Let
\(C_{k,i}= \frac{S_{k,i}}{(k-1)\mu}\), hence, (
2) is equivalent to
$$ \lim_{n\rightarrow\infty}\frac{1}{D_{n}}\sum_{k=1}^{n} d_{k}\mathrm{I}\Biggl(\frac{\mu}{\beta V_{k}}\sum _{i=1}^{k}\log C_{k,i}\leq{x}\Biggr)= \Phi(x) \quad \mbox{a.s.} $$
(13)
So we only need to prove (
13), for a fixed
k,
\(1\leq k\leq n\) and
\(\forall\varepsilon>0\); we have
$$\lim_{k\rightarrow\infty}P\Biggl\{ \bigcup_{m=k}^{\infty} \biggl( \biggl\vert \frac {X_{i}}{m} \biggr\vert \geq\varepsilon\biggr) \Biggr\} =\lim_{k\rightarrow\infty}P\biggl\{ \biggl\vert \frac{X_{i}}{k} \biggr\vert \geq\varepsilon \biggr\} =\lim_{k\rightarrow\infty}P\bigl\{ | X_{1}|\geq \varepsilon k\bigr\} =0, $$
therefore, by Theorem 1.5.2 in [
14], we have
$$\frac{X_{i}}{k}\rightarrow0\quad \mbox{a.s. }k\rightarrow\infty, $$
on the unanimous establishment of
i.
By Lemma
2.1, for some
\(\frac{4}{3}< p<2 \), and enough large
k, we have
$$\begin{aligned} \sup_{1\leq i\leq k} \vert C_{k,i}-1 \vert \leq& \biggl\vert \frac{\sum_{j=1}^{k}(X_{j}-\mu)}{(k-1)\mu} \biggr\vert +\sup_{1\leq i\leq k} \biggl\vert \frac{X_{i}}{(k-1)\mu} \biggr\vert +\frac {1}{k-1} \\ \leq& \biggl\vert \frac{S_{k}-k\mu}{k^{\frac{1}{p}}}\cdot\frac{k^{\frac {1}{p}}}{(k-1)\mu} \biggr\vert \leq Ck^{\frac{1}{p}-1}, \end{aligned}$$
by
\(\log(1+x)=x+O(x^{2})\),
\(x\rightarrow0\), we get
$$\begin{aligned} &\Biggl\vert \frac{\mu}{\beta\delta_{k}\sqrt{(1\pm\varepsilon)k}}\sum_{i=1}^{k} \ln{C}_{k,i}-\frac{\mu}{\beta\delta_{k}\sqrt{(1\pm\varepsilon )k}}\sum_{i=1}^{k}(C_{k,i}-1) \Biggr\vert \\ &\quad \leq\frac{{C}\mu}{\beta\delta_{k}\sqrt{(1\pm\varepsilon)k}}\sum_{i=1}^{k}(C_{k,i}-1)^{2} \\ &\quad \leq\frac{C}{\sqrt{k}}k^{\frac{2}{p}-1}\rightarrow0\quad \mbox{a.s.}, k \rightarrow \infty, \end{aligned}$$
and then, for
\(\delta>0\) and every
ω, there exists
\(k_{0}=k_{0}(\omega,\delta,x)\); when
\(k>k_{0}\), we have
$$\begin{aligned} &\mathrm{I}\Biggl\{ \frac{\mu}{\beta\delta_{k}\sqrt{(1\pm\varepsilon)k}}\sum_{i=1}^{k}(C_{k,i}-1) \leq{x}-\delta\Biggr\} \\ &\quad \leq\mathrm{I}\Biggl\{ \frac{\mu}{\beta\delta_{k}\sqrt{(1\pm\varepsilon )k}}\sum _{i=1}^{k}\log{C}_{k,i}\leq{x}\Biggr\} \\ &\quad \leq\mathrm{I}\Biggl\{ \frac{\mu}{\beta\delta_{k}\sqrt{(1\pm\varepsilon )k}}\sum_{i=1}^{k}(C_{k,i}-1) \leq{x}+\delta\Biggr\} , \end{aligned}$$
(14)
under the condition
\(|X_{i}-\mu|\leq\sqrt{k}\),
\(1\leq i\leq k\), we have
