dynare/dynare++/kord/first_order.cc

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// Copyright 2004, Ondra Kamenik
#include "kord_exception.hh"
#include "first_order.hh"
#include <dynlapack.h>
double qz_criterium = 1.000001;
/* This is a function which selects the eigenvalues pair used by
|dgges|. See documentation to DGGES for details. Here we want
to select (return true) the pairs for which $\alpha<\beta$. */
lapack_int
order_eigs(const double *alphar, const double *alphai, const double *beta)
{
return (*alphar **alphar + *alphai **alphai < *beta **beta * qz_criterium * qz_criterium);
}
/* Here we solve the linear approximation. The result are the matrices
$g_{y^*}$ and $g_u$. The method solves the first derivatives of $g$ so
that the following equation would be true:
$$E_t[F(y^*_{t-1},u_t,u_{t+1},\sigma)] =
E_t[f(g^{**}(g^*(y_{t-1}^*,u_t,\sigma), u_{t+1}, \sigma), g(y_{t-1}^*,u_t,\sigma),
y^*_{t-1},u_t)]=0$$
where $f$ is a given system of equations.
It is known that $g_{y^*}$ is given by $F_{y^*}=0$, $g_u$ is given by
$F_u=0$, and $g_\sigma$ is zero. The only input to the method are the
derivatives |fd| of the system $f$, and partitioning of the vector $y$
(from object). */
void
FirstOrder::solve(const TwoDMatrix &fd)
{
JournalRecordPair pa(journal);
pa << "Recovering first order derivatives " << endrec;
::qz_criterium = FirstOrder::qz_criterium;
// **********************
// solve derivatives |gy|
// **********************
/* The derivatives $g_{y^*}$ are retrieved from the equation
$F_{y^*}=0$. The calculation proceeds as follows:
\orderedlist
\li For each variable appearing at both $t-1$ and $t-1$ we add a dummy
variable, so that the predetermined variables and forward looking would
be disjoint. This is, the matrix of the first derivatives of the
system written as:
$$\left[\matrix{f_{y^{**}_+}&f_{ys}&f_{yp}&f_{yb}&f_{yf}&f_{y^*_-}}\right],$$
where $f_{ys}$, $f_{yp}$, $f_{yb}$, and $f_{yf}$ are derivatives wrt
static, predetermined, both, forward looking at time $t$, is rewritten
to the matrix:
$$\left[
\matrix{f_{y^{**}_+}&f_{ys}&f_{yp}&f_{yb}&0&f_{yf}&f_{y^*_-}\cr
0 &0 &0 &I &-I&0 &0}
\right],$$
where the second line has number of rows equal to the number of both variables.
\li Next, provided that forward looking and predetermined are
disjoint, the equation $F_{y^*}=0$ is written as:
$$\left[f_+{y^{**}_+}\right]\left[g_{y^*}^{**}\right]\left[g_{y^*}^*\right]
+\left[f_{ys}\right]\left[g^s_{y^*}\right]
+\left[f_{y^*}\right]\left[g^*_{y^*}\right]
+\left[f_{y^{**}}\right]\left[g^{**}_{y^*}\right]+\left[f_{y^*_-}\right]=0$$
This is rewritten as
$$\left[\matrix{f_{y^*}&0&f_{y^{**}_+}}\right]
\left[\matrix{I\cr g^s_{y^*}\cr g^{**}_{y^*}}\right]\left[g_{y^*}^*\right]+
\left[\matrix{f_{y^*_-}&f_{ys}&f_{y^{**}}}\right]
\left[\matrix{I\cr g^s_{y^*}\cr g^{**}_{y^*}}\right]=0
$$
Now, in the above equation, there are the auxiliary variables standing
for copies of both variables at time $t+1$. This equation is then
rewritten as:
$$
\left[\matrix{f_{yp}&f_{yb}&0&f_{y^{**}_+}\cr 0&I&0&0}\right]
\left[\matrix{I\cr g^s_{y^*}\cr g^{**}_{y^*}}\right]\left[g_{y^*}^*\right]+
\left[\matrix{f_{y^*_-}&f_{ys}&0&f_{yf}\cr 0&0&-I&0}\right]
\left[\matrix{I\cr g^s_{y^*}\cr g^{**}_{y^*}}\right]=0
$$
The two matrices are denoted as $D$ and $-E$, so the equation takes the form:
$$D\left[\matrix{I\cr g^s_{y^*}\cr g^{**}_{y^*}}\right]\left[g_{y^*}^*\right]=
E\left[\matrix{I\cr g^s_{y^*}\cr g^{**}_{y^*}}\right]$$
\li Next we solve the equation by Generalized Schur decomposition:
$$
\left[\matrix{T_{11}&T_{12}\cr 0&T_{22}}\right]
\left[\matrix{Z_{11}^T&Z_{21}^T\cr Z_{12}^T&Z_{22}^T}\right]
\left[\matrix{I\cr X}\right]\left[g_{y^*}^*\right]=
\left[\matrix{S_{11}&S_{12}\cr 0&S_{22}}\right]
\left[\matrix{Z_{11}^T&Z_{21}^T\cr Z_{12}^T&Z_{22}^T}\right]
\left[\matrix{I\cr X}\right]
$$
We reorder the eigenvalue pair so that $S_{ii}/T_{ii}$ with modulus
less than one would be in the left-upper part.
