On 01/22/2016 11:17 AM, Degang WU wrote:
Hi, I want to simulate a VAR process using the following code
library(dse) AR <- array(c(1, .5, .3, 0, .2, .1, 0, .2, .05, 1, .5,
.3) ,c(3,2,2)) VAR <- ARMA(A=AR, B=diag(1,2)) print(VAR) simData <-
simulate(VAR) Inside dse::simulate:
if (p == 1) invA0 <- matrix(1/A[1, , ], 1, 1) else invA0 <-
solve(A[1, , ]) for (l in 1:a) A[l, , ] <- invA0 %*% A[l, , ] for (l
in 1:b) B[l, , ] <- invA0 %*% B[l, , ]
Where A[,,,] are the coefficient matrix for the process. I have no
idea why the inverse of A[1, ,] is involved in the simulation.
Degang
In response to your previous question I said the dse specification is
y_t + A_1 y_{t-1} + A_2 y_{t-2} + .. A_p y_{t-p} = e_t (2)
but that is the form corresponding to your specification. In general,
VAR models can be written
A_0 y_t + A_1 y_{t-1} + A_2 y_{t-2} + .. A_p y_{t-p} = e_t (3)
In your specification A_0 is the identity matrix. For simulation it is
convenient to re-write (3) as
y_t = (A_0)^-1 (-A_1 y_{t-1} - A_2 y_{t-2} - ... - A_p y_{t-p} + e_t)
R indexes arrays starting with 1, so A_0 is stored in A[1,,]. If A_0 is
the identity, as in your specification, the inverse will also be
identity and the multiplication is not necessary.
Paul
Thanks!
Regards, Degang [[alternative HTML version deleted]]
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