Raising the bar

Vector autoregression (VAR)

Suppose we are observing two stocks and their respective returns are $latex x_{t},y_{t}.$ To take into account their interdependence, we consider a vector autoregression

(1) $latex \begin{array}{c}
x_{t}=a_{1}x_{t-1}+b_{1}y_{t-1}+u_{t} \\
y_{t}=a_{2}x_{t-1}+b_{2}y_{t-1}+v_{t}\end{array}.$

Try to repeat for this system the analysis from Section 3.5 (Application to an AR(1) process) of the Guide by A. Patton and you will see that the difficulties are insurmountable. However, matrix algebra allows one to overcome them, with proper adjustment.

Problem

A) Write this system in a vector format

(2) $latex Y_{t}=\Phi Y_{t-1}+U_{t}.$

What should be $latex Y_{t},\Phi ,U_{t}$ in this representation?

B) Assume that the error $latex U_{t}$ in (1) satisfies

(3) $latex E_{t-1}U_{t}=0,\ EU_{t}U_{t}^{T}=\Sigma ,~EU_{t}U_{s}^{T}=0$ for $latex t\neq
s$ with some symmetric matrix $latex \Sigma =\left(\begin{array}{cc}\sigma _{11} & \sigma _{12} \\\sigma _{12} & \sigma _{22}
\end{array}\right) .$

What does this assumption mean in terms of the components of $latex U_{t}$ from (2)? What is $latex \Sigma $ if the errors in (1) satisfy

(4) $latex E_{t-1}u_{t}=E_{t-1}v_{t}=0,~Eu_{t}^{2}=Ev_{t}^{2}=\sigma ^{2},$ $latex Eu_su_t=Ev_sv_t=0$ for $latex t\neq s,$ $latex Eu_sv_t=0$ for all $latex s,t?$

C) Suppose (1) is stationary. The stationarity condition is expressed in terms of eigenvalues of $latex \Phi $ but we don't need it. However, we need its implication:

(5) $latex \det \left( I-\Phi \right) \neq 0$.

Find $latex \mu =EY_{t}.$

D) Find $latex Cov(Y_{t-1},Y_t$. $latex Cov(Y_{t-1},U_{t}).$

E) Find $latex \gamma _0 = V(Y_t).$

F) Find $latex \gamma _1=Cov(Y_t,Y_{t-1}).$

G) Find $latex \gamma _2.$

Solution

A) It takes some practice to see that with the notation

$latex Y_{t}=\left(\begin{array}{c}x_{t} \\ y_{t}\end{array}\right) ,$ $latex \Phi =\left(\begin{array}{cc}
a_{1} & b_{1} \\ a_{2} & b_{2}\end{array}\right) ,$ $latex U_{t}=\left(
\begin{array}{c}u_{t} \\ v_{t}\end{array}\right) $

the system (1) becomes (2).

B) The equations in (3) look like this:

$latex E_{t-1}U_{t}=\left(\begin{array}{c}E_{t-1}u_{t} \\
E_{t-1}v_{t}\end{array}\right) =0,$ $latex EU_{t}U_{t}^{T}=\left(
\begin{array}{cc}Eu_{t}^{2} & Eu_{t}v_{t} \\ Eu_{t}v_{t} & Ev_{t}^{2}
\end{array}\right) =\left(\begin{array}{cc}
\sigma _{11} & \sigma _{12} \\
\sigma _{12} & \sigma _{22}\end{array}
\right) ,$

$latex EU_{t}U_{s}^{T}=\left(\begin{array}{cc}
Eu_{t}u_{s} & Eu_{t}v_{s} \\ Ev_{t}u_{s} & Ev_{t}v_{s}
\end{array}\right) =0.$

Equalities of matrices are understood element-wise, so we get a series of scalar equations $latex E_{t-1}u_{t}=0,...,Ev_{t}v_{s}=0$ for $latex t\neq s.$

Conversely, the scalar equations from (4) give

$latex E_{t-1}U_{t}=0,\ EU_{t}U_{t}^{T}=\left(\begin{array}{cc}
\sigma ^{2} & 0 \\ 0 & \sigma ^{2}\end{array}
\right) ,~EU_{t}U_{s}^{T}=0$ for $latex t\neq s$.

C) (2) implies $latex EY_{t}=\Phi EY_{t-1}+EU_{t}=\Phi EY_{t-1}$ or by stationarity $latex \mu =\Phi \mu $ or $latex \left( I-\Phi \right) \mu =0.$ Hence (5) implies $latex \mu =0.$

D) From (2) we see that $latex Y_{t-1}$ depends only on $latex I_{t}$ (information set at time $latex t$). Therefore by the LIE

$latex Cov(Y_{t-1},U_{t})=E\left( Y_{t-1}-EY_{t-1}\right) U_{t}^{T}=E\left[ \left(
Y_{t-1}-EY_{t-1}\right) E_{t-1}U_{t}^{T}\right] =0,$

$latex Cov\left( U_{t},Y_{t-1}\right) =\left[ Cov(Y_{t-1},U_{t})\right] ^{T}=0.$

E) Using the previous post

$latex \gamma _{0}= V( \Phi Y_{t-1}+U_{t}) =\Phi V(
Y_{t-1}) \Phi ^T+Cov( U_{t},Y_{t-1}) \Phi ^T+\Phi
Cov(Y_{t-1},U_{t})+V(U_{t}) $

$latex =\Phi \gamma _{0}\Phi ^{T}+\Sigma $

(by stationarity and (3)). Thus, $latex \gamma _0-\Phi \gamma _0\Phi
^T=\Sigma $ and $latex \gamma _0=\sum_{s=0}^{\infty }\Phi ^s\Sigma( \Phi
^T) ^s$ (see previous post).

F) Using the previous result we have

$latex \gamma _{1}=Cov(Y_{t},Y_{t-1})=Cov(\Phi Y_{t-1}+U_{t},Y_{t-1})=\Phi
Cov(Y_{t-1},Y_{t-1})+Cov(U_{t},Y_{t-1})$

$latex =\Phi Cov(Y_{t-1},Y_{t-1})=\Phi \gamma _{0}=\Phi \sum_{s=0}^{\infty }\Phi
^{s}\Sigma( \Phi ^{T}) ^{s}.$

G) Similarly,

$latex \gamma _{2}=Cov(Y_{t},Y_{t-2})=Cov(\Phi Y_{t-1}+U_{t},Y_{t-2})=\Phi
Cov(Y_{t-1},Y_{t-2})+Cov(U_{t},Y_{t-2})$

$latex =\Phi Cov(Y_{t-1},Y_{t-2})=\Phi \gamma _{1}=\Phi ^{2}\sum_{s=0}^{\infty
}\Phi ^{s}\Sigma( \Phi ^{T}) ^{s}.$

Autocorrelations require a little more effort and I leave them out.