\li The Blanchard--Kahn stability argument implies that the pairs
with modulus less that one will be in and only in $S_{11}/T_{11}$.
The exploding paths will be then eliminated when
$$
\left[\matrix{Z_{11}^T&Z_{21}^T\cr Z_{12}^T&Z_{22}^T}\right]
\left[\matrix{I\cr X}\right]=
\left[\matrix{Y\cr 0}\right]
$$
From this we have, $Y=Z_{11}^{-1}$, and $X=Z_{21}Y$, or equivalently
$X=-Z_{22}^{-T}Z_{12}^T$. From the equation, we get
$\left[g_{y^*}^*\right]=Y^{-1}T_{11}^{-1}S_{11}Y$, which is
$Z_{11}T_{11}^{-1}S_{11}Z_{11}^{-1}$.
\li We then copy the derivatives to storage |gy|. Note that the
derivatives of both variables are in $X$ and in
$\left[g_{y^*}^*\right]$, so we check whether the two submatrices are
the same. The difference is only numerical error.
\endorderedlist */
// setup submatrices of |f|
/* Here we setup submatrices of the derivatives |fd|. */
int off = 0;
ConstTwoDMatrix fyplus(fd, off, ypart.nyss());
off += ypart.nyss();
ConstTwoDMatrix fyszero(fd, off, ypart.nstat);
off += ypart.nstat;
ConstTwoDMatrix fypzero(fd, off, ypart.npred);
off += ypart.npred;
ConstTwoDMatrix fybzero(fd, off, ypart.nboth);
off += ypart.nboth;
ConstTwoDMatrix fyfzero(fd, off, ypart.nforw);
off += ypart.nforw;
ConstTwoDMatrix fymins(fd, off, ypart.nys());
off += ypart.nys();
ConstTwoDMatrix fuzero(fd, off, nu);
off += nu;
// form matrix $D$
lapack_int n = ypart.ny()+ypart.nboth;
TwoDMatrix matD(n, n);
matD.zeros();
matD.place(fypzero, 0, 0);
matD.place(fybzero, 0, ypart.npred);
matD.place(fyplus, 0, ypart.nys()+ypart.nstat);
for (int i = 0; i < ypart.nboth; i++)
matD.get(ypart.ny()+i, ypart.npred+i) = 1.0;
// form matrix $E$
TwoDMatrix matE(n, n);
matE.zeros();
matE.place(fymins, 0, 0);
matE.place(fyszero, 0, ypart.nys());
matE.place(fyfzero, 0, ypart.nys()+ypart.nstat+ypart.nboth);
for (int i = 0; i < ypart.nboth; i++)
matE.get(ypart.ny()+i, ypart.nys()+ypart.nstat+i) = -1.0;
matE.mult(-1.0);
// solve generalized Schur
TwoDMatrix vsl(n, n);
TwoDMatrix vsr(n, n);
lapack_int lwork = 100*n+16;
Vector work(lwork);
auto *bwork = new lapack_int[n];
lapack_int info;
lapack_int sdim2 = sdim;
dgges("N", "V", "S", order_eigs, &n, matE.getData().base(), &n,
matD.getData().base(), &n, &sdim2, alphar.base(), alphai.base(),
beta.base(), vsl.getData().base(), &n, vsr.getData().base(), &n,
work.base(), &lwork, bwork, &info);
if (info)
{
throw KordException(__FILE__, __LINE__,
"DGGES returns an error in FirstOrder::solve");
}
sdim = sdim2;
bk_cond = (sdim == ypart.nys());
delete[] bwork;
// make submatrices of right space
/* Here we setup submatrices of the matrix $Z$. */
ConstGeneralMatrix z11(vsr, 0, 0, ypart.nys(), ypart.nys());
ConstGeneralMatrix z12(vsr, 0, ypart.nys(), ypart.nys(), n-ypart.nys());
ConstGeneralMatrix z21(vsr, ypart.nys(), 0, n-ypart.nys(), ypart.nys());
ConstGeneralMatrix z22(vsr, ypart.nys(), ypart.nys(), n-ypart.nys(), n-ypart.nys());
// calculate derivatives of static and forward
