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Spectrum
Posted: Sat Jul 17, 2010 2:12 am
by ivory4
According to BN decomposition, Cycle is defined to be Psi(L)(et)
Psi(L) is defined to be (P(L)-P(1))/(1-L), P(L)=Theta(L)/Phi(L), Thetas are MA coefficients and Phis are AR coefficients.
C_t=Psi(L)e_t
Spe(C_t)=|PSI(L)|^2Sig_e^2/2PI
I tried to draw Spectrum of C_t and delta(C_t)
The latter seems to be
Code: Select all
equation(noconstant,coeffs=||-%beta(4), -%beta(5),-%beta(2), -%beta(3)||) arma22 y 2 2
* %beta() is from Boxjenk function, I switch the order of AR and MA coefficients in order to get Theta(z)/Phi(z) directly
frequency 2 512
trfunc(equation=arma22) 1
cset 2 = %z(t,1)-%z(1,1)
cmult(scale=(2*%pi)/scale) 2 2
ctor 1 256
# 2
# ab
set bn_spect 1 256 = ab
But How to draw C_t?
Code: Select all
* continue from previous case
cset 1= 2/(1-%zlag(t,1)
cmult(scale=(2*%pi)/scale) 1 1
?
I end up with C_t's spectrum at zero frequency extremely high which is consistent with deltaCt's zero spectrum at zero frequency.
Is there a way that I use spectrum at some smal positive frequency to guess this zero frequency?
Re: Spectrum
Posted: Sun Jul 18, 2010 5:02 pm
by TomDoan
ivory4 wrote:According to BN decomposition, Cycle is defined to be Psi(L)(et)
Psi(L) is defined to be (P(L)-P(1))/(1-L), P(L)=Theta(L)/Phi(L), Thetas are MA coefficients and Phis are AR coefficients.
C_t=Psi(L)e_t
Spe(C_t)=|PSI(L)|^2Sig_e^2/2PI
I tried to draw Spectrum of C_t and delta(C_t)
The latter seems to be
Code: Select all
equation(noconstant,coeffs=||-%beta(4), -%beta(5),-%beta(2), -%beta(3)||) arma22 y 2 2
* %beta() is from Boxjenk function, I switch the order of AR and MA coefficients in order to get Theta(z)/Phi(z) directly
frequency 2 512
trfunc(equation=arma22) 1
cset 2 = %z(t,1)-%z(1,1)
cmult(scale=(2*%pi)/scale) 2 2
ctor 1 256
# 2
# ab
set bn_spect 1 256 = ab
But How to draw C_t?
Code: Select all
* continue from previous case
cset 1= 2/(1-%zlag(t,1)
cmult(scale=(2*%pi)/scale) 1 1
?
I end up with C_t's spectrum at zero frequency extremely high which is consistent with deltaCt's zero spectrum at zero frequency.
Is there a way that I use spectrum at some smal positive frequency to guess this zero frequency?
Obviously, there is a problem evaluating this directly at 0 frequency:
(P(L)-P(1))/(1-L)
since it's 0/0. Because of round-off errors, you can get almost anything when you try to compute it. You
can apply L'Hopital's rule to the z transform to evaluate it exactly at z=0, though you can also just use the limit of nearby values.
Re: Spectrum
Posted: Sat Jul 31, 2010 1:16 pm
by ivory4
TomDoan wrote:
since it's 0/0. Because of round-off errors, you can get almost anything when you try to compute it. You can apply L'Hopital's rule to the z transform to evaluate it exactly at z=0, though you can also just use the limit of nearby values.
The nearby ones, I think at least many of them are extremely high. If zero frequency is estimated using L'Hopital'rule, then a hump near zero frequency would be created. To adjust them using analytical spectrum (after L'Hopital's rule applied to the z transform to evaluate it exactly at z=0) or it is better to use right side values ( I mean frequency corresponding bigger w)
Re: Spectrum
Posted: Mon Aug 02, 2010 9:40 am
by TomDoan
Since L'Hopital's rule is exact, you would prefer that over an eyeball limit from nearby values.
Re: Spectrum
Posted: Sat Dec 25, 2010 12:15 pm
by ivory4
If I would like to draw Theoretical spectrum of various ARIMA model, is it better to use @specth ?
I check the source and found that it requires ACFMORIN.SRC instead of ACF.SRC
And also the default for NCOV is 20 instead of 200.
I add this one to the original src and the example in the file, but RATS still says that ACFMORIN is not a procedure (did you forget to source?)
Code: Select all
ALL 250
DECL VECT a(2) b(1) c(1)
INPUT(unit=keyboard) a b c
-.3 .2
.8
.5
SOURCE(NOECHO) POLYMULT.SRC
@POLYMULT(span=12) a b ab
DECL vect invec(15)
DO i = 1,14
COMPUTE invec(i) = ab(i)
END DO
COMPUTE invec(15) = c(1)
SOURCE(noecho) ACFMORIN.SRC
SOURCE(noecho) SPECTH.SRC
@SPECTH(nfreq=201,ncov=250,ma=1) invec specs freqs
Re: Spectrum
Posted: Sat Dec 25, 2010 2:26 pm
by TomDoan
Use ARMASPECTRUM rather than SPETHE.
Re: Spectrum
Posted: Sat Dec 25, 2010 3:10 pm
by ivory4
For comparison purpose;
Code: Select all
all 1000
set y = 0.0
equation(variance=1.0, coeffs=||0.0,1.4,-0.8||) arma y 2 0
simulate(from=3, step=998) 1
# arma y
@spectrum y
@armaspectrum y
When I try to simulate data with non-stationary coeffs like AR(1)=1.4 or AR(2)=1.9, -0.5; SIMULATE does not generate data but instead filled y with "F" ?