$$ \mu\sum_{i=1}^{k}(C_{k,i}-1)=\sum _{i=1}^{k}\frac{S_{k,i}-(k-1)\mu }{k-1}=\sum _{i=1}^{k}Y_{i}=\sum _{i=1}^{k}\bar{Y}_{ki}={T}_{k,i}, $$
(15)
furthermore, by (
14) and (
15), for any given
\(0<\varepsilon<1\),
\(\delta >0\), when
\(k>k_{0}\), we obtain
$$\begin{aligned}& \mathrm{I}\Biggl(\frac{\mu}{\beta V_{k}}\sum_{i=1}^{k} \log C_{k,i}\leq x\Biggr) \\& \quad \leq\mathrm{I}\biggl(\frac{{T}_{k,i}}{\delta_{k}\beta\sqrt{k(1+\varepsilon )}}\leq x+ \delta\biggr) +\mathrm{I}\bigl(\bar{V}_{k}^{2}>(1+ \varepsilon)k\delta_{k}^{2}\bigr) \\& \qquad {}+\mathrm{I}\Biggl(\bigcup _{i=1}^{k}\bigl( \vert X_{i}-\mu \vert >\sqrt{k}\bigr)\Biggr),\quad x\geq0, \\& \mathrm{I}\Biggl(\frac{\mu}{\beta V_{k}}\sum_{i=1}^{k} \log C_{k,i}\leq x\Biggr) \\& \quad \leq\mathrm{I}\biggl(\frac{{T}_{k,i}}{\delta_{k}\beta\sqrt{k(1-\varepsilon )}} \leq x+ \delta\biggr) +\mathrm{I}\bigl(\bar{V}_{k}^{2}< (1- \varepsilon)k\delta_{k}^{2}\bigr) \\& \qquad {}+\mathrm{I}\Biggl(\bigcup _{i=1}^{k}\bigl( \vert X_{i}-\mu \vert >\sqrt{k}\bigr)\Biggr),\quad x< 0, \\& \mathrm{I}\Biggl(\frac{\mu}{\beta V_{k}}\sum_{i=1}^{k} \log C_{k,i}\leq x\Biggr) \\& \quad \geq\mathrm{I}\biggl(\frac{{T}_{k,i}}{\delta_{k}\beta\sqrt{k(1-\varepsilon )}}\leq x- \delta\biggr) -\mathrm{I}\bigl(\bar{V}_{k}^{2}< (1- \varepsilon)k\delta_{k}^{2}\bigr) \\& \qquad {}-\mathrm{I}\Biggl(\bigcup _{i=1}^{k}\bigl( \vert X_{i}-\mu \vert >\sqrt{k}\bigr)\Biggr),\quad x\geq0, \\& \mathrm{I}\Biggl(\frac{\mu}{\beta V_{k}}\sum_{i=1}^{k} \log C_{k,i}\leq x\Biggr) \\& \quad \geq\mathrm{I}\biggl(\frac{{T}_{k,i}}{\delta_{k}\beta\sqrt{k(1+\varepsilon )}}\leq x- \delta\biggr) -\mathrm{I}\bigl(\bar{V}_{k}^{2}>(1+ \varepsilon)k\delta_{k}^{2}\bigr) \\& \qquad {}-\mathrm{I}\Biggl(\bigcup _{i=1}^{k}\bigl( \vert X_{i}-\mu \vert >\sqrt{k}\bigr)\Biggr),\quad x< 0. \end{aligned}$$
Therefore, to prove (
13), for any
\(0<\varepsilon<1\),
\(\delta_{1}>0\), it suffices to prove
$$\begin{aligned}& \lim_{n\rightarrow\infty}\frac{1}{D_{n}}\sum_{k=1}^{n}d_{k} \mathrm{I} \biggl( \frac{{T}_{k,i}}{\beta\delta_{k}\sqrt{k}}\leq\sqrt{1\pm\varepsilon }x\pm \delta_{1} \biggr)=\Phi(\sqrt{1\pm\varepsilon}x\pm\delta_{1}) \quad \mbox{a.s.} , \end{aligned}$$
(16)
$$\begin{aligned}& \lim_{n\rightarrow\infty}\frac{1}{D_{n}}\sum_{k=1}^{n}d_{k} \mathrm{I}\Biggl(\bigcup_{i=1}^{k}\bigl( \vert X_{i}-\mu \vert >\sqrt{k}\bigr)\Biggr)=0\quad \mbox{a.s.} , \end{aligned}$$
