I have read the most popular books in statistical learning
1- The elements of statistical learning.
2- An introduction to statistical learning.
Both mention that ridge regression has two formulas that are equivalent. Is there an understandable mathematical proof of this result?
I also went through Cross Validated, but I can not find a definite proof there.
Furthermore, will LASSO enjoy the same type of proof?
The classic Ridge Regression (Tikhonov Regularization) is given by:
$$ \arg \min_{x} \frac{1}{2} {\left\| x - y \right\|}_{2}^{2} + \lambda {\left\| x \right\|}_{2}^{2} $$
The claim above is that the following problem is equivalent:
$$\begin{align*} \arg \min_{x} \quad & \frac{1}{2} {\left\| x - y \right\|}_{2}^{2} \\ \text{subject to} \quad & {\left\| x \right\|}_{2}^{2} \leq t \end{align*}$$
Let's define $ \hat{x} $ as the optimal solution of the first problem and $ \tilde{x} $ as the optimal solution of the second problem.
The claim of equivalence means that $ \forall t, \: \exists \lambda \geq 0 : \hat{x} = \tilde{x} $.
Namely you can always have a pair of $ t $ and $ \lambda \geq 0 $ such the solution of the problem is the same.
How could we find a pair?
Well, by solving the problems and looking at the properties of the solution.
Both problems are Convex and smooth so it should make things simpler.
The solution for the first problem is given at the point the gradient vanishes which means:
$$ \hat{x} - y + 2 \lambda \hat{x} = 0 $$
The KKT Conditions of the second problem states:
$$ \tilde{x} - y + 2 \mu \tilde{x} = 0 $$
and
$$ \mu \left( {\left\| \tilde{x} \right\|}_{2}^{2} - t \right) = 0 $$
The last equation suggests that either $ \mu = 0 $ or $ {\left\| \tilde{x} \right\|}_{2}^{2} = t $.
Pay attention that the 2 base equations are equivalent.
Namely if $ \hat{x} = \tilde{x} $ and $ \mu = \lambda $ both equations hold.
So it means that in case $ {\left\| y \right\|}_{2}^{2} \leq t $ one must set $ \mu = 0 $ which means that for $ t $ large enough in order for both to be equivalent one must set $ \lambda = 0 $.
On the other case one should find $ \mu $ where:
$$ {y}^{t} \left( I + 2 \mu I \right)^{-1} \left( I + 2 \mu I \right)^{-1} y = t $$
This is basically when $ {\left\| \tilde{x} \right\|}_{2}^{2} = t $
Once you find that $ \mu $ the solutions will collide.
Regarding the $ {L}_{1} $ (LASSO) case, well, it works with the same idea.
The only difference is we don't have closed for solution hence deriving the connection is trickier.
Have a look at my answer at StackExchange Cross Validated Q291962 and StackExchange Signal Processing Q21730 - Significance of $ \lambda $ in Basis Pursuit.
Remark
What's actually happening?
In both problems, $ x $ tries to be as close as possible to $ y $.
In the first case, $ x = y $ will vanish the first term (The $ {L}_{2} $ distance) and in the second case it will make the objective function vanish.
The difference is that in the first case one must balance $ {L}_{2} $ Norm of $ x $. As $ \lambda $ gets higher the balance means you should make $ x $ smaller.
In the second case there is a wall, you bring $ x $ closer and closer to $ y $ until you hit the wall which is the constraint on its Norm (By $ t $).
If the wall is far enough (High value of $ t $) and enough depends on the norm of $ y $ then i has no meaning, just like $ \lambda $ is relevant only of its value multiplied by the norm of $ y $ starts to be meaningful.
The exact connection is by the Lagrangian stated above.
I found this paper today (03/04/2019):
A less mathematically rigorous, but possibly more intuitive, approach to understanding what is going on is to start with the constraint version (equation 3.42 in the question) and solve it using the methods of "Lagrange Multiplier" (https://en.wikipedia.org/wiki/Lagrange_multiplier or your favorite multivariable calculus text). Just remember that in calculus $x$ is the vector of variables, but in our case $x$ is constant and $\beta$ is the variable vector. Once you apply the Lagrange multiplier technique you end up with the first equation (3.41) (after throwing away the extra $-\lambda t$ which is constant relative to the minimization and can be ignored).
This also shows that this works for lasso and other constraints.
