Deterministic equivalent for the bounded complex sample-covariance resolvent

Fix $n,z$, suppress the index $n$, write $N=N_n$, $s_j=T^{1\mathrm{/}2}{X}_{\cdot j}\mathrm{/}\sqrt{N}$, $R=(A-zI{)}^{-1}$, $R_j=(A-s_j{s}_j^{\ast }-zI{)}^{-1}$, and ${\beta }_j=(1+s_j^{\ast }{R}_j{s}_j{)}^{-1}$. The companion matrix $\tilde{A}=N^{-1}{X}^{\ast }TX$ is positive semidefinite by [F4], so $q=N^{-1}\mathbb{E}\mathrm{tr}(\tilde{A}-zI{)}^{-1}$ is the Stieltjes transform of an expected probability measure on $[0,\mathrm{\infty })$ by [F5] and has positive imaginary part. If $q=\int (\lambda -z{)}^{-1}dG(\lambda )$, then for every $t\ge 0$, $\mathrm{Im}\{z(1+tq)\}=\eta +t\int \frac{\lambda \eta }{\mathrm{\mid }\lambda -z{\mathrm{\mid }}^2}dG(\lambda )\ge \eta ,$ so $C:=(I+qT{)}^{-1}$ exists and $\mathrm{\parallel }C{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le \mathrm{\mid }z\mathrm{\mid }\mathrm{/}\eta$. The same spectral calculation applied to $A-s_j{s}_j^{\ast }$ gives $\mathrm{\mid }{\beta }_j\mathrm{\mid }\le \mathrm{\mid }z\mathrm{\mid }\mathrm{/}\eta$.

Set $b_j=(1+N^{-1}\mathrm{tr}(T{R}_j){)}^{-1}$ and ${\xi }_j=s_j^{\ast }{R}_j{s}_j-N^{-1}\mathrm{tr}(T{R}_j)$. The trace in $b_j$ is a Stieltjes transform with nonnegative weights: in an eigenbasis of $A-s_j{s}_j^{\ast }$ its weights are $⟨v_k,T{v}_k⟩\mathrm{/}N\ge 0$. The inequality in step 1.1 therefore also gives $\mathrm{\mid }{b}_j\mathrm{\mid }\le \mathrm{\mid }z\mathrm{\mid }\mathrm{/}\eta$. Conditional on all columns except $j$, [F1] and $\mathrm{\parallel }{T}^{1\mathrm{/}2}{R}_j{T}^{1\mathrm{/}2}{\mathrm{\parallel }}_F^2\le n{M}^2\mathrm{/}{\eta }^2$ give ${\mathbb{E}}_j\mathrm{\mid }{\xi }_j{\mathrm{\mid }}^2\le C_{1}n\mathrm{/}{N}^2=O(N^{-1})$, where ${\mathbb{E}}_j$ integrates column $j$. For any deterministic unit $x$, set ${\alpha }_j=x^{\ast }C{s}_j{s}_j^{\ast }{R}_{j}x-N^{-1}{x}^{\ast }CT{R}_{j}x\mathrm{.}$ This is a centered quadratic form in column $j$. Its coefficient matrix is $N^{-1}(T^{1\mathrm{/}2}{R}_{j}x)(x^{\ast }C{T}^{1\mathrm{/}2}),$ so it has rank at most one and Frobenius norm at most $M\mathrm{\parallel }C{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\mathrm{/}(N\eta )$. Thus [F1] gives ${\mathbb{E}}_j\mathrm{\mid }{\alpha }_j{\mathrm{\mid }}^2=O(N^{-2})$. Also ${\mathbb{E}}_j{\alpha }_j=0$ and ${\beta }_j-b_j=-{\beta }_j{b}_j{\xi }_j$.

For all sufficiently large $n$ we have $N=N_n\ge 2$. Put $Y=[s_2\cdots s_N]$, $D=Y^{\ast }Y-z{I}_{N-1}$, and $C_{-}=Y{Y}^{\ast }-z{I}_n=A-s_1{s}_1^{\ast }-z{I}_n$. The matrix $\tilde{A}-zI$ has block form $\left(\begin{array}{cc}{s}_1^{\ast }{s}_1-z& s_1^{\ast }Y\\ Y^{\ast }{s}_1& D\end{array}\right)\mathrm{.}$ For a block matrix $\left(\begin{array}{cc}a& b^{\ast }\\ b& D\end{array}\right)$ with invertible $D$, solving its product with $(u,v{)}^T=(1,0{)}^T$ gives $v=-D^{-1}bu$ and $(a-b^{\ast }{D}^{-1}b)u=1$. Hence its inverse's upper-left entry is $(a-b^{\ast }{D}^{-1}b{)}^{-1}$. Here $(C_{-})Y=YD$, so $(C_{-})(I-Y{D}^{-1}{Y}^{\ast })=C_{-}-Y{Y}^{\ast }=-zI$ and $I-Y{D}^{-1}{Y}^{\ast }=-z{C}_{-}^{-1}=-z{R}_1$. The Schur complement is therefore $s_1^{\ast }(I-Y{D}^{-1}{Y}^{\ast })s_1-z=-z(1+s_1^{\ast }{R}_1{s}_1)$, proving $(\tilde{R}{)}_{11}=-{\beta }_1\mathrm{/}z$.

