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    $\begingroup$ @jeza, As I wrote above, for any $ t $ there is $ \lambda \geq 0 $ (Not necessarily equal to $ t $ but a function of $ t $ and the data $ y $) such that the solutions of the two forms are the same. $\endgroup$ Commented May 28, 2018 at 4:07
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    $\begingroup$ @jeza, both $\lambda$ & $t$ are essentially free parameters here. Once you specify, say, $\lambda$, that yields a specific optimal solution. But $t$ remains a free parameter. So at this point the claim is that there can be some value of $t$ that would yield the same optimal solution. There are essentially no constraints on what that $t$ must be; it's not like it has to be some fixed function of $\lambda$, like $t=\lambda/2$ or something. $\endgroup$ Commented May 29, 2018 at 20:46
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    $\begingroup$ @Royi, I have a few things. 1- what do you mean by "move form my πœ† to the OP πœ† by a factor of two. 2" not clear to me yet. 2- I am still cannot understand that logic. 3- I am confused because you introduce "mu" with "t" and "lambda", also "‖𝑦‖^2_2≀𝑑 one must set πœ‡=0 which means that for 𝑑 large enough in order for both to be equivalent one must set πœ†=0. 𝑦𝑑(𝐼+2πœ‡πΌ)βˆ’1(𝐼+2πœ‡πΌ)βˆ’1𝑦=𝑑. This is basically when β€–π‘₯Μƒ β€–22=𝑑" not clear to me. 4- I expect to see that "t" = "lambda" then you mentioned "mu" = "lambda". Could you please help me by elaborating or editing the answer. Thx $\endgroup$ Commented Apr 2, 2019 at 23:17
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    $\begingroup$ @Royi, I think the question will be duplicated! $\endgroup$ Commented Apr 4, 2019 at 10:14
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    $\begingroup$ The most confusing part is that the relation between $\lambda$ and $t$ is data-dependent. @Royi has mentioned this twice in the comments, but IMO this is an important point that deserves to be put in bold in the answer. Although this fact is somewhat obvious (for given $\lambda, t$ we scale $y$ away from the regression line, so that the solution penalised by $\lambda$ doesn't fit into the constraint set by $t$), the phrasing "equivalent way to write the problem" is misleading, as in order to write the problem in the equivalent way we are required to first solve it. $\endgroup$ Commented May 10, 2022 at 23:58