A randomly weighted estimate of the population mean
Consider the model: \[ Y_{ij}= \beta_ 0+ \sigma_ i e_{ij}, \qquad j=1,\dots, N_ i, \quad i=1,\dots, n,\dots, \] where \(e_{ij}\), \(j=1,\dots, N_ i\), \(i=1,2,\dots\) are i.i.d. with mean zero and variance one. Put \[ \overline {Y}_ i= (Y_{i1}+ \cdots+ Y_{iN_ i} )/N_ i, \qquad S_ i^ 2= \sum_{j=1}^{N_ i} (Y_{ij}- \overline {Y}_ i )^ 2/ (N_ i- 1). \] An estimate \(\beta_ n\) of \(\beta_ 0\) is obtained by replacing \(\sigma^ 2_ i\) by \(S^ 2_ i\) in \[ \widehat {\beta}_ n= \sum_{i=1}^ n N_ i \sigma_ i^{-2} \overline {Y}_ i \biggl/ \sum_{i=1}^ n N_ i \sigma_ i^{-2}, \] thus resulting in \[ \beta_ n= \sum_{i=1}^ n N_ i S_ i^{-2} \overline {Y}_ i \biggl/ \sum_{i=1}^ n N_ i S_ i^{-2}. \] Note that \(\widehat {\beta}_ n\) is the Best Linear Unbiased Estimate of \(\beta_ 0\) when \(\sigma_ 1^ 2, \sigma^ 2_ 2, \dots\) are known. As we see, \(\beta_ n\) is a randomly-weighted average of \(\{\overline {Y}_ i\}\). Sections 2 and 3 consider the case where \(e_{ij}\sim N(0,1)\): Section 2 studies the consistency of \(\beta_ 0\), and in Section 3 we establish the normal approximation of the distribution of \(\beta_ n\), based on which a large-sample confidence interval of \(\beta_ 0\) can be constructed. The final section takes up the general (non-normal) case.
scientific article; zbMATH DE number 713758