Algorithm 2: Trans-GLM

$$\begin{array}{cc}\hfill & \mathrm{Input}:\text{target data}({\mathit{X}}^{(0)},{\mathit{y}}^{(0)}),\text{all source data}\{({\mathit{X}}^{(k)},{\mathit{y}}^{(k)}){\}}_{k=1}^K,\text{a constant}{C}_0>0,\hfill \\ \hfill & \text{penalty parameters}\{\{{\lambda }^{(k)[r]}{\}}_{k=0}^K{\}}_{r=1}^3\hfill \\ \hfill & \mathrm{Output}:\text{the estimated coefficient vector}\mathit{\beta },\text{and the determind transferring set}𝒜\hfill \\ \hfill & 1\mathbf{Transferable}\mathbf{source}\mathrm{detection}:\text{Randomly divide}({\mathit{X}}^{(0)},{\mathit{y}}^{(0)})\text{into three sets of equal}\hfill \\ \hfill & \text{size as}\{({\mathit{X}}^{(0)[i]},{\mathit{y}}^{(0)[i]}){\}}_{i=1}^3\hfill \\ \hfill & 2\mathbf{for}r=1\mathbf{to}3\mathbf{do}\hfill \\ \hfill & 3{\mathit{\beta }}^{(0)[r]}\leftarrow \text{fit the Lasso on}\{({\mathit{X}}^{(0)[i]},{\mathit{y}}^{(0)[i]}){\}}_{i=1}^3\setminus ({\mathit{X}}^{(0)[r]},{\mathit{y}}^{(0)[r]})\text{with penalty parameter}\hfill \\ \hfill & {\lambda }^{(0)[r]}\hfill \\ \hfill & 4{\mathit{\beta }}^{(k)[r]}\leftarrow \text{run step}1\text{in Algorithm}1\text{with}\hfill \\ \hfill & (\{({\mathit{X}}^{(0)[i]},{\mathit{y}}^{(0)[i]}){\}}_{i=1}^3\setminus ({\mathit{X}}^{(0)[r]},{\mathit{y}}^{(0)[r]}))\cup ({\mathit{X}}^{(k)},{\mathit{y}}^{(k)})\text{and penalty parameter}{\lambda }^{(k)[r]}\text{for all}\hfill \\ \hfill & k\ne 0\hfill \\ \hfill & 5\text{Calculate the loss function}{\widehat{L}}_0^{[r]}({\mathit{\beta }}^{(k)[r]})\text{on}({\mathit{X}}^{(0)[r]},{\mathit{y}}^{(0)[r]})\text{for}k=1,\dots ,K\hfill \\ \hfill & 6\mathbf{end}\hfill \\ \hfill & 7{\widehat{L}}_0^{(k)}\leftarrow \sum _{r=1}^3{\widehat{L}}_0^{[r]}({\mathit{\beta }}^{(k)[r]})∕3,{\widehat{L}}_0^{(0)}\leftarrow \sum _{r=1}^3{\widehat{L}}_0^{[r]}({\mathit{\beta }}^{(0)[r]})∕3,\widehat{\sigma }=\sqrt{\sum _{r=1}^3({\widehat{L}}_0^{[r]}({\mathit{\beta }}^{(0)[r]})-{\widehat{L}}_0^{(0)}{)}^2∕2}\hfill \\ \hfill & 8𝒜\leftarrow \{k\ne 0:{\widehat{L}}_0^{(k)}-{\widehat{L}}_0^{(0)}\le C_0(\widehat{\sigma }\vee 0.01)\}\hfill \\ \hfill & 9𝒜-\mathbf{Trans}-\mathrm{GLM}:\mathit{\beta }\leftarrow \text{run Algorithm}1\text{using}\{({\mathit{X}}^{(k)},{\mathit{y}}^{(k)})\}k\in \{0\}{\cup }^{𝒜}\hfill \\ \hfill & 10\text{Output}\mathit{\beta }\text{and}𝒜\hfill \end{array}$$