Signs of the two score components for OOD detection

[j-cb]

Hi,

in the scoring formula on page 7 in the paper, shouldn't the KL divergence of the classifier prediction from uniform be small for OOD inputs, and the rotation CE be large on OOD since the rotation head has not been trained to predict the original rotation on OOD inputs? I.e. one of the terms should have a minus sign, right?

If I read it correctly, the code uses different signs for those terms:

ss-ood/multiclass_ood/test_auxiliary_ood.py

Lines 182 to 185 in 2a284be

	classification_loss = -1 * kl_div(class_uniform_dist, classification_smax)
	rot_one_hot = torch.zeros_like(rot_smax).scatter_(1, target_rots.unsqueeze(1).cuda(), 1)
	rot_loss = kl_div(rot_one_hot, rot_smax)

where KL is positive CE minus the constant entropy of U.

[lygjwy]

Hi,

in the scoring formula on page 7 in the paper, shouldn't the KL divergence of the classifier prediction from uniform be small for OOD inputs, and the rotation CE be large on OOD since the rotation head has not been trained to predict the original rotation on OOD inputs? I.e. one of the terms should have a minus sign, right?

If I read it correctly, the code uses different signs for those terms:

ss-ood/multiclass_ood/test_auxiliary_ood.py

Lines 182 to 185 in 2a284be

	classification_loss = -1 * kl_div(class_uniform_dist, classification_smax)
	rot_one_hot = torch.zeros_like(rot_smax).scatter_(1, target_rots.unsqueeze(1).cuda(), 1)
	rot_loss = kl_div(rot_one_hot, rot_smax)

where KL is positive CE minus the constant entropy of U.

I think the code is correct. For OOD inputs, the whole loss, i.e., kl_div(rot_one_hot, rot_smax) -1 * kl_div(class_uniform_dist, classification_smax) is bigger than ID inputs.

[zjysteven]

The code is correct while the formulation in the paper (Sec. 4.1 Method) is not.