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I would like to reproduce the result of multiple linear regression with Predict. In approach 1, I used matrix computation and obtained 3.2, while in approach 2, I used Predict directly and obtained 4.05.

Upon checking with other statistical software, 3.2 should be correct. May I ask why approach 2 is wrong, and how can I fix the problem?

data1 = {1, 2, 3};
data2 = {2, 3, 5};
data3 = {3, 5, 8};
data4 = {4, 5, 8.5};
X = {data1, data2, data3, data4};
Y = {1.2, 2.5, 3.3, 4.9};
ToBePredicted = {3, 6, 9};

Approach 1 - Matrix Computation - Result:3.2 (correct)

X1 = ArrayFlatten[{{X, ConstantArray[{1}, 4]}}];
X1T = Transpose[X1];
β = Inverse[X1T.X1].X1T.Y;
β.Append[ToBePredicted, 1] 

Approach 2 - Predict - Result:4.05 (wrong)

asso = MapThread[#1 -> #2 &, {X, Y}];
p = Predict[asso, Method -> "LinearRegression"];
p[ToBePredicted]
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    $\begingroup$ p2 = Predict[asso, Method -> {"LinearRegression","L2Regularization"->0}]; p2[ToBePredicted] gives the same result as β.Append[ToBePredicted, 1] $\endgroup$
    – kglr
    May 1, 2018 at 1:17
  • 1
    $\begingroup$ as stated in Documentation >> Predict >> Properties & Relations : The linear regression predictor without regularization and LinearModelFit can train equivalent models. $\endgroup$
    – kglr
    May 1, 2018 at 1:18

1 Answer 1

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Using the suboption "L2Regularization" -> 0 for "LinearRegression" in Method option setting gives the same result as the matrix computation in OP:

p2 = Predict[asso, Method -> {"LinearRegression", "L2Regularization" -> 0}];
p2[ToBePredicted]

2.8

which is the same as

β.Append[ToBePredicted, 1]

2.8

Documentation >> Predict >> Properties & Relations :

The linear regression predictor without regularization and LinearModelFit can train equivalent models.

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