Matrix concentration
Posted: 2016-03-13 , Modified: 2016-03-13
Tags: random matrix
Posted: 2016-03-13 , Modified: 2016-03-13
Tags: random matrix
See wikipedia.
Theorem (Matrix Bernstein): Let \(\{X_k\}_{k=1}^n\) be a sequence of independent random \(d\times d\) matrices with \[ \E X_k = 0, \quad \la_{\max}(X_k)\le R\text{ a.s.}\] Then for all \(t\ge 0\), \[ \Pj \ba{\la_{\max}\pa{\sumo kn X_k}\ge t} \le de^{-\fc{t^2}{2\si^2+\fc 23 Rt}}\] where \(\si^2 = \ve{\sumo kn \E(X_k^2)}\).
(Note that requiring \(X_i\le R\) can be restrictive. If this doesn’t hold use the following instead.)
Reference: THE EXPECTED NORM OF A SUM OF INDEPENDENT RANDOM MATRICES: AN ELEMENTARY APPROACH
Theorem (Matrix Rosenthal): Let \(S_1,\ldots, S_n\) be random \(d_1\times d_2\) complex-valued matrices with \(\E S_i=O\), and \(Z:=\sumo in S_i\). Let \[\begin{align} v(X):&= \max\bc{\ve{\sum_i \E[S_iS_i^*]}, \ve{\sum_i \E [S_i^*S_i]}}\\ L:&= \pa{\E \max_i \ve{S_i}^2}^{\rc 2}\\ C(d_1,d_2):&= 4(1+2\ce{\ln(d_1+d_2)})\\ c:&=\rc4. \end{align}\]Then \[\sqrt{c v(Z)} + cL \le \pa{\E \ve{Z}^2}^{\rc 2} \le \sqrt{C(d) v(Z)} + C(d)L \]
Proof: pg. 9 of NB 13.