TITLE: Why are additional constraint and penalty term equivalent in ridge regression?
QUESTION [31 upvotes]: Tikhonov regularization (or ridge regression) adds a constraint that $\|\beta\|^2$, the $L^2$-norm of the parameter vector, is not greater than a given value (say $c$). Equivalently, it may solve an unconstrained minimization of the least-squares penalty with $\alpha\|\beta\|^2$ added, where $\alpha$ is a constant (this is the Lagrangian form of the constrained problem). 

The above is from Wikipedia. Why is the unconstrained LS  with $\alpha\|\beta\|^2$ added to the cost equivalent to the LS problem with an additional constraint that $\|\beta\|^2 \leq c$?
What is the relation between $\alpha$ and $c$?
Thanks!

REPLY [2 votes]: You can do this directly if you want to. To solve the optimization problem
\begin{align}
\min_{\beta} ~&\Vert y-X\beta\Vert^2\\
\mathrm{s.t.}~&\Vert \beta\Vert^2\le c\ ,
\end{align}
as in the standard primal-dual procedure, first let
\begin{align}
g(\lambda)=&\inf_\beta\mathcal{L}(\beta,\lambda)\\
=&\inf_\beta\Vert y-X\beta\Vert^2+\lambda (\Vert \beta\Vert^2- c)\\
=& \Vert y-X(X^\mathrm{T}X+\lambda I)^{-1}X^\mathrm{T}y\Vert^2 + \lambda (\Vert(X^\mathrm{T}X+\lambda I)^{-1}X^\mathrm{T}y\Vert^2-c)\ ,
\end{align}
then solve $\max_{\lambda\ge 0} g(\lambda)$. You will find that
$$
\frac{\partial g}{\partial\lambda}=y^\mathrm{T}X(X^\mathrm{T}X+\lambda I)^{-2}X^\mathrm{T}y-c=0\iff c=\Vert\beta^*_{\mathrm{ridge}}(\lambda)\Vert^2\ .
$$
The matrix derivatives
\begin{align}
\frac{\partial AU(x)B}{\partial x} = & A\frac{\partial U(x)}{\partial x}B\\
\frac{\partial U(x)^{-1}}{\partial x} = &-U(x)^{-1} \frac{\partial U(x)}{\partial x}U(x)^{-1}
\end{align}
will be helpful.
Update:
By the way you can prove when $\lambda$ increases, $c$ doesn't increase. More generally, Let $L(x;\lambda)=f(x)+\lambda g(x)$, and $x_i^*=\mathrm{arg\,min}_xL(x;\lambda_i)\,(i=1,2)$. Suppose $\lambda_2>\lambda_1$ and $g(x_2^*)>g(x_1^*)$, we have
\begin{align}
&(\lambda_2-\lambda_1)(g(x_2^*)-g(x_1^*))>0\\
\Longrightarrow & \lambda_1g(x_1^*)+\lambda_2g(x_2^*)>\lambda_1g(x_2^*)+\lambda_2g(x_1^*)\\
\Longrightarrow & [f(x_1^*)+\lambda_1g(x_1^*)]+[f(x_2^*)+\lambda_2g(x_2^*)]>[f(x_2^*)+\lambda_1g(x_2^*)]+[f(x_1^*)+\lambda_2g(x_1^*)] \ge [f(x_1^*)+\lambda_1g(x_1^*)]+[f(x_2^*)+\lambda_2g(x_2^*)]
\end{align}
which is a contradiction, so $g(x^*)$ doesn't increase when $\lambda$ increases. In the context of OP's problem, $c$ doesn't increase when $\lambda$ increases.