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Can someone please explain me how the optimization of 6.12 leads to 6.14 and the optimization of 6.13 leads to 6.15?

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For the first equation, it's the result of zero gradient; $$ \begin{aligned} S &= \sum_{j=1}^p (y_j-\beta_j)^2 +\lambda\sum_{j=1}^p\beta_j^2\\ \end{aligned} $$ at extrema, $$ \begin{aligned} \frac{\partial S}{\partial \beta_j} &=0\\ -2(y_j -\beta_j) +2\lambda\beta_j &= 0\\ \beta_j &= \frac{y_j}{1+\lambda}. \end{aligned} $$

I think you should be able to derive the other expression using the same technique shown above and use the fact that $$ \vert \beta_j \vert = \begin{cases} \beta_j \ \text{if} \ \beta_j > 0\\ -\beta_j \ \text{if} \ \beta_j < 0\end{cases}. $$

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The main technique has two steps: 1) take derivative w.r.t $\beta_j$ and 2) set it to $0$. I will outline the steps to help go from 6.12 to 6.14 and leave the other one.

Step 1): Combining the summation and opening the quadratic expression of 6.12 gives $$ \sum_{j=1}^p y_j^2 - 2y_j \times \beta_j + (1 + \lambda) \beta_j^2.$$ Now we take partial derivative of this entire summation w.r.t $\beta_j$ and get $$ \frac{\partial}{\partial \beta_j} = -2 y_j + 2(1+\lambda) \beta_j. $$ Note the other $j'\neq j$ do not appear since they do no affect $\beta_j$.

Step 2): We set the partial derivative to $0$ to find $\hat{\beta}_j^R$. Normally, we need to check the second order derivative (make sure it is positive) so that we know have found a minimizer. But we don't have to here because the quadratic form in $\beta_j$ is convex in $\beta_j$.

Barring some additional careful checking of the conditions on $y_j$, the Lasso regression (from 6.13 to 6.15) is very similar.

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