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The output of the auto.arima function gives log-likelihood as 93.69. However I calculate it as 99.40152 by using the residuals. What causes the difference? 

resid=c( 0.0009133036, -0.0024778952,  0.0016247257, -0.0262748348,-0.0245270803,  0.0416803278,  0.0283238192,-0.0200986496, -0.0637956188,-0.0304700821, -0.0389081231, -0.0066333658, -0.0079335582, -0.0327841692, 0.0422120965, -0.0107629969,  0.0141159479,  0.0033650571, -0.0195425507, -0.0016719690, -0.0092509480,-0.0173254345,  0.0233666990, -0.0063053729, -0.0249087321,  0.0175185423,  0.0090894124,  0.0094325355,-0.0170129762, -0.0368547384,  0.0118255009, -0.0728012098,  0.0069336386,  0.0052472716, -0.0028184078, 0.0052579484,  0.0117586931,  0.0079340426, -0.0270301760,  -0.0590942951,  0.0448756767,  0.0175497969,-0.0178870348, -0.0310982909,  0.0033326883)


L=length(resid)*log(1)-length(resid)*log(sd(resid)*sqrt(2*pi))-(sum(resid^2)/(2*var(resid)))

Any df for std and var of residuals is not considered.

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  • $\begingroup$ What does your last sentence mean? $\endgroup$ – Richard Hardy Oct 20 '15 at 15:47
  • $\begingroup$ In the above given calculation, degrees of freedom is not considered for the calculation of standard deviation and the variance of the residuals. The residuals are from an ARIMA(1,2,1) model. $\endgroup$ – Dirk Oct 20 '15 at 18:22

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