Vector autoregression (VAR)
Suppose we are observing two stocks and their respective returns are 
(1)
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)
What should be 
B) Assume that the error 
(3) 
What does this assumption mean in terms of the components of 
(4) 
C) Suppose (1) is stationary. The stationarity condition is expressed in terms of eigenvalues of 
(5) 
Find
D) Find
E) Find
F) Find
G) Find
Solution
A) It takes some practice to see that with the notation
the system (1) becomes (2).
B) The equations in (3) look like this:
Equalities of matrices are understood element-wise, so we get a series of scalar equations 
Conversely, the scalar equations from (4) give
C) (2) implies 
D) From (2) we see that 
![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,](https://s0.wp.com/latex.php?latex=Cov%28Y_%7Bt-1%7D%2CU_%7Bt%7D%29%3DE%5Cleft%28+Y_%7Bt-1%7D-EY_%7Bt-1%7D%5Cright%29+U_%7Bt%7D%5E%7BT%7D%3DE%5Cleft%5B+%5Cleft%28++Y_%7Bt-1%7D-EY_%7Bt-1%7D%5Cright%29+E_%7Bt-1%7DU_%7Bt%7D%5E%7BT%7D%5Cright%5D+%3D0%2C&bg=ffffff&fg=000&s=0&c=20201002)
![Cov\left( U_{t},Y_{t-1}\right) =\left[ Cov(Y_{t-1},U_{t})\right] ^{T}=0.](https://s0.wp.com/latex.php?latex=Cov%5Cleft%28+U_%7Bt%7D%2CY_%7Bt-1%7D%5Cright%29+%3D%5Cleft%5B+Cov%28Y_%7Bt-1%7D%2CU_%7Bt%7D%29%5Cright%5D+%5E%7BT%7D%3D0.&bg=ffffff&fg=000&s=0&c=20201002)
E) Using the previous post
(by stationarity and (3)). Thus, 
F) Using the previous result we have
G) Similarly,
Autocorrelations require a little more effort and I leave them out.
is one way to take into account their interdependence. This theory is undeservedly omitted from the Guide by A. Patton.
that holds for numbers.
is square, it is invertible if and only if 
(this is an equivalent of the condition 
for numbers).
for 
where 
are known. Assuming that 
to obtain 
(Because of lack of commutativity, we need to keep the order of the factors). Using intuitive properties 
and 
we obtain the solution: 
In particular, we see that if 
then the equation 
has a unique solution 
be two vectors. 
are assumed to be known and 
is unknown. We want to check that 
solves the equation 
(Note that for this equation the trick used to solve 
variance becomes
are constant matrices. We want an analog of 
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}](https://s0.wp.com/latex.php?latex=V%5Cleft%28+AX%2BBY%5Cright%29+%3DE%5Cleft%5B+AX%2BBY-E%5Cleft%28+AX%2BBY%5Cright%29+%5Cright%5D+%5Cleft%5B++AX%2BBY-E%5Cleft%28+AX%2BBY%5Cright%29+%5Cright%5D+%5E%7BT%7D&bg=ffffff&fg=000&s=0&c=20201002)
![=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}](https://s0.wp.com/latex.php?latex=%3DE%5Cleft%5B+A%5Cleft%28+X-EX%5Cright%29+%2BB%5Cleft%28+Y-EY%5Cright%29+%5Cright%5D+%5Cleft%5B+A%5Cleft%28++X-EX%5Cright%29+%2BB%5Cleft%28+Y-EY%5Cright%29+%5Cright%5D+%5E%7BT%7D&bg=ffffff&fg=000&s=0&c=20201002)
![=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}](https://s0.wp.com/latex.php?latex=%3DE%5Cleft%5B+A%5Cleft%28+X-EX%5Cright%29+%5Cright%5D+%5Cleft%5B+A%5Cleft%28+X-EX%5Cright%29+%5Cright%5D++%5E%7BT%7D%2BE%5Cleft%5B+B%5Cleft%28+Y-EY%5Cright%29+%5Cright%5D+%5Cleft%5B+A%5Cleft%28+X-EX%5Cright%29+%5Cright%5D++%5E%7BT%7D&bg=ffffff&fg=000&s=0&c=20201002)
![+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}](https://s0.wp.com/latex.php?latex=%2BE%5Cleft%5B+A%5Cleft%28+X-EX%5Cright%29+%5Cright%5D+%5Cleft%5B+B%5Cleft%28+Y-EY%5Cright%29+%5Cright%5D++%5E%7BT%7D%2BE%5Cleft%5B+B%5Cleft%28+Y-EY%5Cright%29+%5Cright%5D+%5Cleft%5B+B%5Cleft%28+Y-EY%5Cright%29+%5Cright%5D++%5E%7BT%7D&bg=ffffff&fg=000&s=0&c=20201002)
)
defined in Chapter 3.
