2006/01/31 by Tiefeng Jiang · 4 citations
Mathematics · #Advanced Algebra and Geometry #Advanced Combinatorial Mathematics #Random Matrices and Applications #math.PR #msc:15A52 #msc:60B10 #msc:60B15 #msc:60F05 #msc:60F99 #msc:62H10
paper · pdf · doi:10.1214/009117906000000205
published as Annals of Probability 2006, Vol. 34, No. 4, 1497-1529 · Published at http://dx.doi.org/10.1214/009117906000000205 in the Annals of Probability (http://www.imstat.org/aop/) by the Institute of Mathematical Statistics (http://www.imstat.org)
openalex publication_date 2006/07/01 · arxiv created 2006/09/25 · arxiv updated 2009/12/01 · openalex created_date 2025/10/10 · openalex updated_date 2026/07/28
We solve an open problem of Diaconis that asks what are the largest orders of pn and qn such that Zn, the pn×qn upper left block of a random matrix Γn which is uniformly distributed on the orthogonal group O(n), can be approximated by independent standard normals? This problem is solved by two different approximation methods. First, we show that the variation distance between the joint distribution of entries of Zn and that of pnqn independent standard normals goes to zero provided pn=o(√(n) ) and qn=o(√(n) ). We also show that the above variation distance does not go to zero if pn=[x√(n) ] and qn=[y√(n) ] for any positive numbers x and y. This says that the largest orders of pn and qn are o(n1/2) in the sense of the above approximation. Second, suppose Γn=(γij)n×n is generated by performing the Gram–Schmidt algorithm on the columns of Yn=(yij)n×n, where yij;1≤i,j≤n are i.i.d. standard normals. We show that ε n(m):=max1≤ i≤ n,1≤ j≤ m|√(n)⋅γij-yij| goes to zero in probability as long as m=mn=o(n/logn). We also prove that ε n(mn)→ 2√(α) in probability when mn=[nα/logn] for any α>0. This says that mn=o(n/logn) is the largest order such that the entries of the first mn columns of Γn can be approximated simultaneously by independent standard normals.