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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