It's perhaps worth reading about Lagrangian duality and a broader relation (at times equivalence) between:
Assume we have some function $f(x,y)$ of two variables. For any $\hat{x}$ and $\hat{y}$, we have:
$$ \min_x f(x, \hat{y}) \leq f(\hat{x}, \hat{y}) \leq \max_y f(\hat{x}, y)$$
Since that holds for any $\hat{x}$ and $\hat{y}$ it also holds that:
$$ \max_y \min_x f(x, y) \leq \min_x \max_y f(x, y)$$
This is known as weak duality. In certain circumstances, you have also have strong duality (also known as the saddle point property):
$$ \max_y \min_x f(x, y) = \min_x \max_y f(x, y)$$
When strong duality holds, solving $\max_y \min_x f(x, y)$ also solves $\min_x \max_y f(x, y)$.
Let me define the function $\mathcal{L}$ as:
$$ \mathcal{L}(\mathbf{b}, \lambda) = \sum_{i=1}^n (y - \mathbf{x}_i \cdot \mathbf{b})^2 + \lambda \left( \sum_{j=1}^p b_j^2 - t \right) $$
The Ridge regression problem subject to hard constraints is:
$$ \min_\mathbf{b} \max_{\lambda \geq 0} \mathcal{L}(\mathbf{b}, \lambda) $$
You pick $\mathbf{b}$ to minimize the objective, cognizant that after you pick $\mathbf{b}$, your opponent will set $\lambda$ to infinity if you chose $\mathbf{b}$ such that the constraint was violated (in this case $\sum_{j=1}^p b_j^2 > t$).
If strong duality holds (which it does here because it's a convex optimization problem where Slater's condition is satisfied for $t>0$), you then achieve the same result by reversing the order:
$$ \max_{\lambda \geq 0} \min_\mathbf{b} \mathcal{L}(\mathbf{b}, \lambda) $$
In this dual problem, your opponent chooses $\lambda$ first! You then choose $\mathbf{b}$ to minimize the objective, already knowing your opponent's choice of $\lambda$. The $\min_\mathbf{b} \mathcal{L}(\mathbf{b}, \lambda)$ part (taking $\lambda$ as given) is equivalent to the 2nd form of your Ridge Regression problem.
As you can see, this isn't a result particular to Ridge Regression. It is a broader concept.
I started this post following an exposition of Rockafellar.
Rockafellar, R.T., Convex Analysis
You might also examine lectures 7 and lecture 8 from Prof. Stephen Boyd's course on convex optimization.
They are not equivalent.
For a constrained minimization problem
$$\min_{\mathbf b} \sum_{i=1}^n (y - \mathbf{x}'_i \cdot \mathbf{b})^2\\ s.t. \sum_{j=1}^p b_j^2 \leq t,\;\;\; \mathbf b = (b_1,...,b_p) \tag{1}$$
we solve by minimize over $\mathbf b$ the corresponding Lagrangean
$$\Lambda = \sum_{i=1}^n (y - \mathbf{x}'_i \cdot \mathbf{b})^2 + \lambda \left( \sum_{j=1}^p b_j^2 - t \right) \tag{2}$$
Here, $t$ is a bound given exogenously, $\lambda \geq 0$ is a Karush-Kuhn-Tucker non-negative multiplier, and both the beta vector and $\lambda$ are to be determined optimally through the minimization procedure given $t$.
Comparing $(2)$ and eq $(3.41)$ in the OP's post, it appears that the Ridge estimator can be obtained as the solution to
$$\min_{\mathbf b}\{\Lambda + \lambda t\} \tag{3}$$
Since in $(3)$ the function to be minimized appears to be the Lagrangean of the constrained minimization problem plus a term that does not involve $\mathbf b$, it would appear that indeed the two approaches are equivalent...
But this is not correct because in the Ridge regression we minimize over $\mathbf b$ given $\lambda >0$. But, in the lens of the constrained minimization problem, assuming $\lambda >0$ imposes the condition that the constraint is binding, i.e that
$$\sum_{j=1}^p (b^*_{j,ridge})^2 = t$$
The general constrained minimization problem allows for $\lambda = 0$ also, and essentially it is a formulation that includes as special cases the basic least-squares estimator ($\lambda ^*=0$) and the Ridge estimator ($\lambda^* >0$).
So the two formulation are not equivalent. Nevertheless, Matthew Gunn's post shows in another and very intuitive way how the two are very closely connected. But duality is not equivalence.
