# Why combining BICs for two paths will never work

Suppose we have the weights $x$ and $y$ of lenghts $n$ and $m$ for two paths,
then
$$BIC(x) = 2n -\sum_i^n pdf(x_i)$$
and
$$BIC(y) = 2m -\sum_i^m pdf(y_i)$$

If we want to combine, for example $[x,x]$ and $[x,y]$, then
$$BIC([x, x]) = 2(n+n) - \sum_i^{n+n} pdf([x,x]_i) = 2n - \sum_i^n pdf(x_i) + 2n - \sum_i^n pdf(x_i)$$
and
$$BIC([x, y]) = 2(n+m) - \sum_i^{n+m} pdf([x, y]_i) = 2n - \sum_i^n pdf(x_i) + 2m - \sum_i^m pdf(y_i)$$

Suppose that $BIC(x) < BIC(y)$, then it follows
$$ 2n - \sum_i^n pdf(x_i) < 2m - \sum_i^m pdf(y_i)$$
if we add the same value $2n - \sum_i^n pdf(x_i)$ to both sides then
$$ 2n - \sum_i^n pdf(x_i) + 2n - \sum_i^n pdf(x_i) < 2m - \sum_i^m pdf(y_i) + 2n - \sum_i^n pdf(x_i)$$
therefore
$$BIC([x, x]) < BIC([x, y])$$