properties of the conventional sequential confidence set estimation procedure with less computation time and only some of the data. We run each procedure independently, and so we first describe below an individual estimation procedure.

At stage $n$ of procedure $j$, let notation $C_{jn} = \{(y_{ji},{\bm{{x}}_{ji}}), i=1,...,n\}$ be the data set with the observations recruited from the original data set—say, $\mathcal D$. Suppose that ${\hat{{\bm{{\beta}}}}_{jn}}$ is a conventional least square estimate (LSE) of ${\bm{{\beta}}_{0}}$, based on $C_{jn}$, of model ; additionally, let $$\hat\sigma^2_{jn}=\frac{1}{n-p}\sum_{i=1}^n (y_{ji}-{\bm{{x}}_{ji}^{\top}} {\hat{{\bm{{\beta}}}}_{jn}})^2$$ be an estimate of $\sigma^2$. With the notations defined above, we now describe the $j$th sequential estimation procedure as follows.

Let ${\hat{{\bm{{\beta}}}}_{jn_0}}$ be an LSE based on the initial data set $C_{jn_0} =  \{(y_{ji},{\bm{{x}}_{ji}}), i=1,\ldots,n_0\}$ of size $n_0$. Let $\mu_{jn}=\lambda_{min}[({\bm{{X}}_{jn}}{\bm{{X}}_{jn}^{\top}})/n]$; ${\bm{{X}}_{jn}}$ is a design matrix based on a set of covariate vectors, $\{{\bm{{x}}_{j1}},...,{\bm{{x}}_{jn}}\}$, of the observations in $C_{jn}$. (Notation $\lambda_{min}(A)$ denotes the minimum eigenvalue of matrix $A$.) Let $a^2$ be the $1-\alpha$ quantile of the chi-square distribution with degree of freedom $p$ and $$\sum_{j=1}^M\tilde{a}_j^2=a^2,$$ where $\tilde{a}_j>0$. For a given $d >0$, define a stopping time $$\begin{aligned}
\label{Ndef}
    N_j=N_{jd}\equiv\inf\left\{n\geq n_0: (\hat\sigma_{jn}^2+\frac{1}{n})\leq \frac{d^2n}{\tilde{a}_j^2\mu_{jn}}\right\}.\end{aligned}$$ Hence, $N_j$ is the smallest number of observations, such that the inequality in is fulfilled. 