yaringal
commented
4 years ago "Under the formula (2) and (3), the probility output has a gaussian distribution."
- no, the random variable y follows a Gaussian distribution under (2) only. It follows a categorical distribution under (3)
"However, the probility can't be a gaussian distribution as it distributed in [0,1] rather than (-infty, +infty)."
- no, the outcomes a random variable can take are in (-infty, +infty) for a Gaussian, and "classes" (categories) for a categorical rv. These are different to the probability of the rv to take these values (density for the continuous case). Also note that the density can't be negative (it is a measure) and that it can be larger than 1 for continuous random variables.
"if we just look at the first line in formula (7), if independent assumption is established, -log p(y1, y2|f(w,x)) = -log p(y1|f(w,x)) - log(y2|f(w,x)); which is just a sum of cross-entropy loss over different tasks."
- no, for Gaussian likelihoods the log Gaussian is a scaled MSE loss (and not cross-entropy).
"This is apparently contradicted with the result under additional gaussian assumption."
- I don't understand this sentence
"Somehow, the paper repalce the cross entrophy loss with mse"
- no, the log of a Gaussian density is the Euclidean distance. Have a look here https://en.wikipedia.org/wiki/Gaussian_function and here John Denker and Yann LeCun. "Transforming neural-net output levels to probability distributions". In Advances in Neural Information Processing Systems 3. 1991
"which finally reach the result that higher loss task should have higher theta weights."
- what's theta?
jiazhiguan
I read the paper carefully, the formula in paper is fundamentally wrong.
Under the formula (2) and (3), the probility output has a gaussian distribution. However, the probility can't be a gaussian distribution as it distributed in [0,1] rather than (-infty, +infty).
Under the independent assumption(formula (4)) and gaussian distribution mentioned above, the formula (7) is correct. However, if we just look at the first line in formula (7), if independent assumption is established, -log p(y1, y2|f(w,x)) = -log p(y1|f(w,x)) - log(y2|f(w,x)); which is just a sum of cross-entropy loss over different tasks. This is apparently contradicted with the result under additional gaussian assumption.
Somehow, the paper repalce the cross entrophy loss with mse which finally reach the result that higher loss task should have higher theta weights. If the paper report is correct, I think the benefit here comes from loss re-balance. Which means, re-balance the task loss will benefit multi-task performance?