Consider the following formula for a loss function used to train a model on ranked lists of outputs, where N is the number of items in a given list Y:

$\mathcal{L} = -\mathbb{E}\left[\frac{1}{N(N-1)}\sum_{\substack{\mathbf{y}_a\in Y, \mathbf{y}_b\in Y \\ \mathbf{y}_a\neq \mathbf{y}_b}} \log\Pr(\mathbf{y}_a \succ \mathbf{y}_b|\mathbf{x})\right]$

What is the primary analytical consequence of including the normalization term $\frac{1}{N(N-1)}$ in this calculation?

Applying the Listwise Loss Summation

Consider the listwise loss formula used for training on ranked preferences:

$\mathcal{L} = -\mathbb{E}\left[\frac{1}{N(N-1)}\sum_{\substack{\mathbf{y}_a, \mathbf{y}_b \in Y \\ \mathbf{y}_a\neq \mathbf{y}_b}} \log\Pr(\mathbf{y}_a \succ \mathbf{y}_b|\mathbf{x})\right]$

True or False: If a model is completely uncertain about the preferences within a ranked list (i.e., it assigns $\Pr(\mathbf{y}_a \succ \mathbf{y}_b|\mathbf{x}) = 0.5$ for all distinct pairs), the contribution of that specific list to the overall loss will be zero.