SPECTRUM in this case still generate graph if ALL 1000 while STEP=100; Otherwise it generates blank graph.
What is wrong here?
Re: Spectrum
Posted: Sun Dec 26, 2010 7:41 am
by TomDoan
ivory4 wrote:For comparison purpose;
Code: Select all
all 1000
set y = 0.0
equation(variance=1.0, coeffs=||0.0,1.4,-0.8||) arma y 2 0
simulate(from=3, step=998) 1
# arma y
@spectrum y
@armaspectrum y
When I try to simulate data with non-stationary coeffs like AR(1)=1.4 or AR(2)=1.9, -0.5; SIMULATE does not generate data but instead filled y with "F" ?
SPECTRUM in this case still generate graph if ALL 1000 while STEP=100; Otherwise it generates blank graph.
What is wrong here?
The SPECTRUM seems to be OK. It's the ARMASPECTRUM, which takes the equation (ARMA) rather than the series Y as the input.
Re: Spectrum
Posted: Sun Jan 23, 2011 10:03 pm
by ivory4
Question when (1-L) is involved
Equation looks like this: yt=PSI(L)(et-et-1) (Yt=Phi*1Y_t-1+Phi2*Y_t-2+e_t - e_t-1)
To draw arma spectrum of yt, which of the following is correct?
Code: Select all
equation(noconstant,coeffs=||@@, @@,-1||) arma21 y 2 1
frequency 1 512
com scale=@@^2
trfunc(equation=arma21) 1
cmult(scale=(2*%Pi)/scale) 1 1
cset 1 =1.0/%z(t,1)
ctor 1 256
# 1
# abc
OR
Code: Select all
equation(noconstant,coeffs=||@@, @@||) arma y 2 0
frequency 1 512
com scale=@@^2
trfunc(equation=arma) 1
cset 1 = (1-%zlag(t,1))/%z(t,1)
cmult(scale=(2*%Pi)/scale) 1 1
ctor 1 256
# 1
# abc
It seems that the graph look the same but the scale is different now. The former seems to be more reasonable.
How to fix it ?
Thanks
Re: Spectrum
Posted: Mon Jan 24, 2011 9:13 am
by TomDoan
In the first case, you're squaring, then inverting; in the second, you're inverting, then squaring. But you're using the same scale factor in the square in both cases, which can't be correct. You need to multiply in one and divide in the other.
Re: Spectrum
Posted: Wed Nov 09, 2011 9:27 pm
by ivory4
Equation: yt=PSI(L)(et-et-1) and assume we do not have MA terms here and only AR(2) terms.
I correct the squaring error in previous post
The spectrum are still not the same. What is the reason? Which calculation is wrong?
ivory4 wrote:
Code: Select all
equation(noconstant,coeffs=||@@, @@,-1||) arma21 y 2 1
frequency 1 512
com scale=@@^2
trfunc(equation=arma21) 1
cmult(scale=(2*%Pi)/scale) 1 1
cset 1 =1.0/%z(t,1)
ctor 1 256
# 1
# abc
*****same as******
cset 1 =(1.0-@@*%zlag(t,1)-@@*%zlag(t,2))/(1.0-%zlag(t,1))
com scale=@@^2
cmult(scale=(2*%Pi)/scale) 1 1
cset 1 =1.0/%z(t,1)
ctor 1 256
# 1
# abc
OR
Code: Select all
equation(noconstant,coeffs=||@@, @@||) arma y 2 0
frequency 1 512
com scale=@@^2
trfunc(equation=arma) 1
cmult(scale=(2*%Pi)/scale) 1 1
cset 1 = (1-%zlag(t,1))/%z(t,1)
ctor 1 256
# 1
# abc
Another problem is when we have MA terms, say et-et-1 follow MA(1), then the first approach (especially the part below ******the same as******) is very different from the second approach.
Code: Select all
****when MA terms exist***********
cset 1 = (1.0-@@*%zlag(t,1)-@*%zlag(t,2))/((1.0+##*%zlag(t,1))*(1.0-%zlag(t,1)))
Or I think the following is correct?
Code: Select all
equation(noconstant,coeffs=||%beta(4),%beta(5),%beta(6)-1.0,-1.0*%beta(6)||) arma y 2 2
frequency 1 512
trfunc(equation=arma) 1
com scale = ###^2
cmult(scale=(2*%pi)/scale) 1 1
cset 1 =1.0/%z(t,1)
ctor 1 256
# 1
# ab
Re: Spectrum
Posted: Thu Nov 10, 2011 12:43 pm
by TomDoan
It might be easier to understand if ARMASPECTRUM had been written like this:
Code: Select all
trfunc(equation=eqn) 1
cmult 1 1
*
* Take the scaled reciprocal and send half of the resulting series back to the
* time domain.
*
cset 1 = (scale/(2*%pi))/%z(t,1)
Multiplying by sigma^2/(2pi) is the
final step after you've finished manipulating the squared transfer function. The spectral density is
sigma^2/(2*pi)*|C(w)|^2
where C(w) is the transfer function from epsilon to y. Since TRFUNC generates the transfer from Y to epsilon (not epsilon to Y), you need the reciprocal of the squared transfer function to make the |C(w)|^2.
Re: Spectrum
Posted: Thu Nov 10, 2011 3:29 pm
by ivory4
What I got wrong is the use of (1-%zlag(t,1) to represent (1-L) part after calculating |C(w)|^2, when I have the equation like (1-Phi(L))yt=THETA(L)(1-L)et
Now I define the EQUATION using parameters in Phi(L) and Theta(L) by expanding Theta(L)(1-L)
Code: Select all
equation(noconstant,coeffs=||phi1, phi2, theta1-1,-theta1||) arma21 y 2 1
Then with TRFUNC and CMULT to square, CSET to invert, it seems correct.