we will calculate the new estimates, ${\hat{{\bm{{\beta}}}}_{jk}}$ and $\hat\sigma^2_{jk}$, based on $C_{j{k}}$. We repeat this sequential recruiting process until the inequality in is satisfied.

Let $N^*=\sum_{j=1}^M N_j$ and $\rho_j=N_j/N^*$, with $j=1, \ldots, M$. When all $M$ procedures are stopped, let $$\begin{aligned}
    &{\hat{{\bm{{\beta}}}}} \equiv {\hat{{\bm{{\beta}}}}_{N^*}}=\sum_{j=1}^M \rho_j{\hat{{\bm{{\beta}}}}_{jN_j}}\nonumber\end{aligned}$$ be a weighted average of $\{{\hat{{\bm{{\beta}}}}_{jN_j}}:j=1\ldots, M\}$, where $\sum_{j=1}^M \rho_j=1$. By definition, we know that $N_j$ is a function of $d$ and $\tilde a$, which are the same for all procedures in our current setup. Hence, $N_j$ values are random variables with the same mean (see Appendix B, for the proof of Theorem \[Th1\]). Thus, the actual number of observations used in each of the various procedures is different. Hence, we take into account the number of real observations used in each procedure, and we define ${\hat{{\bm{{\beta}}}}}$ as the weighted average of estimates obtained from each procedure with weights equal to $\rho_j=N_j/N^*$.

Then, based on the total number of observations $N^*$, we can define a confidence set for ${\bm{{\beta}}_{0}}$ as follows: $$\begin{aligned}
\label{FRn}
    R_{N^*}\!\! =\left\{{\bm{{z}}}\in R^p: ({\bm{{z}}} - {\hat{{\bm{{\beta}}}}})^\top \left\{\sum_{j=1}^M \rho_j^2({\bm{{X}}_{jN_j}}{\bm{{X}}_{jN_j}^{\top}})^{-1}\right\}^{-1}({\bm{{z}}} - {\hat{{\bm{{\beta}}}}})\leq \frac{N^*d^2}{\mu_{N^*}}\right\},\end{aligned}$$ where $\mu_{N^*}=\sum_{j=1}^M \rho_j\mu_{jN_j}$, $\mu_{jN_j}=\lambda_{min}[({\bm{{X}}_{jN_j}}{\bm{{X}}_{jN_j}^{\top}})/N_j]$, $j=1,...,M$. We show that $R_{N^*}$ has the desired coverage probability and estimation precision. (Detailed arguments are provided 