class, we conside optimizing an Square Error (MSE) loss. practice, there are other choices of loss functions as well. For this problem, we will consider linear regression using Mean Absolute Error (MAE) as our loss function. Specifically, the MAE loss is given as: JMAE (0) = n 22 i=1 xy). (1) 1 (a) [4 points] Derive the partial derivative MAE() Hint: For this question, you don't need to worry about J=0. You can use the sign function s.t. sign(y-y) +1 y, otherwise." Solution: -1 . (0) 0, 21 = - 00; n 00; n sign(0x)-y)(0x) - y()) sign(0x-y) (b) [2 points] Write the vectorized solution for the gradient of the loss function, i.e., VeJMAE (0). Solution: VeJMAE(0) = 1 sign(x)-y)x() (c) [4 points] Given the following dataset of 8 points, what is the value of of JMAE(0) and JMSE (0) at 0 [1.0, 1.0, 1.0]? How about their gradients Ve.JMAE(0) and V.JMSE (0)? 2 3 4 5 6 7 8 [4,0] [1,1] [0,1] -2,-2] -2,1] [1,0] [1,0] [5,2] [3,0] 12 3 1 6 3 6 8 7 Solution: JMAR (A) 3.375 MSE(A) 10.3125.