[R] Estimating MA parameters through arima or through package "dlm"

[Stefano Sofia]

Dear list users,
I want to use apply a MA(2) process (x=beta1*epsilon_(t-1) + beta2*epsilon_(t-1) + epsilon_(t)) to a given time series (x), and I want to estimate the two parameters beta1, beta2 and the variance of the random variable epsilon_(t).

If I use
MA2_1 <- Arima(x, order=c(0,0,2))
I get the following result

[1] "MA2_1"
Series: x
ARIMA(0,0,2) with non-zero mean

Coefficients:
          ma1     ma2  intercept
      -0.0279  0.0783     5.3737
s.e.   0.0667  0.0622     0.0245

sigma^2 estimated as 0.1284:  log likelihood=-92.63
AIC=193.25   AICc=193.43   BIC=207.11
[1] 0 2 0 0 1 0 0

From this straightforward analysis V[epsilon]=0.1284, beta1=-0.0279 and beta2=0.0783.

I also tried to use a DLM representation of ARIMA models and estimate the unknown parameters by maximum likelihood through the dlm package (in particular applying the example at section 3.2.6, page 115, of "Dynamic Linear Models with R" by Petris, Petrone and Campagnoli:

arma_parameters <- function(x)
{
  buildGap <- function(u)
  {
    gap <- dlmModARMA(ma = u[2 : 3], sigma2 = u[1])
    return(gap)
   }
   init <- c(0.005, 0.004, 0.003)
   outMLE <- dlmMLE(x, init, buildGap)
   dlmGap <- buildGap(outMLE$par)
}

and this gives:
[1] "outMLE"
$par
[1] 1.00816794 0.02349296 0.02364788

$value
[1] 3089.196

$counts
function gradient
      10       10

$convergence
[1] 0

$message
[1] "CONVERGENCE: REL_REDUCTION_OF_F <= FACTR*EPSMCH"

[1] "dlmGap"
$FF
     [,1] [,2] [,3]
[1,]    1    0    0

$V
     [,1]
[1,]    0

$GG
     [,1] [,2] [,3]
[1,]    0    1    0
[2,]    0    0    1
[3,]    0    0    0

$W
           [,1]         [,2]         [,3]
[1,] 1.00816794 0.0236848488 0.0238410337
[2,] 0.02368485 0.0005564272 0.0005600964
[3,] 0.02384103 0.0005600964 0.0005637899

$m0
[1] 0 0 0

$C0
      [,1]  [,2]  [,3]
[1,] 1e+07 0e+00 0e+00
[2,] 0e+00 1e+07 0e+00
[3,] 0e+00 0e+00 1e+07

In this case
V[epsilon]=W[1,1]=1.00816794
beta1=W[2,1]/W[1,1]=0.02349296
beta2=W[3,1]/W[1,1]=0.02364788

I presume that these two approaches should give comparable results, but this does not happen.
Is the model that I used correct? And does it make sense to perform this kind of comparison?

This is the log of a rainfall time series (which has already been deseasonalised):
[1] 6.014937 4.978801 5.654592 5.616771 5.612398 5.837147 5.121580 5.832176
[9] 5.205654 5.355642 5.405376 6.257859 5.516247 5.500850 4.708629 5.482304
[17] 5.689684 5.727824 4.779123 5.289277 5.217107 5.976351 4.630838 5.683240
[25] 5.345678 5.906179 5.605434 5.497578 5.898801 5.660875 5.111988 5.571013
[33] 5.949340 5.374352 4.841033 5.995706 5.661223 5.458734 4.454347 5.795754
[41] 5.995706 5.596939 5.399971 5.908898 5.282696 5.438514 5.528635 6.022721
[49] 5.524257 5.519459 4.957235 5.547518 5.080783 5.411200 5.056883 5.798183
[57] 5.086361 5.536547 5.220356 5.141664 5.847017 5.052417 5.734635 5.340419
[65] 5.724238 5.634432 5.685958 5.307773 5.817706 5.134032 4.987708 5.110179
[73] 5.423628 5.347108 4.859037 5.556828 5.487283 5.661223 5.732370 5.469325
[81] 5.726848 5.419207 5.172187 5.608006 5.130490 5.586874 5.171052 5.683240
[89] 4.674696 5.286245 5.342813 5.370638 5.432411 5.748118 6.355239 5.557986
[97] 5.399067 5.222516 5.279644 5.425390 5.540871 5.917818 5.132853 5.689007
[105] 5.900993 5.007296 5.102911 5.778271 5.318120 5.927726 5.066385 5.716699
[113] 5.511815 4.714921 5.383577 5.319100 5.269403 5.354698 5.145749 5.204556
[121] 5.878296 5.070161 5.441552 5.213304 5.450180 5.695750 4.893352 5.425390
[129] 5.682559 5.487283 4.213608 5.751620 5.432411 5.379436 5.700444 5.580484
[137] 5.357529 5.319100 4.532599 5.603225 5.208393 5.254888 5.017280 5.349961
[145] 4.374498 5.187944 5.585374 5.716370 3.561046 5.119789 5.163070 5.422745
[153] 5.863915 5.651436 4.762174 5.655642 4.797442 5.735927 4.911183 5.240688
[161] 5.148076 5.477300 4.572647 5.493473 5.437644 4.854371 4.908233 4.755313
[169] 5.582744 5.527841 5.613128 5.211124 5.275049 5.462984 5.016617 5.981919
[177] 5.566817 5.094364 5.314191 5.712742 5.299317 5.452325 4.691348 5.851628
[185] 5.410753 5.488938 5.660179 5.900993 5.380819 5.256453 4.781641 5.531807
[193] 5.497578 5.274537 4.325456 5.271973 5.077047 5.258536 5.280662 5.247024
[201] 5.995208 4.700480 4.991113 5.457029 5.194622 5.487283 5.197391 5.747161
[209] 5.842094 5.372497 5.306781 5.641907 5.565286 5.259057 5.241218 4.759607
[217] 4.550714 5.230574 4.470495 5.664348 4.846547 5.771130 4.823502 5.598422
[225] 5.627621 5.547518 5.596939 5.468482 5.536940 5.606170 5.281680 5.656691
[233] 5.283204 5.752255 5.192401 4.550714


Thank you for your attention and your help
Stefano


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