = cov(\hat y_t, a_{t-k}) / \sigma_a^2 = corr(\hat y_t, a_{t-k})\frac{se(\hat y_t)}{se(a_t)}\propto corr(\hat y_t, a_{t-k})$$

Therefore, we can test the statistical significance of $v_k$ by examining the statistical significance of $corr(\hat y_t, a_{t-k})$

state the steps of using Box-Tiao transformation to estimate $v_j$

The steps of the estimation procedures

1. Run the OLS regression on $y_t = \sum_{j=1}^s v_j x_{t-j} + e_t$ to collect the residuals $\{\hat e_t\}$
2. Identify an ARMA model for $\hat e_t$
3. Apply Box-Tiao transformation to filter $y_t, x_t$
4. Run regression on the transformed equation
5. check the correlation of regression residuals

Find the l-ahead optimal forecast of $y_{t+l}, \hat y_t(l)$ with $a_t,e_t$.

Since $v(B)=\sum_0^k v_i B^i$ has finite terms, it can be transformed $v(B)=\delta(B)/w(B)$, then

$$y_t = \frac{\delta(B) \theta(B)}{w(B)\phi_x(B)}a_t + \frac{\theta_n(B)}{\phi_n(B)}e_t$$

$$y_t = u(B)a_t + \psi(B)e_t$$

all of them are finite order with max K of polynomials in B. Therefore,

$$y_{t+l} = \sum_0^k u_i a_{t+l-i} + \psi_i e_{t+l-i}$$

$$\hat y_t(l) = \sum_0^k u_{i+l}^* a_{t-i} + \psi_{i+l}^* e_{t-i}$$

Derive the MSE

Consider $y_{t-l} - \hat y_t(l)$, which equals to
$=\sum_0^{l-1}u_i a_{t+l-i} + \psi_i e_{t+l - i} (i)$ after time lag $t$
$-\sum_0^k (u_{i+l}^* - u_{i+l})a_{t-i} (ii)$
$-(\psi_{i+l}^* - \psi_{i+l})e_{t-i} (iii)$ up to time $t$

$E(y_l - \hat y_t(l))^2 = E(i)^2 + E(ii)^2 + E(iii)^2$
$=\sum_0^{l-1} u_i^2 \sigma_a^2 + \psi_i^2 \sigma_e^2$
$+ \sum_0^k \sigma_a^2(u_{i+l^*}- u_{i+l})$
$+ \sum_0^k \sigma_e^2(\psi_{i+l^*}- \psi_{i+l})$

minimized when $u^* = u, \psi^* = \psi$

VAR

Consider a VAR(p) model of 2-d variables, i.e. $y_t = [y_{i,t}, y_{2,t}]^T$ $y_t = \sum_1^p A_i y_{t-i} + a_t$, and each $A_i = \begin{bmatrix}\phi_{i,11}&\phi_{i,12}\\\phi_{i,21}&\phi_{i,22}\end{bmatrix}$

state how to check the stationarity

All the roots of $\det(I_k - A_1B-...-A_pB^p) = 0$ must lie outside of the unit circle, or the companion form $\xi_t = A\xi_{t-1}+v_t$ must have the moduli of the eigenvalues of $A$ being <1.

Describe the methods to select the order for Equation (1)

• Selection by information criteria, for example, BIC, AIC, DIC, HQ, SC, FPE
• Using LRT for VAR(p) vs. VAR(p-1)

State how to test Granger causality for that $X_{1t}$ Granger causes $X_{2t}$ but not the other way around. Basd on the same condition, express $X_{2t}$ as the TFN model of $X_{1t}$

$$X_{2,t} = \sum_1^p \phi_{i,21}X_{1,t-i} + \phi_{i,22}X_{2,t-i} + a_{2,t}$$

$$X_{2,t} - \sum_1^p \phi_{i,22}X_{2,t-i} = \sum_1^p \phi_{i,21}X_{1,t-i} + a_{2,t}$$

$$\Phi_{22}(B)X_{2,t} = \Phi_{21}(B)X_{1,t}+a_{2,t}$$

$$X_{2,t} = \frac{\Phi_{21}(B)}{\Phi_{22}(B)}X_{1,t} + \frac{a_{2,t}}{\Phi_{22}(B)}$$

Let $v(B) = \frac{\Phi_{21}(B)}{\Phi_{22}(B)}, N_t = \frac{a_{2,t}}{\Phi_{22}(B)}$, the TFN model is

$$X_{2,t}=v(B)X_{1,t} + N_t$$

To see whether $X_{1,t}$ Granger causes $X_{2,t}$, if not, then all $\phi_{i,21} = 0$

Describe how to test Granger causality using univariate approach. Suppose ($\Phi_1(B)X_{1,t}=\Theta_1(B)a_{1,t}, \Phi_2(B)X_{2,t}=\Theta_2(B)a_{2,t}$\)

Using the Portmanteau test,

$$\rho_{a_1a_2}(k) = \frac{E(a_{1,t}, a_{2,t+k})}{\sqrt{\sigma_1^2\sigma_2^2}}$$

If $\forall k < 0. \rho(k)=0$, then $X_2$ does not Granger cause $X_1$.