satisfies 
(regressors are not collinear) and the error 
satisfies
is estimated by 
Denote 
Using (1) we see that 
and the residual 
is a projector and has properties which are derived 
is an eigenvalue of 
then multiplying 
by 
we get 
Hence eigenvalues of 
or 
and the remaining 
are zeros. Let 
be the 
Here 
is an 
is a diagonal matrix with eigenvalues of 
is distributed as 
2) The estimators 
and 
are independent.
From (2) and (5)
is normal as a linear transformation of a normal vector. It follows that 
where 
is a standard normal vector with independent standard normal coordinates 
Hence, (6) implies
are independent. Since they are normal, their independence follows from
This allows us to show that 
:
Assuming that 
, on Chart 1 we see that 
Hence, using 
for all 
, which is maintained for continuous random variables throughout the guide by Abdey.
By the 
is an increasing differentiable function with the inverse 
and 
answer questions similar to 1 and 2.
Just applying (2) to 
and 
we get
the 
with independent standard normals 
we can write 
where the chi-squared variables on the right are independent and all have one degree of freedom. This is because deterministic (here quadratic) functions of independent variables are independent.
Then by the 
have densities 
Then they are independent if and only if their joint density 
is a product of individual densities: 
are normal, then they are independent if and only if they are uncorrelated: 
be independent standard normal variables. A standard normal variable is 
have the same density. We achieve independence, according to Theorem 1, by defining their joint density to be a product of individual densities.
is defined by
because all of 
for 
by Theorem 2 and 
and vector 
of compatible dimensions a normal vector is defined by 
and
converges to 1 in probability.
we have 
and 
(both properties can be verified in Mathematica). Hence, 
and
(c) the regressors are 
should be estimated and don't pay attention to estimation of 
Partly this is because 
does not appear in the regression and partly because the result on estimation of error variance is more complex than the result on the OLS estimator of 
be the 
is called the fitted value and 
is called the residual.
and 
show that 
and 
Since 
we have further 
and 
would be easy to find from (2). However, we need to find 
where there is an obstructing 
(
is a scalar, so its trace is equal to itself)
(applying 
(the regressors and hence 
)
(applying (2)) 
because this is the 
Therefore 
Thus, 
and 
![(Px)\cdot (Qx)=(Px)\cdot[(I-P)x]=x\cdot[(P-P^2)x]=0.](https://s0.wp.com/latex.php?latex=%28Px%29%5Ccdot+%28Qx%29%3D%28Px%29%5Ccdot%5B%28I-P%29x%5D%3Dx%5Ccdot%5B%28P-P%5E2%29x%5D%3D0.&bg=ffffff&fg=000&s=0&c=20201002)
(the dependent vector) and vectors 
(independent vectors or regressors) are given. The OLS estimator is defined as that vector 
which minimizes the total sum of squares 
we see that 
and that finding the OLS estimator means approximating 
with vectors from the image 
should be linearly independent, otherwise the solution will not be unique.
Since 
belongs to the image of 
Denoting also 
(by Exercise 1)
is a lower bound for 
This lower bound is achieved when the second term is made zero. From![P(y-X\beta)=Py-X\beta =X(X^TX)^{-1}X^Ty-X\beta=X[(X^TX)^{-1}X^Ty-\beta]](https://s0.wp.com/latex.php?latex=P%28y-X%5Cbeta%29%3DPy-X%5Cbeta+%3DX%28X%5ETX%29%5E%7B-1%7DX%5ETy-X%5Cbeta%3DX%5B%28X%5ETX%29%5E%7B-1%7DX%5ETy-%5Cbeta%5D&bg=ffffff&fg=000&s=0&c=20201002)
where 
for all 
.
for all 
are the same but it's better to use the verbal definition.
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