Calculation of AICc NonLinearModelFit

Question

I have been using NonlinearModelFit in a current study. Outputs from this procedure include AIC and AICc.

AICc can be calculated from AIC via the following relation:

$$\text{AICc} = \text{AIC} + \frac{2k(k+1)}{n-k-1}$$

If a model with $3$ parameters is considered, then $k = 3$, and if $5$ data points are used in the fitting then $n = 5$. With these assumptions,

$$\frac{2k(k+1)}{n-k-1}=\frac{2\times 3(3+1)}{5-3-1}=24/1=24$$

The denominator of this expression would be zero for $4$ data points.

I have performed an analysis with 5 data points. The value of AICc should be AIC + 24. In this instance NonlinearModelFit calculated AICc = Infinity.

Is this an error or am I missing something?

Answer 1

You're not counting all of the parameters as you need to include the error variance. So with $n=5$ data points and 3 coefficients and one error variance the number of parameters is $k=4$ and $n-k-1=0$, hence the resulting infinity. (Variances are parameters, too!)

See Hurvich & Tsai (1989) (Biometrika (1989), 76, 2, pp. 297-307).

Update

So for a standard regression model with just a single error term Mathematica uses the Hurvich & Tsai (1989) formula:

$$AIC_c=AIC+{{2(k+1)(k+2)}\over{n-k-2}}$$

where $k$ is the number of parameters in the model (not counting the error variance). (For a linear model $k$ is the rank of the design matrix.) This gives $\infty$ when $k=n-2$ as you found.

For whatever it's worth PROC MIXED in SAS with METHOD=ML (ML for maximum likelihood rather than REML for restricted maximum likelihood) with a single error variance uses a slightly different formula for $AIC_c$ to avoid dividing by zero when $k=n-2$:

$$AIC_c=AIC+{{2(k+1)(k+2)}\over{\max{(n,k+3)}-k-2}}$$