[FIXED] How to forecast a Time Series Regression Models with Distributed Lag (using dLagM)?

Issue

I am trying to forecast a time series with distributed lag (using dLagM). I guess I can fit the model properly, it shows all the expected results. But I am unable to forecast any value. The error, at least for me, is opaque.

I guess it is something to do with my dummy variables and its lags, but I cannot figure it out by myself, so after a couple of days stranded I call for help!

Here is a reproducible example. It uses the dummies and the lags proposed by previous work.

# data
df <- dplyr::tribble(
     ~y ,    ~x,   ~dummy1, ~dummy2,
   207.414  , 59.717     ,  0    ,  0    , 
   177.416  , 59.576     ,  0    ,  0    , 
   245.526  , 63.288     ,  0    ,  0    , 
   276.641  , 61.801     ,  0    ,  0    , 
   371.803  , 58.529     ,  0    ,  0    , 
   519.777  , 56.790     ,  1    ,  0    , 
   430.641  , 54.012     ,  0    ,  1    , 
   251.612  , 57.151     ,  0    ,  0    , 
   269.787  , 57.480     ,  0    ,  0    , 
   230.034  , 60.042     ,  0    ,  0    , 
   202.376  , 60.280     ,  0    ,  0    , 
   253.497  , 61.323     ,  0    ,  0    , 
   239.166  , 61.235     ,  0    ,  0    , 
   272.894  , 60.206     ,  0    ,  0    , 
   293.951  , 62.020     ,  0    ,  0    , 
   278.437  , 61.393     ,  0    ,  0    , 
   424.190  , 58.876     ,  0    ,  0    , 
   652.256  , 56.978     ,  1    ,  0    , 
   536.587  , 56.381     ,  0    ,  1    , 
   263.116  , 61.193     ,  0    ,  0    , 
   289.288  , 60.123     ,  0    ,  0    , 
   227.690  , 60.957     ,  0    ,  0    , 
   234.306  , 62.563     ,  0    ,  0    , 
   293.728  , 61.540     ,  0    ,  0     )

# new auxiliary data to be used as input to forecast y for 12 periods
newdata <- dplyr::tribble(
  ~x,   ~dummy1, ~dummy2,
  61.903     ,  0    ,  0    , 
  60.594     ,  0    ,  0    , 
  63.358     ,  0    ,  0    , 
  65.178     ,  0    ,  0    , 
  64.275     ,  0    ,  0    , 
  59.872     ,  1    ,  0    , 
  59.273     ,  0    ,  1    , 
  59.665     ,  0    ,  0    , 
  58.643     ,  0    ,  0    , 
  63.354     ,  0    ,  0    , 
  65.743     ,  0    ,  0    , 
  65.158     ,  0    ,  0    )



# Model ARDL(1,4)
model = dLagM::ardlDlm(formula = y ~ x + dummy1 + dummy2 ,
                     data = df, 
                     p = 1 , # lag; given by previous analysis
                     q = 4, # order of autoregressive process; given by previous analysis
                     remove = list(p = list(dummy1 = c(1:1), 
                                            dummy2 = c(1:1)))
                    )

# transposed (for dLagM::forecast)
transposed_newdata <- t(newdata)

# forecasting
fLeves <- dLagM::forecast(model,
                   x = transposed_newdata, 
                   h = nrow(newdata),
                   interval = TRUE, 
                   level = 0.95 , 
                   nSim = 100)

# Error
# Error in if (n == 0) return(v) : missing value where TRUE/FALSE needed

Any help is much appreciated!

Solution

Thanks for posting this. There was a bug in the code related to the variable names. I fixed the issue in the new version of dLagM 1.1.8. Also, please note that you need to avoid nested variable names working with version 1.1.8. For example, if you have "x1" as a variable, avoid using a name that has "x1" in it, such as "x11". I'll fix this issue in the next version.

Answered By - Haydar

Answer Checked By - Senaida (PHPFixing Volunteer)

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[FIXED] How should I get the forecasted values in R using Arima Model?

Issue

I am new to R so doesn't know how exactly the forecasting works in R using Arima Model. I have a dataset is like: 92 Aug-17 9533 93 Sep-17 8718 94 Oct-17 2035 95 Nov-17 2539 96 Dec-17 1333 97 Jan-18 2444 98 Feb-18 9371 99 Mar-18 9697 100 Apr-18 3989 101 May-18 4061 102 Jun-18 2797 103 Jul-18 2949

I want see the forecasted values for next 12 months by using arima model in R. So how should I achieve this.

Thanks in Advance. this is a dataset

Solution

Suppose you have fitted your ARIMA model in fit

so you can use the following method for predicting values. Here n is the number of values you want to predict. In your case, n will be 12.

pred = predict(fit,n.ahead = 1*n)

If you are taking log of values during ARIMA modeling so you use the following method for actual output.

values = 2.718^pred$pred

Answered By - girijesh96

Answer Checked By - Timothy Miller (PHPFixing Admin)

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