Formula:
Memory Index called delta?in Parzen(1983); see pdf attachment p.536
?
Code:
?
##########################################################################
# I am using a simulated long memories time series X1 of length 2000;??? #
# I actually used d=.25 for AFRIMA (0,.25,0)???????????????????????????? #
# and I am trying to estimate d through the memory index discussed in??? #
# Parzen(1983) on p.536 . I am in need of an assessment of my code for?? #
# the Parzen window as well as the choice of k and n. in my code I used? #
# k to be 999 and n to be 2000. I am not confortable with the memory???? #
#? index estimator and I will appreciate some help on the the code.????? #
#?????????????????????????? Thank you!?????????????????????????????????? #
##########################################################################
?
Pt <- acf(X1,2000)
n <- length(X1)
vv <- 1:(n-1)
T <- 2000
MT <- T/2
MT2 <- MT%/%2
## Parzen window formula on p.536
M_vT <- KK <- as.numeric(0)
M_vT = vv/MT
for (v in vv) {
??????? K[v] <- if (v <= MT2)
??????????? 1 - 6 * M_vT[v]^2 * (1 - M_vT[v])
??????? else if ( v <= MT)
??????????? 2 * (1 - M_vT[v])^3
??????? else 0
??? }
## Non-parametric kernel spectral density estimator formula on p.536
?p? = Pt$acf
?P = g = 0
?for (v in 1:999) {
?g = g + (K[v]*p[v])
?P[v] = g
?}
w? <- seq(.005, 1, by = .005)
i.c <- sqrt(as.complex(-1))
g.w <- 0
f.w <- function(w){
?for (v in 1:999) {
?g.w = g.w+ P[v]*exp(-2*pi*i.c*w*v)
????}
?g.w
?}
# f.w(.015) for w=.015 for instance
## memory index delta formula on p.536
g.d = 0
j = 1:999
j1 = j/n
j2 = 1000/n
f1 = f.w(j1)
f2 = f.w(j2)
delta = 0
deltak = 0
for (i in 1:999){?
?g.d = g.d + (log(f1[i]) - log(f2))
???}
??delta = g.d
?
?deltak = delta/999
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