# About the derivation of group Lasso

I've been reading the paper of group lasso, "Model selection and estimation in regression with grouped variables". http://www.stat.washington.edu/courses/stat527/s13/readings/yuanlin07.pdf

In page 53 of the above paper, I don't know how to obtain Eqn 2.3 and derive Eqn. 2.4 from Eqn. 2.2:

The following is my derivation, from Eqn. 2.2: $$-X_j^TY+X_j^TX_j\beta_j+\sum_{i\neq j}X_j^TX_i\beta_i + \frac{\lambda \beta_j \sqrt{p_j}}{\|\beta_j\|}=0$$ then, we have $$\left( X_j^TX_j+\frac{\lambda \sqrt{p_j}}{\|\beta_j\|} \right)\beta_j=X_j^T(Y-\sum_{i\neq j}X_i^T\beta_i)=S_j$$ So, $$\beta_j={\left( X_j^TX_j+\frac{\lambda\sqrt{p_j}}{\|\beta_j\|} \right)}^{-1}S_j$$ The result above has a little difference from Eqn. 2.4 in the paper. Could you please indicate what's wrong with my derivation? In addition, in another document http://statweb.stanford.edu/~tibs/ftp/sparse-grlasso.pdf, the result of following is also different.

UPDATED: Yes! The first problem is solved. $$S_j=X_j^T(Y-X\beta_{-j})=X_j^Tr_j$$ then, $$\|\beta_j\|=\|S_j\|{\left( 1+\frac{\lambda\sqrt{p_j}}{\|\beta_j\|} \right)}^{-1}$$ Then, we could get $$\|\beta_j\|=\|S_j\|-\lambda\sqrt{p_j}$$ then re-substitute into the formulation above above, then we get $$\beta_j={\left( 1-\frac{\lambda\sqrt{p_j}}{\|S_j\|} \right)}S_j$$ But, where is the $+$ in the above formula?

1. $\frac{\beta_j}{\|\beta_j\|}$ when $\beta_j \ne 0$.
2. any vector with $\| \beta_j \| \le 1$ when $beta_j = 0$.