# [R] Fractional degree of differencing, d

From: Fologo Dubois <fologodubois_at_yahoo.com>
Date: Thu, 03 Mar 2011 04:07:41 -0800 (PST)

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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Received on Thu 03 Mar 2011 - 15:13:17 GMT

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