wrong proof of theorem 1 in the paper?

[linzhiqiu]

I was reading the CVPR paper and confused by the following highlighted statements in proof of theorem 1:

Could you explain why there must exist some x ′ != x such that p(x ′ ) < q(x ′ ) with f(x ′ ) = f(x) and g(x ′ ) = g(x)?

Especially, I don't think we can ensure g() will ever map two inputs to the same vector. I feel like some assumptions are missing here?

[m-Just]

Given f and g, the joint distribution of y and z is determined by the distribution of x. Suppose an x maps to (y, z) and p(x) > a > q(x) for some value a between 0 and 1, then we have p(y, z) >= p(x) > a. If x is unique so there's no other x' such that q(x') > 0 and it maps to the same (y, z), then we have q(y, z) = q(x) < a. It follows that p(y, z) > q(y, z).

I don't quite understand the second issue you raised. Could you elaborate?

[linzhiqiu]

I would like to first get some clarification, because I believe:

1. g() is a feature mapping function that maps an image x to an embedding z. However, the paper didn't define this mapping, only stated it will be trained end-to-end. Therefore, I think it is okay for g() to be an identity function in the context of theorem 1.
2. qhat(y, z) and phat(y, z) (of theorem 1) are different from the actual distribution p(y, z) and q(y, z)?

[m-Just]

1. Yes, we didn't define this mapping explicitly. It stands for the mapping induced by the feature extractor, and the mapping changes as we train the extractor. The point is that when the network attains the theoretically optimal performance on the environment classification task, the extractor g() must have learned some particular mapping so that p(y, z) != q(y, z).
2. Yes, they are different because z is the extracted feature, which is undefined in the data distributions, p and q. Here I abused the notation for simplicity.