General-population sample-covariance law for bounded entries

Write $c_n=n\mathrm{/}{N}_n$, let ${\bar{F}}_n(B)=\mathbb{E}{F}_n(B)$, and let ${\bar{G}}_n$ be defined similarly for ${\tilde{A}}_n=N_n^{-1}{X}_n^{\ast }{T}_n{X}_n$. By [F14], each ordered eigenvalue is continuous in the matrix entries, so these setwise expectations are defined for Borel $B$. For disjoint Borel sets, countable additivity of the random ESDs followed by [F13] gives countable additivity of ${\bar{F}}_n$ and ${\bar{G}}_n$; their total masses are one. Independence and unit second moment give $\int td{\bar{F}}_n(t)=n^{-1}\mathrm{tr}{T}_n$ and $\int td{\bar{G}}_n(t)=N_n^{-1}\mathrm{tr}{T}_n$. Both first moments are uniformly bounded because $c_n\to c>0$ and $\mathrm{\parallel }{T}_n{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$ is uniformly bounded. By [F4], their tails above $R$ are at most the corresponding first moment divided by $R$, so both expected-measure sequences are tight on $[0,\mathrm{\infty })$. The half-line is Polish: it is a closed subset of the complete real line and $\mathbb{Q}\cap [0,\mathrm{\infty })$ is dense. By [A1] and [F5], every subsequence of ${\bar{F}}_n$ has a further weakly convergent subsequence.

For each fixed $z\in {\mathbb{C}}^{+}$, [F7] gives $\int (t-z{)}^{-1}d{F}_n(t)-\int (t-z{)}^{-1}d{\bar{F}}_n(t)\to 0$ almost surely. Since all $X_n$ are on the common probability space in the Statement, intersect these events over the countable dense set $\{a+ib:a,b\in \mathbb{Q},b>0\}$.

Fix $z\in {\mathbb{C}}^{+}$ and take any such subsequence with ${\bar{F}}_n\Rightarrow G$. Let $m_G(z)=\int (t-z{)}^{-1}dG(t)$ and $q_n(z)=\int (t-z{)}^{-1}d{\bar{G}}_n(t)$. By [F11], the two sample-covariance matrices have the same nonzero eigenvalues, so their transforms satisfy exactly $q_n=-(1-c_n)\mathrm{/}z+c_n\int (t-z{)}^{-1}d{\bar{F}}_n(t)$. Hence $q_n\to u:=-(1-c)\mathrm{/}z+c{m}_G(z)$. Moreover $\mathrm{Im}{m}_G(z)>0$. The first moments of ${\bar{G}}_n$ are bounded, so at least half of each ${\bar{G}}_n$ lies in some fixed interval $[0,L]$; consequently $\mathrm{Im}{q}_n(z)\ge \eta \mathrm{/}(2(L+\mathrm{\mid }z\mathrm{\mid }{)}^2)$ and $\mathrm{Im}u>0$.

Apply [F6] to an orthonormal eigenbasis of $T_n$. Its uniform quadratic-form error implies $\int (t-z{)}^{-1}d{\bar{F}}_n(t)=-\frac{1}{zn}\mathrm{tr}(I+q_n{T}_n{)}^{-1}+o(1)=-\frac{1}{z}\int \frac{1}{1+q_{n}t}d{H}_n(t)+o(1)\mathrm{.}$ Since $H_n\Rightarrow H$, all $H_n$ are supported in one compact interval, $q_n\to u$, and $\mathrm{Im}u>0$, the integrands converge uniformly on that interval to $-1\mathrm{/}[z(1+ut)]$. Thus $m_G(z)=\int [-z(1+ut){]}^{-1}dH(t)$. The relation for $u$ in step 2.1 gives $-z(1+ut)=t(1-c-cz{m}_G(z))-z$, so $m_G$ satisfies the fixed-point equation in the Statement. The argument applies to every $z\in {\mathbb{C}}^{+}$ and every weak subsequential limit $G$. By [F8], all such limits have the same transform; [F9] makes them equal. At least one such limit exists by step 1.1, so call it $F_{c,H}$. Tightness then implies ${\bar{F}}_n\Rightarrow F_{c,H}$.

On the same event, [F10] gives $\mathrm{\parallel }{A}_n{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le L_1$ eventually for a finite deterministic $L_1$, so the random measures $F_n$ are eventually supported on $[0,L_1]$ and are tight. The half-line is Polish by the metric facts in step 1.1 and [F15]. By [A1] and [F5], every subsequence has a weakly convergent further subsequence. For each dense-set $z$, its transform equals that of $F_{c,H}$ by steps 3.1 and 1.2, since the real and imaginary parts of $(t-z{)}^{-1}$ are bounded continuous and [F16] applies to each. For any probability law $\mu$, its transform satisfies $\mathrm{\mid }{m}_{\mu }(z)-m_{\mu }(w)\mathrm{\mid }\le \mathrm{\mid }z-w\mathrm{\mid }\mathrm{/}(\mathrm{Im}z\mathrm{Im}w)$, so equality on the dense set extends to all ${\mathbb{C}}^{+}$; [F9] identifies the subsequential limit as $F_{c,H}$. Every subsequence has the same further limit, and [F16] gives $F_n\Rightarrow F_{c,H}$ almost surely.