(2 points) Let S and S+ be the set of all indices where the residual is...

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(2 points) Let S and S+ be the set of all indices where the residual is negative or positive, respectively: S = {i: Yi <ŷi}, and S {i: Yi > ŷi}, and let #(S) and #(S+) represent the sizes of the two sets. Then any MAE- minimizing solution for â, b must approximately solve the two equations, = #(S¯) = #(S+), and Στ= iES- More precisely, we must have |#(S) - #(S¹)| ≤ #(Sº), and Σ x₁ = Σx₁. ieS+ Στ-Σ Σ lies- ieS+ Xi lieS⁰ where Sº is the set of data points with zero one or two data points, whereas S and S the bounds in (2) are typically small compared to the quantities in (1). (1) (2) residual. This would typically be only each have about n/2 data points, so You do not need to show the more complicated result (2). Instead just show that, if there is any MAE-minimizing solution with S0 = 0 (no residuals are exactly zero), then we have the two equalities in (1). (2 points) Let S and S+ be the set of all indices where the residual is negative or positive, respectively: S = {i: Yi <ŷi}, and S {i: Yi > ŷi}, and let #(S) and #(S+) represent the sizes of the two sets. Then any MAE- minimizing solution for â, b must approximately solve the two equations, = #(S¯) = #(S+), and Στ= iES- More precisely, we must have |#(S) - #(S¹)| ≤ #(Sº), and Σ x₁ = Σx₁. ieS+ Στ-Σ Σ lies- ieS+ Xi lieS⁰ where Sº is the set of data points with zero one or two data points, whereas S and S the bounds in (2) are typically small compared to the quantities in (1). (1) (2) residual. This would typically be only each have about n/2 data points, so You do not need to show the more complicated result (2). Instead just show that, if there is any MAE-minimizing solution with S0 = 0 (no residuals are exactly zero), then we have the two equalities in (1).

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