Conditional IID bootstrap CLT for an empirical law

Put $Y_i=a^T(X_i-\mu )$ and ${\bar{Y}}_n=n^{-1}{\sum }_i{Y}_i$. Its variance is ${\sigma }^2$. By [F2], the empirical variance $s_n^2=\frac{1}{n}{\sum }_{i=1}^n(Y_i-{\bar{Y}}_n{)}^2=a^T{\widehat{\mathrm{\Sigma }}}_{n}a$ converges in probability to ${\sigma }^2>0$, while ${\bar{Y}}_n\to 0$ in probability and $n^{-1}{\sum }_i{Y}_i^2\to {\sigma }^2$ in probability.

For $t>0$, define the unnormalized empirical tail quantity $T_n(t)=\frac{1}{n}{\sum }_{i=1}^n(Y_i-{\bar{Y}}_n{)}^{2}1\{\mathrm{\mid }{Y}_i-{\bar{Y}}_n\mathrm{\mid }>t\sqrt{n}\}\mathrm{.}$ On $\mathrm{\mid }{\bar{Y}}_n\mathrm{\mid }\le t\sqrt{n}\mathrm{/}2$, the indicator implies $\mathrm{\mid }{Y}_i\mathrm{\mid }>t\sqrt{n}\mathrm{/}2$, and $T_n(t)\le 2\frac{1}{n}{\sum }_i{Y}_i^{2}1\{\mathrm{\mid }{Y}_i\mathrm{\mid }>t\sqrt{n}\mathrm{/}2\}+2{\bar{Y}}_n^2\frac{1}{n}{\sum }_{i}1\{\mathrm{\mid }{Y}_i\mathrm{\mid }>t\sqrt{n}\mathrm{/}2\}\mathrm{.}$ The expectation of the first term tends to zero by [F7], so [F8] makes it converge to zero in probability. The second empirical tail fraction is at most $4(t^{2}n{)}^{-1}(n^{-1}{\sum }_i{Y}_i^2)$; hence its product with $2{\bar{Y}}_n^2$ tends to zero in probability by [F2]. The complement event has probability tending to zero by [F2], so $T_n(t)\to 0$ in probability for every $t>0$.

Conditionally on the data, the row variables $Z_{n\mathrm{\ell }}^{\ast }=n^{-1\mathrm{/}2}{a}^T(X_{\mathrm{\ell }}^{\ast }-{\bar{X}}_n)$, $1\le \mathrm{\ell }\le n$, are independent and centered by [F1]. They are finite conditional rows of the type in [F3], with total variance $s_n^2$. By [F4], when $s_n>0$ their normalized Lindeberg quantity at threshold $\epsilon >0$ is $s_n^{-2}{T}_n(\epsilon s_n)$. On $s_n\ge \sigma \mathrm{/}2$, this is at most $4{\sigma }^{-2}{T}_n(\epsilon \sigma \mathrm{/}2)$, which tends to zero in probability by step 1.2; also $s_n^2\to {\sigma }^2$ in probability by [F2].

Take any subsequence, indexed $(n_k{)}_{k\ge 1}$. Recursively, after $k_0=0$, let $k_j$ be the least integer $k>k_{j-1}$ for which $\mathbb{P}(\mathrm{\mid }{s}_{n_k}^2-{\sigma }^2\mathrm{\mid }>1\mathrm{/}j)+{\sum }_{r=1}^j\mathbb{P}(T_{n_k}(1\mathrm{/}r)>1\mathrm{/}j)<2^{-j}$. For fixed $j$, each of these finitely many probabilities tends to zero as $k\to \mathrm{\infty }$ by [F2] and step 1.2, so the qualifying set is nonempty; [F12] gives its least index.

Let $E_j$ be the union of the variance-deviation event and the $j$ tail events at index $k_j$. By [F9], $\mathbb{P}(E_j)<2^{-j}$ and $\mathbb{P}({\bigcup }_{j\ge m}{E}_j)\le {\sum }_{j\ge m}{2}^{-j}=2^{1-m}$. The event that infinitely many $E_j$ occur is contained in every such tail union, so its probability is zero. Off this null event, $s_{n_{k_j}}^2\to {\sigma }^2$ and, for every fixed $r\ge 1$, $T_{n_{k_j}}(1\mathrm{/}r)\to 0$. For any $t>0$, choose $r$ with $1\mathrm{/}r<t$ by [F13]; since $T_n$ is nonincreasing, $0\le T_{n_{k_j}}(t)\le T_{n_{k_j}}(1\mathrm{/}r)\to 0$.

Along almost every selected data path, $s_{n_{k_j}}\to \sigma >0$, and for every $\epsilon >0$ the Lindeberg quantity is eventually bounded by $4{\sigma }^{-2}{T}_{n_{k_j}}(\epsilon \sigma \mathrm{/}2)\to 0$. Thus the deterministic conditional rows satisfy [F5], giving ${\sum }_{\mathrm{\ell }=1}^{n_{k_j}}{Z}_{n_{k_j},\mathrm{\ell }}^{\ast }\mathrm{/}{s}_{n_{k_j}}\Rightarrow N(0,1)$. Every starting subsequence has a further subsequence with this almost-sure conditional convergence.

Along each data sequence from step 4.1, put $m_j=n_{k_j}$ and let $H_j$ be the conditional CDF of ${\sum }_{\mathrm{\ell }=1}^{m_j}{Z}_{m_j,\mathrm{\ell }}^{\ast }\mathrm{/}{s}_{m_j}$, discarding finitely many rows before $s_{m_j}>0$. By [F6], $H_j(t)\to \mathrm{\Phi }(t)$ for every real $t$. Fix $x\in \mathbb{R}$ and $h>0$. Since $s_{m_j}\to \sigma >0$, eventually $x\mathrm{/}\sigma -h\le x\mathrm{/}{s}_{m_j}\le x\mathrm{/}\sigma +h$, so monotonicity gives $H_j(x\mathrm{/}\sigma -h)\le F_{m_j}^{\ast }(x)=H_j(x\mathrm{/}{s}_{m_j})\le H_j(x\mathrm{/}\sigma +h)$. Letting $j\to \mathrm{\infty }$ and then $h\downarrow 0$, continuity in [F10] yields $F_{m_j}^{\ast }(x)\to \mathrm{\Phi }(x\mathrm{/}\sigma )={\mathrm{\Phi }}_{\sigma }(x)$. To make this uniform, fix $\eta >0$, choose $R$ with ${\mathrm{\Phi }}_{\sigma }(-R)<\eta$ and $1-{\mathrm{\Phi }}_{\sigma }(R)<\eta$, and partition $[-R,R]$ into finitely many intervals on which ${\mathrm{\Phi }}_{\sigma }$ varies by less than $\eta$, using [F11]. Monotonicity of each conditional CDF bounds its error between grid points by the endpoint errors plus $\eta$; outside $[-R,R]$ the two tail probabilities add at most $2\eta$ plus endpoint errors. The maximum grid error tends to zero, and then $\eta \downarrow 0$ proves almost-sure uniform CDF convergence along the selected subsequence. Since $D_n$ is measurable by [F14], this is almost-sure convergence of real random variables.

If $D_n$ did not converge to zero in probability, there would be $\delta ,\epsilon >0$ and a subsequence on which $\mathbb{P}(D_n>\delta )\ge \epsilon$. Applying the subsequence construction in step 2.2 and its consequences in steps 3.1–5.1 gives a further subsequence with $D_n\to 0$ almost surely by step 5.1. By [F15], this is convergence in probability, contradicting the lower bound.