By exchangeability of the columns and step 2.2, $zq=-N^{-1}{\sum }_{j=1}^N\mathbb{E}{\beta }_j$. The identity $R^{-1}={\sum }_j{s}_j{s}_j^{\ast }-zI$ gives ${\sum }_j{s}_j{s}_j^{\ast }R=I+zR$. The rank-one formula [F2] gives $s_j{s}_j^{\ast }R={\beta }_j{s}_j{s}_j^{\ast }{R}_j$. Substituting these identities and the formula for $zq$ yields $\mathbb{E}R+\frac{1}{z}C=\frac{1}{z}{\sum }_{j=1}^N\mathbb{E}\left[{\beta }_j(C{s}_j{s}_j^{\ast }{R}_j-N^{-1}CT\mathbb{E}R)\right]\mathrm{.}$ Testing against $x$ and adding then subtracting $N^{-1}{x}^{\ast }CT{R}_{j}x$ and $N^{-1}{x}^{\ast }CTRx$ decomposes the right side into three terms ${\delta }_1+{\delta }_2+{\delta }_3$.

For the first term, ${\delta }_1=(N\mathrm{/}z)\mathbb{E}({\beta }_1{\alpha }_1)$. Since $b_1$ is independent of column 1 and ${\mathbb{E}}_1{\alpha }_1=0$, we have $\mathbb{E}(b_1{\alpha }_1)=0$. Therefore step 2.1, Cauchy–Schwarz, and the uniform bounds on ${\beta }_1,b_1$ imply $\mathrm{\mid }{\delta }_1\mathrm{\mid }\le C_{2}N(\mathbb{E}\mathrm{\mid }{\xi }_1{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}(\mathbb{E}\mathrm{\mid }{\alpha }_1{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}=O(N^{-1\mathrm{/}2})\mathrm{.}$

For the second term, $R_1-R={\beta }_1{R}_1{s}_1{s}_1^{\ast }{R}_1$. Conditional second moments of the two linear forms $x^{\ast }CT{R}_1{s}_1$ and $s_1^{\ast }{R}_{1}x$ are each $O(N^{-1})$, uniformly in the other columns. Cauchy–Schwarz and boundedness of ${\beta }_1$ yield $\mathrm{\mid }{\delta }_2\mathrm{\mid }=O(N^{-1})$.

The third term is ${\delta }_3=z^{-1}\mathbb{E}[{\beta }_1{x}^{\ast }CT(R-\mathbb{E}R)x]$. For any deterministic sequence of unit vectors $x=x_n$, put $u=(CT{)}^{\ast }x$ and $v=x$; their norms are uniformly bounded by step 1.1. For $Q(w)=w^{\ast }(R-\mathbb{E}R)w$, direct expansion gives $B_n:=u^{\ast }(R-\mathbb{E}R)v=\frac{1}{4}\{Q(u+v)-Q(u-v)-iQ(u+iv)+iQ(u-iv)\}\mathrm{.}$ Rescale each nonzero vector to unit length; zero vectors contribute zero. For each fixed $t>0$, [F3] and the union bound imply $\mathbb{P}(\mathrm{\mid }{B}_n\mathrm{\mid }>t)\to 0$. Also $\mathrm{\mid }{B}_n\mathrm{\mid }\le 2\mathrm{\parallel }u\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }\mathrm{/}\eta \le L$ for a fixed finite $L$. Thus $\mathbb{E}\mathrm{\mid }{B}_n{\mathrm{\mid }}^2\le t^2+L^2\mathbb{P}(\mathrm{\mid }{B}_n\mathrm{\mid }>t)$, so first letting $n\to \mathrm{\infty }$ and then $t\downarrow 0$ gives $\mathbb{E}\mathrm{\mid }{B}_n{\mathrm{\mid }}^2\to 0$. Boundedness of ${\beta }_1$ and Cauchy–Schwarz yield ${\delta }_3\to 0$. This holds for every deterministic unit-vector sequence, hence uniformly over unit $x$: otherwise a sequence of deterministic directions with $\mathrm{\mid }{\delta }_3\mathrm{\mid }$ bounded away from zero along a subsequence would contradict the conclusion just proved. No shrinking-threshold tail estimate is used.

The bounds in steps 4.1, 4.2, and 4.3 tend to zero uniformly over deterministic unit $x$. The tested resolvent identity in step 3.1 therefore proves the asserted supremum limit.