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LinearModelFit[{m,v}] will return a coefficients list $\beta$ from the design matrix $m$ and response vector $v$, where $m.\beta$ is fitted to $v$. However, the parameters in $\beta$ could be negative. For example,

LinearModelFit[{{{1, 2, 3, 4}, {2, 3, 7, 6}}, {2, 3}}]["BestFitParameters"]

will return {0.0381679, 0.198473, -0.00763359, 0.396947}, where the third parameter is negative.

How could I force LinearModelFit[{m,v}] to fit with only positive parameters? Is there an Option I can set constrain on coefficients?

I use LinearModelFit[{m,v}] because the length of $m$ (the number of variable) could vary from case to case.

Seems I raise a bad example. In reality the number of unknowns is less than the number of equations, that is Length@m[[1]]<Length@v.

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  • $\begingroup$ An easy yet inefficient route: NonlinearModelFit[Append @@@ Transpose[{{{1, 2, 3, 4}, {2, 3, 7, 6}}, {2, 3}}], {Array[c, 4].Array[x, 4], And @@ Thread[Array[c, 4] > 0]}, Array[c, 4], Array[x, 4]] $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Sep 21, 2020 at 8:42
  • $\begingroup$ something like Array[x, 4] could be used as four variables..OMG.... $\endgroup$
    – Harry
    Commented Sep 21, 2020 at 9:02

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Consider this picture: enter image description here

What would the best approximation to "vec" be, when the condition: "y component of vec is zero" is imposed? The best approx. under this condition is setting the y component to zero.

Back to your question. You simply need to set the negative component to zero.

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  • $\begingroup$ This is genius....wait...suppose the formula is $C_x x+C_y y$, if now $C_y$ is setted to 0 from negative value, shouldn't $C_x$ decrease a bit to balance (given xs, ys are all positive) and achieve better fitting? $\endgroup$
    – Harry
    Commented Sep 21, 2020 at 8:25
  • $\begingroup$ The picture is illuminating....but is somewhere a proof make sure "set the negative component to zero" gives the best approx? seems it makes a vector closest to the solution with no constrain....emmm.... $\endgroup$
    – Harry
    Commented Sep 21, 2020 at 8:32
  • $\begingroup$ Think about it this way: you are searching a vector vec1 so that the distance vec1-vec is minimal. In the above example this vector is the projection of vec on the x axis.In the general case, it is the projection to the subspace with y=0 $\endgroup$
    – Daniel Huber
    Commented Sep 21, 2020 at 8:53
  • $\begingroup$ NMinimize evaluates a much better solution than the projective one: NMinimize[{#.# &[{{1, 2, 3, 4}, {2, 3, 7, 6}}.{b1, b2, b3, b4} - {2, 3}], b1 > 0, b2 > 0, b3 > 0, b4 > 0}, {b1, b2, b3, b4}] (*{3.35725*10^-9, {b1 -> 0.000104177, b2 -> 0.502257, b3 -> 0.0000209469, b4 -> 0.248818}}*) $\endgroup$
    – Ulrich Neumann
    Commented Sep 21, 2020 at 9:12
  • $\begingroup$ You are partially right. The main problem is that we have 4 unknowns and only 2 equations. Therefore, there are an infinity of solutions. the simplest one would be: {0,1,0,0}. Now, what I did is to approximate the vector: {0.0381679, 0.198473, -0.00763359, 0.396947} as well as possible with positive coefficients. Now, the question arise why neither LinearModelFit nor NMinimize gives a correct result. I can only guess that numerical errors with real numbers are the reason. $\endgroup$
    – Daniel Huber
    Commented Sep 21, 2020 at 9:48

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