Difference Between LASSO and $ {L}_{1} $ Norm Minimization Problems

277 Views Asked by At

Let $\mathbf{x} \in \mathbb{R}^N$ and $\Phi \in \mathbb{R}^{M \times N}$. Let $\mathbf{y} \in \mathbb{R}^{M}$, where $\mathbf{y} = \Phi \mathbf{x} + \mathbf{w}$ and $\mathbf{w}$ is additive white Gaussian noise.

The LASSO problem can be written as:

\begin{align} \min_{\mathbf{x}} || \mathbf{y} - \Phi \mathbf{x} ||_2^2 + \lambda ||\mathbf{x}||_1 \quad \mbox{subject to} \quad \lambda > 0. \end{align}

The $\ell_1$-norm minimization problem can be written as: \begin{align} \min_{\mathbf{x}} || \mathbf{y} - \Phi \mathbf{x} ||_2^2 \quad \mbox{subject to} \quad ||\mathbf{x}||_1 \le \tau. \end{align}

Are the solutions for the above two convex optimizaton problems the same? If yes, under what conditions are they same.

1

There are 1 best solutions below

0
On BEST ANSWER

Yes. The solution for both of the problems are the same. Following is a more general proof:
Let $f$ and $g$ be convex functions, $$\ w_1^*= arg\underset{x}min \ f(x)+\lambda g(x) \quad \mbox{subject to} \quad \lambda > 0 $$ $$\ w_2^*= arg\underset{x}min \ f(x) \quad \mbox{subject to } \ g(x) \leq \tau $$ Writing Lagrangian for the second problem, $$\ \mathcal{L}(x, \lambda) = f(x)+\lambda_L\left(g(x) - \tau \right) \ \text{where } \lambda_L \text{ is the multiplier} $$ From FONC for the second problem, $$\ \nabla \mathcal{L}(w_2^*, \lambda^*) = \nabla f(w_2^*) +\lambda^*\left(\nabla g(w_2^*)\right) = 0 \ \text{where } \lambda^* = 0 \text{ if } g(w_2^*) < \tau $$ For the first problem FONC gives, $$\ \nabla f(w_1^*)+\lambda\left(\nabla g(w_1^*) \right)=0 $$ Since $f$ and $g$ are convex, we can conclude that both solutions are the same for $$\ \tau=g(w_1^*) \text{ and corresponding multiplier } \lambda^*=\lambda $$