Home → Techniques and Tips → @RISK Distribution Fitting → 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)/(n–k–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