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Is there an analytical solution for the following optimization problem:

$b, y \in \mathbb{R}^n$ (vectors), $b^* = \arg min_{b \in \mathbb{R}^n} f(b)$, where $f(b) = ||b-y||_{2}^{2} + \varphi ||b||_{1}$ where $\varphi > 0$

Its so close to $L_1$ regression, but I cant solve this even in that way.

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  • $\begingroup$ Welcome to CV. It would all depend on what "$\varphi$" means: that detail is essential for answering your question. $\endgroup$
    – whuber
    Feb 12 '20 at 13:53
  • $\begingroup$ $\phi$ is a constant which greater than 0 $\endgroup$
    – Leeds
    Feb 14 '20 at 10:57
  • $\begingroup$ Ah--so $\phi$ is not a function. This problem is called the "Lasso." See stats.stackexchange.com/search?q=lasso+closed+form for the general problem. Since yours is so specific, it is plausible it would have a specific answer (+1). $\endgroup$
    – whuber
    Feb 14 '20 at 15:49
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Using vector math re-writing your problem, also replacing b with X: $$f(X)=(X-Y)^T(X-Y)+\varphi |X|^T 1$$ First Order Conditions: $$f(X)'=0$$ $$2(X-Y)+\varphi \space\mathrm{sign}(X)=0$$ Solution: $$X=Y-\mathrm{sign}(X)\varphi/2$$

  • when $|Y_i|\ge\varphi/2$: $$X=Y-\mathrm{sign}(Y)\varphi /2$$
  • when $|Y_i|<\varphi/2$: $$X=0$$
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  • $\begingroup$ +1. But what does "F.O.C." mean? $\endgroup$
    – whuber
    Feb 14 '20 at 17:47
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    $\begingroup$ @whuber, I spelled it out, 1st order conditions $\endgroup$
    – Aksakal
    Feb 14 '20 at 18:11

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