ZachariahRosenberg
commented
3 years ago Is this so that the "2" is removed from the derivative i.e. f(x) = [(value_error) **2] / 2, so therefore f'(x) = value_error ?
Closed ZachariahRosenberg closed 2 years ago
ZachariahRosenberg
commented
3 years ago Is this so that the "2" is removed from the derivative i.e. f(x) = [(value_error) **2] / 2, so therefore f'(x) = value_error ?
In the VPG implementation, the value loss is calculated,
value_loss = value_error.pow(2).mul(0.5).mean()Isn't the value loss simply the MSE, so just
value_error.pow(2).mean()? Why the additional multiplication of 0.5?Thank you!