Can't understand Mean Absolute Error (MAE) formulation I'm reading Taylor Cross Entropy Loss paper and came across the formulation of Mean Absolute Error (MAE), which is described as following:
$$
\mathcal{L}_{MAE}(f({\bf x}), y) = \Vert{e_y - f({\bf x})}\Vert_{1} = 2 - 2f_y({\bf x})
$$
As mentioned in the linked paper's section 3.1, $e_y$ is one hot encoded vector having same dimension as $f({\bf x})$.
What I don't understand is that from where does the constant 2 come from in above formulation? I'm referring to this formulation if MAE on Wikipedia.
Any hint/reference would be appreciated. Thanks.
Edit
As requested in the comment, here is the context of the above formulation.

*

*The linked paper talks about training a deep neural network for k class classification. ${\bf x}$ is a feature vector (e.g. image of cat) and $y$ is ground truth label.

*The neural network is represented as an unknown complex function $f$.

*$e_y$ is one hot encoded vectors of ground truths. For example, if k=2 (cat and dog), then $e_y = [0, 1]$ for cat and $e_y = [1, 0]$ for dog.

*$f({\bf x})$ is what predicted by neural network. It can be probabilities for each class. For example, a well trained neural network will output $f({\bf x}) = [0.05, 0.95]$ for an image of cat.

*The mean absolute difference between ground truth labels and the predicted outputs for all images is what I refer to MAE.

 A: Suppose both $e_y$ and $f(x)$ are vectors where a single element has value $1$ and all other elements have value $0$. Therefore, $\Vert{e_y - f({\bf x})}\Vert_{1}=0$ if the position of the $1$ is the same in both vectors, and $\Vert{e_y - f({\bf x})}\Vert_{1}=2$ if the position is not the same. For $e_y$, the $1$ is at position $y$, so if $(f({\bf x}))_y$ is also $1$, the error is $0$.
Interestingly, the formula also holds  when the assumption on $f(x)$ is relaxed to assuming that the elements are nonnegative and sum to $1$:
\begin{align}\Vert{e_y - f({\bf x})}\Vert_{1} &= \sum_j |(e_y)_j - (f(x))_j| \\
&= |(e_y)_y - (f(x))_y| + \sum_{j:j\neq y} |(e_y)_j - (f(x))_j| \\
&= |1 - (f(x))_y| + \sum_{j:j\neq y} |0 - (f(x))_j| \\
&= 1 - (f(x))_y + \sum_{j:j\neq y} (f(x))_j \\
&= 1 - 2(f(x))_y + \sum_j (f(x))_j \\
&= 2 - 2(f(x))_y
\end{align}
A: Thanks to @LinAlg, I got the answer. I'm writing it for 2 classes and can be generalized further.
Consider the predictions $ f({\bf x}) = [\epsilon, 1-\epsilon]$ and ground truths $e_y = [1, 0]$. The mean absolute error will be
$$MAE = |1-\epsilon| + |-1+\epsilon| = 2 - 2\epsilon$$
This will be same even if $e_y = [0, 1]$.
