Skip to main content

Question

There is no closed-form solution to (3.59), and hence an iterative method is needed to perform the optimization. Next, we will rewrite (3.59) into an equivalent quadratic programming (QP) problem, which can be plugged into a standard solver (e.g., quadprog in MATLAB).

(b) First, we rewrite θ\theta as a difference between two vectors with positive entries.

θ=θ+−θ−,(3.60)\theta = \theta^+ - \theta^-, \tag{3.60} θ+≥0,θ−≥0.(3.61)\theta^+ \ge 0, \quad \theta^- \ge 0. \tag{3.61}

The original optimization problem (3.59) can now be rewritten as

θ^=argminθ+,θ−12∥y−ΦT(θ+−θ−)∥2+λ∑i∣θi+−θi−∣,(3.62)\hat{\theta} = \text{argmin}_{\theta^+, \theta^-} \frac{1}{2} \|y - \Phi^T (\theta^+ - \theta^-)\|^2 + \lambda \sum_i |\theta_i^+ - \theta_i^-|, \tag{3.62} s.t. θ+≥0,θ−≥0.\text{s.t. } \theta^+ \ge 0, \quad \theta^- \ge 0.

Using a bit of optimization theory "magic", we can rewrite (3.62) as

θ^=argminθ+,θ−12∥y−ΦT(θ+−θ−)∥2+λ∑i(θi++θi−),(3.63)\hat{\theta} = \text{argmin}_{\theta^+, \theta^-} \frac{1}{2} \|y - \Phi^T (\theta^+ - \theta^-)\|^2 + \lambda \sum_i (\theta_i^+ + \theta_i^-), \tag{3.63} s.t. θ+≥0,θ−≥0.\text{s.t. } \theta^+ \ge 0, \quad \theta^- \ge 0.

Why is the optimization problem in (3.63) equivalent to that of (3.62)? Hint: at the optimum, what can we say about the values of the pair {θi+,θi−}\{\theta_i^+, \theta_i^-\}?