Why doesn't the formula for the BIC include $\frac{\lambda}{\pi}$ in it? I've been going through the derivation of the BIC. In Schwarz's original paper (linked below) he arrives at
\begin{align*}
nA-\frac{1}{2}k_j\log\left(\frac{n\lambda}{\pi}\right)+\log\alpha_j
\end{align*}
for the penalty term, where $n$ is the sample size, $A$ is a constant (I think), $k_j$ is the dimension of the parameter space of model $j$, $\lambda$ is a constant and $\alpha_j$ is the prior probability of model $j$. I think I can follow up to here, but from here onwards I'm struggling a bit. Firstly, I'm not sure how he justifies ignoring $nA$. From the central limit theorem, we can assume the likelihood function is approximately a normal likelihood and then I believe we have
\begin{align*}
nA=n\log\left(\frac{1}{\sqrt{2\pi}|\Sigma|}\right),
\end{align*}
which is independent of the model if $\Sigma$ is independent of the model. Then the $nA$ term can be ignored. $\alpha_j$ is the prior probability of model $j$, so if you make all the prior probabilities equal (questionable, but it leads to the standard form of the BIC) then this term can be ignored. Then you are left with
\begin{align*}
-\frac{1}{2}k_j\log\left(\frac{n\lambda}{\pi}\right).
\end{align*}
Clearly in the limit we can approximate this with
\begin{align*}
-\frac{1}{2}k_j\log{}n
\end{align*}
since $k_j$ and $\lambda$ are constants. However, my question is why not leave $\lambda$ and $\pi$ in the equation? If we assume a normal likelihood from the central limit theorem then $\lambda=\frac{1}{2}$ and we can evaluate the more general equation. Since $k_j$ depends on the model, then $k_j\log\left(\frac{\lambda}{\pi}\right)$ also depends on the model. Although it is $O\left(1\right)$, $k_j\log{}n$ is $O\left(\log{}n\right)$, so the divergence is very slow and it seems like $-\frac{1}{2}k_j\log{}n$ would often be a poor approximation to $-\frac{1}{2}k_j\log\left(\frac{n\lambda}{\pi}\right)$ for finite $n$.
Thanks.
https://projecteuclid.org/euclid.aos/1176344136
 A: The simple answer to your question is right in the abstract:

These terms are a valid large-sample criterion beyond the Bayesian context, since they do not depend on the a priori distribution.

Schwarz wants a result that does not depend on the prior distribution. Notice that he never assumes that the prior probabilities of each model $\alpha_j$ are equal, only that the conditional priors given that the $j^{\text{th}}$ model is true are locally bounded away from zero. Other people might routinely assume a uniform prior over models ($\alpha_i = \alpha_j$ for all $i, j$) when using the BIC, but he does not.
This means that Schwarz is dealing with with terms $\alpha_j$ in a different way entirely than those who assume a uniform prior over models, and he deals with those terms by considering only the limit as $n \to \infty$. This yields the main result, that as $n \to \infty$, we have:
$$
\begin{align*}
S(Y, n, j) & = nA - \frac{1}{2} k_j \log\left(\frac{\lambda}{\pi} n\right) + \log(\alpha_j) + R_0 \\
& = nA - \frac{1}{2} k_j \log(n) - \frac{1}{2} k_j \log\left(\frac{\lambda}{\pi}\right) + \log(\alpha_j) + R_0 \\
& = nA - \frac{1}{2} k_j \log(n) + R
\end{align*}
$$
where $R_0$ and $R$ are bounded with respect to $n$.
Essentially, because Schwarz wants a result that is true even if the prior is not assumed to be uniform over models, he has to derive an asymptotic result to eliminate the $\log(\alpha_j)$ term. But if you are only deriving an asymptotic result, you can throw away the $- \frac{1}{2} k_j \log\left(\frac{\lambda}{\pi}\right)$ term as well. He is not considering the case of finite $n$ at all.
Schwarz does, however, make assumptions that imply $A$ is constant. Because $Y$ and $b$ are constant (main proposition states that $Y$ is fixed, and $b$ does not appear in the arguments of $S$), the supremum can be inferred to be over $\theta$. Then $A = \sup_{\theta} Y \circ \theta - b(\theta)$ is constant. Thus Schwarz is not considering the case when $A$ would differ between models. Since the ultimate goal is model comparison, and the term $nA$ is the same for both models, it can be ignored.
If you make different assumptions than Schwarz, such as assuming a finite $n$, a uniform prior over models, and perhaps even that the $A$'s could differ, then of course you would derive a different result. If your real question is why people use (or perhaps "abuse") the BIC for finite $n$ without the $\lambda/\pi$ term, I cannot answer, but I hope that Schwarz's original argument is clear.
