For a phenotype of SNP heritability 0.5, with effective number of independent markers (independent genomic segments), Meff, of ~ 60,000 and sample size, N, of 500,000, \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$R_{{\bf{b}},\widehat {\bf{b}}_{{\mathrm{SBLUP}}}}^2$$\end{document}Rb,b^SBLUP2 from summary statistics in an independent reference sample will be 91% of the value of \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$R_{{\bf{b}}_k,\widehat {\bf{b}}_{{\mathrm{BLUP}}_k}}^2$$\end{document}Rbk,b^BLUPk2 if individual-level data were available. Likewise, for a two-trait example where both traits have h2 = 0.5 and N = 500,000, the accuracy of the multi-trait SBLUP predictor will also be 91% of the accuracy of the multi-trait BLUP predictor.