Using the BIC to select a model I have a data set (with $n$ points) and there are two models which I think will fit the data well. The first model is a simple power law and the second model is a power law plus a linear term and an intercept;
\begin{eqnarray}
y&=&ax^b\\
y&=&ax^b+cx+d.
\end{eqnarray}
I would like to use the Bayesian Information Criterion (BIC) to see if using the second model is really worth it. My question is, what exactly is the likelihood $L$ for both of these models? How exactly do I compute $L$ so that I can estimate the BIC?
 A: Actually, a term is missing in your (two) models: the error term. Let's play with the first one:
$y_i = ax_i^b + u_i$ for $i=1,...,n$
The (distribution-specific) likelihood function used in the Bayesian Information Criterion (BIC) is that of $\boldsymbol{u}$ (of the residual vector $\boldsymbol{y} - \widehat{a}\boldsymbol{x}^\widehat{b}=\widehat{\boldsymbol{u}}$ in practice). 
This means that you first need to assume a probability density function for $\boldsymbol{u}$, say, $f(\boldsymbol{u},(a,b))$.
Then you can compute -- not to say maximize -- the corresponding log-likelihood function, $\ln(\widehat{L})=\sum_{i=1}^n\ln(f(\widehat{u_i},(\widehat{a},\widehat{b})))$.
Finally, $\text{BIC} = k \ln(n) -2 \ln(\widehat{L})$.

In the (first model) case you describe, $k=2$ and the sensible part is about specifying the probability density function. With your normally distributed residuals, one has
$f(\widehat{u_i},(\widehat{a},\widehat{b})) = (\widehat{\sigma}^22\pi)^{-.5} e^{-.5\widehat{u_i}/\widehat{\sigma}}$
