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4.11. Discrepancy in AIC Calculation?

Applies to: @RISK 6.x/7.x, Professional and Industrial Editions

I fitted {1,2,3,4,5} to a RiskIntUniform distribution. I used the formula

AIC = 2k – 2×ln(L)

where k = 2 is the number of parameters and L is the likelihood.

For a uniform integer distribution fitted to a sample of n = 5 points, every point has probability of 1/5, and so ln(L) = 5 ln(1/5). I computed

AIC = 2k – 2×ln(L) = 2×2 – 10×ln(1/5) = about 20.0944

But @RISK gives 26.0944 in the Fit Results window. How do you reconcile this?

@RISK actually computes AICc, which includes a correction for finite sample sizes.

The formula is AICc = AIC + 2k(k+1)/(nk–1)

for a distribution with k parameters fitted to n data points.

With k = 2 for a RiskIntUniform and n = 5 data points,

AICc = AIC + 2×2×3/(5–2–1) = AIC + 6 = 26.0944

just as shown by @RISK.

The finite-sample correction is important for very small samples, but much less important for samples of reasonable size. For example, for fitting a 2-parameter distribution to 30 data points, the correction would be 12/27, about 0.4444.

Additional keywords: Distribution fitting, IntUniform

Last edited: 2015-06-19

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