Closed cshjin closed 4 years ago
The standard variance is also consistent with the accuracy distribution figures. We would like to treat each device equally, such that each user's experiences are the same (even if they have different numbers of training samples, etc).
My concern is for a higher acc in a device with a smaller sample size won't reflect its fairness in the variance. And vice versa.
As in the eq (2), the objective has the weights involved, then the variance (fairness) could also have the weights involved.
A larger loss potentially means small accuracy. Why won't it affect the fairness/variance?
Sorry, revised as accuracy. Consider a toy example: three devices (acc, # sample) (0.5, 10), (0.5, 10), (0.5, 10000) -- low variance (0.1, 10), (0.1, 10), (0.9, 10000) -- high variance
Which one is better in terms of "fairness"?
Sorry I missed your previous question... By our definition (Def 1 in https://arxiv.org/pdf/1905.10497.pdf), the first one is more fair and in your second example, the model seems to overfit to the 'dominant' devices with 10,000 samples, which is what we are trying to mitigate.
Yes, thanks for your reply. I see the published work on ICLR. Congrats. I will follow up later.
Hi Hongwei,
do you have any additional questions/comments regarding this issue?
Not at this moment, thanks. Issue can be closed.
I see you have the fairness definition according to the variance of accuracies across m devices.
However, the variance you compared https://github.com/litian96/fair_flearn/blob/5f174cba7521df606fc946f40a0b5fef09b546e3/plot_fairness.py#L51 has the assumption that the accuracies are equally like the same.
My question is, instead of using the standard the variance, why not calculate the variance in a weighted fashion?
A revised variance, either in the weights of the sample distributions or the q-federated weights makes much more sense to me. Is that true?