Suppose $X_{1,t}, X_{2,t}$ not weakly stationary. How do you model the join dynamics of $\{X_{1,t}, X_{2,t}\}$ using co-integration.

• differencing the two time series individually until each are stationary
• use VAR(p) to fit the two stationary process and test the Granger causality
• If $(X_{1,t}, X_{2,t})$ are cointegrated, use Error correction model to include lagged disequilibrium terms as explanatory variables.

Discuss the reasons why we have to choose different models based on the condition of integration

If cointegration exists, if we use VAR(p) model directly fitted to the differenced stationary processes, the model will be misspecified.

Discuss the Engle-Granger approach for modeling cointegrated $X_{1t}$ and $X_{2t}$

1. test whether $X_t,Y_t$ are I(1) using unit root test
2. if both I(1), regress one against the other using least squares
3. run a unit root test on regression residuals. If residuals are stationary, these two series are cointegrated.
4. Where the regression line indicate the long-run equilibrium relationship between two variables, the disequilirium term is simply the regression residuals.
5. Finally, consider the ECM

$$\Delta X_t = c_1 + \rho_1(Y_{t-1} - \hat a X_{t-1}) + \beta_{x,1}\Delta X_{t-1} + ... + \beta_{y,1}\Delta Y_{t-1}+...+\epsilon_{x,t}$$

$$\Delta Y_t = c_2 + \rho_2(Y_{t-1} - \hat a X_{t-1}) + \gamma_{x,1}\Delta X_{t-1} + ... + \gamma_{y,1}\Delta Y_{t-1}+...+\epsilon_{y,t}$$

Discuss the implication of Granger representation theorem

We have and should use ECM to model non-stationary time series

Bootstrap

Given an AR(2) model

$$y_t = \mu + \phi y_{t-1} + \phi_2 y_{t-2} + a_t$$

(unconditional) parametric bootstrap

$E(Y_t) = \mu + E(\phi_1 Y_{t-1}) + E(\phi_2 Y_{t-2}) + E(a_t)$
$E(Y)=\mu + \phi_1 E(Y) + \phi_2 E(Y)$
$E(Y)=\frac{\mu}{1-\phi_1-\phi_2}$

$var(Y) = 0 + \phi_1^2 var(Y) + \phi_2^2 var(Y) + E(a)^2$
$var(Y) = \frac{\sigma^2}{1-\phi_1^2 - \phi_2^2}$

1. The unconditional distribution of $Y_t\sim N(\frac{\mu}{1-\phi_1-\phi_2}, \frac{\sigma^2}{1-\phi_1^2 - \phi_2^2})$
2. simulate $y_0, y_1$ by drawing a random number from $Y$.
3. recursively simulate $y_2$ by the model, recursively

pros easy to compute
cons don't know whether it is correlated

GARCH

ARCH process

ARCH(1) The first order of autoregressive conditional heteroskedastic process is $e_t \sim N(0, \sigma_t^2)$ where $\sigma_t^2 = a_0 + a_1 e_{t-1}^2$.
On defining $v_t = e_t^2 - \sigma_t^2$, the model can also be written as

$$e_t^2 = a_0 + a_1 e_{t-1}^2 + v_t$$

Since $E(v_t \mid x_{t-1}, x_{t-2},...)=0$, the model corresponds directly to an AR(1) model for the squared error $e_t^2$.

ARCH(q) Then, the ARCH(q) process is defined as

with

\[a_0 \geq 0, a_i > 0. \sum_{i=1}^q a_i < 1\]

or

\[e_t^2 = a_0 + \sum_1^q a_i e_{t-i}^2 + v_t\]

\[e_t^2 = a_0 + a(B)e_{t-1}^2 + v_t\]

GARCH(p,q)

\[e_t^2 = a_0 + \sum_1^q \beta_i \sigma_{t-i}^2 + \sum_1^q a_i e_{t-i}^2\]

\[\sigma_t^2 = a_0 + a(B)e_{t-1}^2 + \beta(B)\sigma_{t-1}^2\]

with \(a_0 \geq 0, a(B)\) and \(\beta(B)\) have no common roots and that the roots of \(1-\beta(B)\) all less than unit root.