5
May 22

## Vector autoregressions: preliminaries

Suppose we are observing two stocks and their respective returns are $x_{t},y_{t}.$ A vector autoregression for the pair $x_{t},y_{t}$ is one way to take into account their interdependence. This theory is undeservedly omitted from the Guide by A. Patton.

### Required minimum in matrix algebra

Matrix notation and summation are very simple.

Matrix multiplication is a little more complex. Make sure to read Global idea 2 and the compatibility rule.

The general approach to study matrices is to compare them to numbers. Here you see the first big No: matrices do not commute, that is, in general $AB\neq BA.$

The idea behind matrix inversion is pretty simple: we want an analog of the property $a\times \frac{1}{a}=1$ that holds for numbers.

Some facts about determinants have very complicated proofs and it is best to stay away from them. But a couple of ideas should be clear from the very beginning. Determinants are defined only for square matrices. The relationship of determinants to matrix invertibility explains the role of determinants. If $A$ is square, it is invertible if and only if $\det A\neq 0$ (this is an equivalent of the condition $a\neq 0$ for numbers).

Here is an illustration of how determinants are used. Suppose we need to solve the equation $AX=Y$ for $X,$ where $A$ and $Y$ are known. Assuming that $\det A\neq 0$ we can premultiply the equation by $A^{-1}$ to obtain $A^{-1}AX=A^{-1}Y.$ (Because of lack of commutativity, we need to keep the order of the factors). Using intuitive properties $A^{-1}A=I$ and $IX=X$ we obtain the solution: $X=A^{-1}Y.$ In particular, we see that if $\det A\neq 0,$ then the equation $AX=0$ has a unique solution $X=0.$

Let $A$ be a square matrix and let $X,Y$ be two vectors. $A,Y$ are assumed to be known and $X$ is unknown. We want to check that $X=\sum_{s=0}^{\infty }A^{s}Y\left( A^{T}\right) ^{s}$ solves the equation $X-AXA^{T}=Y.$ (Note that for this equation the trick used to solve $AX=Y$ does not work.) Just plug $X:$

$\sum_{s=0}^{\infty }A^{s}Y\left( A^{T}\right) ^{s}-A\sum_{s=0}^{\infty }A^{s}Y\left( A^{T}\right) ^{s}A^{T}$ $=Y+\sum_{s=1}^{\infty }A^{s}Y\left(A^{T}\right) ^{s}-\sum_{s=1}^{\infty }A^{s}Y\left( A^{T}\right) ^{s}=Y$

(write out a couple of first terms in the sums if summation signs frighten you).

Transposition is a geometrically simple operation. We need only the property $\left( AB\right) ^{T}=B^{T}A^{T}.$

### Variance and covariance

Property 1. Variance of a random vector $X$ and covariance of two random vectors $X,Y$ are defined by

$V\left( X\right) =E\left( X-EX\right) \left( X-EX\right) ^{T},$ $Cov\left( X,Y\right) =E\left( X-EX\right) \left( Y-EY\right) ^{T},$

respectively.

Note that when $EX=0,$ variance becomes

$V\left( X\right) =EXX^{T}=\left( \begin{array}{ccc}EX_{1}^{2} & ... & EX_{1}X_{n} \\ ... & ... & ... \\ EX_{1}X_{n} & ... & EX_{n}^{2}\end{array}\right) .$

Property 2. Let $X,Y$ be random vectors and suppose $A,B$ are constant matrices. We want an analog of $V\left( aX+bY\right) =a^{2}V\left( X\right) +2abcov\left( X,Y\right) +b^{2}V\left( X\right) .$ In the next calculation we have to remember that the multiplication order cannot be changed.

$V\left( AX+BY\right) =E\left[ AX+BY-E\left( AX+BY\right) \right] \left[ AX+BY-E\left( AX+BY\right) \right] ^{T}$

$=E\left[ A\left( X-EX\right) +B\left( Y-EY\right) \right] \left[ A\left( X-EX\right) +B\left( Y-EY\right) \right] ^{T}$

$=E\left[ A\left( X-EX\right) \right] \left[ A\left( X-EX\right) \right] ^{T}+E\left[ B\left( Y-EY\right) \right] \left[ A\left( X-EX\right) \right] ^{T}$

$+E\left[ A\left( X-EX\right) \right] \left[ B\left( Y-EY\right) \right] ^{T}+E\left[ B\left( Y-EY\right) \right] \left[ B\left( Y-EY\right) \right] ^{T}$

(applying $\left( AB\right) ^{T}=B^{T}A^{T}$)

$=AE\left( X-EX\right) \left( X-EX\right) ^{T}A^{T}+BE\left( Y-EY\right) \left( X-EX\right) ^{T}A^{T}$

$+AE\left( X-EX\right) \left( Y-EY\right) ^{T}B^{T}+BE\left( Y-EY\right) \left( Y-EY\right) ^{T}B^{T}$

$=AV\left( X\right) A^{T}+BCov\left( Y,X\right) A^{T}+ACov(X,Y)B^{T}+BV\left( Y\right) B^{T}.$