# AIC for a model with non-normal residuals

I have a model comparison I'm doing using AIC, only the residuals of my models are not normally distributed. I know that in this case, the simple formula for AIC does not work.

I'm thinking I may need to use bootstrapping to calculate the likelihood of various models. However, I see several pitfalls in this approach, the biggest being that the bootstrapped likelihood will, in essence, be coming up with a different distribution of residuals for each model fitted, totally invalidating the comparison. How do I do this bootstrapping properly?

• What simple formula? The definition of AIC still works, for example. Nov 16, 2017 at 13:35
• $\textrm{AIC} = 2k- 2\hat{L}$, where $\hat{L}$ is the maximum value of of the log-likelihood of the model. In least squares with gaussian IID residuals, we have $\hat{L} = \frac{n}{2} \ln(\frac{\textrm{RSS}}{n}) + C$. Substitution leads directly to $\textrm{AIC} = 2k-n\cdot \ln(\frac{\textrm{RSS}}{n}) + C$. This is the simple formula I was talking about. Nov 16, 2017 at 16:59
• The first formula works all the time, the last one is indeed only suitable for the Gaussian case (or maybe also some related cases, I am not sure). Nov 16, 2017 at 18:04

• I believe this will cause problems in my case. Suppose that my data is generated by a process $y(x) = a\cdot x+b+e(x)$ where $e(x)$ is IID and sampled from a symmetrical distribution, but not any familiar distribution, even in approximation, and certainly not a normal distribution. Suppose further that I fit two models to my data, $y(x) = a\cdot x + b+e(x)$ and $y(x) = b+e(x)$, and then try to compare them with AIC. $e(x)$ is unknown so I must bootstrap the model likelihood, but the two models will generate different bootstrap distributions for $e(x)$ and non-comparable model likelihoods Nov 16, 2017 at 17:20