/* Here we calculate $X=-Z_{22}^{-T}Z_{12}^T$, where $X$ is |sfder| in the
code. */
GeneralMatrix sfder(z12, "transpose");
z22.multInvLeftTrans(sfder);
sfder.mult(-1);
// calculate derivatives of predetermined
/* Here we calculate
$g_{y^*}^*=Z_{11}T^{-1}_{11}S_{11}Z_{11}^{-1}
=Z_{11}T^{-1}_{11}(Z_{11}^{-T}S^T_{11})^T$. */
ConstGeneralMatrix s11(matE, 0, 0, ypart.nys(), ypart.nys());
ConstGeneralMatrix t11(matD, 0, 0, ypart.nys(), ypart.nys());
GeneralMatrix dumm(s11, "transpose");
z11.multInvLeftTrans(dumm);
GeneralMatrix preder(dumm, "transpose");
t11.multInvLeft(preder);
preder.multLeft(z11);
// copy derivatives to |gy|
gy.place(preder, ypart.nstat, 0);
GeneralMatrix sder(sfder, 0, 0, ypart.nstat, ypart.nys());
gy.place(sder, 0, 0);
GeneralMatrix fder(sfder, ypart.nstat+ypart.nboth, 0, ypart.nforw, ypart.nys());
gy.place(fder, ypart.nstat+ypart.nys(), 0);
// check difference for derivatives of both
GeneralMatrix bder((const GeneralMatrix &)sfder, ypart.nstat, 0, ypart.nboth, ypart.nys());
GeneralMatrix bder2(preder, ypart.npred, 0, ypart.nboth, ypart.nys());
bder.add(-1, bder2);
b_error = bder.getData().getMax();
// **********************
// solve derivatives |gu|
// **********************
/* The equation $F_u=0$ can be written as
$$
\left[f_{y^{**}_+}\right]\left[g^{**}_{y^*}\right]\left[g_u^*\right]+
\left[f_y\right]\left[g_u\right]+\left[f_u\right]=0
$$
and rewritten as
$$
\left[f_y +
\left[\matrix{0&f_{y^{**}_+}g^{**}_{y^*}&0}\right]\right]g_u=f_u
$$
This is exactly done here. The matrix
$\left[f_y +\left[\matrix{0&f_{y^{**}_+}g^{**}_{y^*}&0}\right]\right]$ is |matA|
in the code. */
GeneralMatrix matA(ypart.ny(), ypart.ny());
matA.zeros();
ConstGeneralMatrix gss(gy, ypart.nstat+ypart.npred, 0, ypart.nyss(), ypart.nys());
GeneralMatrix aux(fyplus, gss);
matA.place(aux, 0, ypart.nstat);
ConstGeneralMatrix fyzero(fd, 0, ypart.nyss(), ypart.ny(), ypart.ny());
matA.add(1.0, fyzero);
gu.zeros();
gu.add(-1.0, fuzero);
ConstGeneralMatrix(matA).multInvLeft(gu);
journalEigs();
if (!gy.isFinite() || !gu.isFinite())
{
throw KordException(__FILE__, __LINE__,
"NaN or Inf asserted in first order derivatives in FirstOrder::solve");
}
}
void
FirstOrder::journalEigs()
{
if (bk_cond)
{
JournalRecord jr(journal);
jr << "Blanchard-Kahn conditition satisfied, model stable" << endrec;
}
else
{
JournalRecord jr(journal);
jr << "Blanchard-Kahn condition not satisfied, model not stable: sdim=" << sdim
<< " " << "npred=" << ypart.nys() << endrec;
}
if (!bk_cond)
{
for (int i = 0; i < alphar.length(); i++)
{
if (i == sdim || i == ypart.nys())
{
JournalRecord jr(journal);
jr << "---------------------------------------------------- ";
if (i == sdim)
jr << "sdim";
else
jr << "npred";
jr << endrec;
}
JournalRecord jr(journal);
double mod = sqrt(alphar[i]*alphar[i]+alphai[i]*alphai[i]);
mod = mod/round(100000*std::abs(beta[i]))*100000;
jr << i << "\t(" << alphar[i] << "," << alphai[i] << ") / " << beta[i]
<< " \t" << mod << endrec;
}
}
}