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Say I have a 4-dimensional data set with two independent variables x1, x2, and two dependent variables y1, y2, i.e. each row is {x1, x2, y1, y2}:

data = RandomReal[{0, 1}, {100, 4}];

How do I feed it to LinearModelFit to fit

$$ \begin{pmatrix} y_1 \\ y_2 \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} c_1 \\ c_2 \end{pmatrix} $$

?

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3 Answers 3

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LinearModelFit doesn't do multivariate regression as far as I'm aware. You can use my repository function BayesianLinearRegression instead. The first example in the "Scope" section shows you how. You can provide the data in the format

data[[All, {1, 2}]] -> data[[All, {3, 4}]]

or

#[[{1, 2}]] -> #[[{3, 4}]]& /@ data

For example:

fitData = ResourceFunction["BayesianLinearRegression"][
  data[[All, {1, 2}]] -> data[[All, {3, 4}]],
  {1, x1, x2}, (* basis functions *)
  {x1, x2} (* independent variables *)
];

You can find the best-fit expression with:

Mean[fitData["Posterior", "PredictiveDistribution"]]

The output should provide you with all the details about the uncertainties of the predictions and the regression coefficients. My blog post has more background information, if you need it.

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  • $\begingroup$ Very good. What error structure does this assume? A bivariate normal? $\endgroup$
    – JimB
    Jul 26, 2020 at 15:56
  • $\begingroup$ @JimB Yes, the errors are assumed to be multivariate normal. I implemented the conjugate prior model from this Wikipedia article: en.wikipedia.org/wiki/Bayesian_multivariate_linear_regression It's the easiest model you can do in closed form. $\endgroup$ Jul 26, 2020 at 16:21
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You can build the regression yourself as an NMinimize of residuals which are squared distances to points.

First let's build some synthetic noisy data:

(* create some noisy data that follows a linear model *)
n = 1000;
datax = RandomReal[{-1, 1}, {n, 2}];
testmtx = {{3, 4}, {1/2, 1/6}};
testoffset = {3/2, 5/7};
fn[{x1_, x2_}] := testmtx.{x1, x2} + testoffset
noise = RandomVariate[NormalDistribution[0, 1/10], {n, 2}];
datay = (fn /@ datax) + noise;

(* this is the noisy 4d data *)
data = MapThread[Join, {datax, datay}];

ListPlot[{datax, datay}, PlotRange -> {{-4, 4}, {-4, 4}}, 
 AspectRatio -> 1, PlotStyle -> PointSize[Small]]

The ideal fit is:

$$ \left( \begin{array}{cc} y_1\\ y_2 \end{array} \right)= \left( \begin{array}{cc} 3 & 4 \\ 1/2 & 1/6 \\ \end{array} \right) \left( \begin{array}{cc} x_1\\ x_2 \end{array} \right) + \left( \begin{array}{cc} 3/2\\ 5/7 \end{array} \right) $$

... but pretend we don't know that and we only work with data from this point. Here's what the $x_1,x_2$ values (blue) vs the noisy $y_1,y_2$ values (orange) look like: noisy data

Then construct a residual function and an objective which is to minimize the total residuals:

matrix = {{a1, a2}, {a3, a4}};
offset = {c1, c2};
sqresidual[{x1_, x2_, y1_, y2_}, mtx_, c_] := 
 SquaredEuclideanDistance[c + mtx.{x1, x2}, {y1, y2}]
objective = Total[sqresidual[#, matrix, offset] & /@ data];

... and finally use NMinimize:

NMinimize[objective, {a1, a2, a3, a4, c1, c2}]
(* result: {19.8142, {a1 -> 2.99722, a2 -> 4.00609, a3 -> 0.498218, 
  a4 -> 0.165467, c1 -> 1.49577, c2 -> 0.7118}} *)

The result is pretty close to ideal!

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    $\begingroup$ very nicely done! One can easily change the distance as well, say sqresidual[{x1_, x2_, y1_, y2_}, mtx_, c_] := Norm[c + mtx.{x1, x2} - {y1, y2}, 1] to make it more robust to outliers. $\endgroup$
    – chris
    Jul 25, 2020 at 16:42
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For simple applications of regression, there is no need to fit multiple dependent variables simultaneously. The results are the same as a regression of each dependent variable separately. There are caveats if you are doing further analysis, but if you just want the basic regression results, you can do separate fits.

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  • $\begingroup$ These are excellent points. "The results are the same" I assume means that the estimates of the coefficients are identical. And the "caveats" include that the measures of precision can be inappropriate (i.e., wrong) when performing separate regressions. $\endgroup$
    – JimB
    Jul 26, 2020 at 16:06
  • $\begingroup$ @JimB The coefficients and their standard errors are the same. $\endgroup$
    – nanoman
    Jul 26, 2020 at 16:25
  • $\begingroup$ Sorry. I wasn't specific enough. I agree that if all of the observations are independent but there is correlation between the two response variables within an observation, the estimates of the coefficients and standard deviations are identical. But if one wanted to compare the difference in the two intercepts, then one would obtain a wrong standard error for the difference if one performed the regressions separately. Your answer could be enhanced by giving examples of when there is a difference. $\endgroup$
    – JimB
    Jul 27, 2020 at 0:26

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