(17)
$$\begin{aligned}& \lim_{n\rightarrow\infty}\frac{1}{D_{n}}\sum_{k=1}^{n}d_{k} \mathrm{I} \bigl(\bar{V}_{k}^{2}>(1+\varepsilon)k \delta_{k}^{2}\bigr)=0 \quad \mbox{a.s.} , \end{aligned}$$
(18)
$$\begin{aligned}& \lim_{n\rightarrow\infty}\frac{1}{D_{n}}\sum_{k=1}^{n}d_{k} \mathrm{I} \bigl(\bar{V}_{k}^{2}< (1-\varepsilon)k \delta_{k}^{2}\bigr)=0 \quad \mbox{a.s.} \end{aligned}$$
(19)
Firstly, we prove (
16), by
\(\mathrm{E}(Y^{2})<\infty\), we know
\(\lim_{x\rightarrow\infty}x^{2}P(|Y|>x)=0\), and by
\(\mathrm{E}(Y)=0\), it follows that
$$\begin{aligned} \bigl\vert \mathrm{E}({T}_{k,i}) \bigr\vert =& \Biggl\vert \mathrm{E}\Biggl(\sum_{i=1}^{k}\bar {Y}_{ki}\Biggr) \Biggr\vert = \vert k\mathrm{E} \bar{Y}_{k1} \vert \\ \leq& k \bigl\vert \mathrm{E}\bigl(Y\mathrm{I}\bigl( \vert Y \vert > \sqrt{k}\bigr)\bigr) \bigr\vert +k^{\frac{3}{2}}\mathrm{E}\bigl(\mathrm{I} \bigl( \vert Y \vert >\sqrt{k}\bigr)\bigr) \\ \leq&\sqrt{k}\mathrm{E}\bigl(Y^{2}\mathrm{I}\bigl( \vert Y \vert >\sqrt{k}\bigr)\bigr)+k^{\frac {3}{2}}P\bigl( \vert Y \vert >\sqrt{k} \bigr)=o(\sqrt{k}), \end{aligned}$$
so, combining with
\(\delta_{k}^{2}\rightarrow\mathrm{E}(Y^{2})<\infty\), for any
\(\alpha>0\), when
\(k\rightarrow\infty\), we have
$$\begin{aligned} &\mathrm{I} \biggl(\frac{{T}_{k,i}-E{T}_{k,i}}{\beta\delta _{k}\sqrt{k}}\leq\sqrt{1\pm\varepsilon} x\pm \delta_{1}-\alpha \biggr) \\ &\quad \leq \mathrm{I} \biggl(\frac{{T}_{k,i}}{\beta\delta_{k}\sqrt{k}} \leq\sqrt {1\pm\varepsilon}x\pm\delta_{1} \biggr) \\ &\quad \leq \mathrm{I} \biggl(\frac{{T}_{k,i}-E{T}_{k,i}}{\beta\delta_{k}\sqrt{k}}\leq \sqrt{1\pm\varepsilon}x\pm \delta_{1}+\alpha \biggr), \end{aligned}$$
thus, by (
4), we get
$$\begin{aligned}& \begin{aligned}[b] &\lim_{n\rightarrow\infty}\frac{1}{D_{n}}\sum _{k=1}^{n}d_{k} \mathrm{I} \biggl( \frac{{T}_{k,i}}{\beta\delta_{k}\sqrt {k}}\leq\sqrt{1\pm\varepsilon}x\pm\delta_{1} \biggr) \\ &\quad \geq \lim_{n\rightarrow\infty}\frac{1}{D_{n}}\sum _{k=1}^{n}d_{k} \mathrm{I} \biggl( \frac{{T}_{k,i}-E{T}_{k,i}}{\beta\delta_{k}\sqrt{k}}\leq \sqrt{1\pm\varepsilon}x\pm\delta_{1}-\alpha \biggr) \\ &\quad \rightarrow\Phi(\sqrt{1\pm\varepsilon}x\pm\delta_{1}-\alpha), \end{aligned} \end{aligned}$$
(20)
$$\begin{aligned}& \begin{aligned}[b] &\lim_{n\rightarrow\infty}\frac{1}{D_{n}}\sum _{k=1}^{n}d_{k} \mathrm{I} \biggl( \frac{{T}_{k,i}}{\beta\delta_{k}\sqrt {k}}\leq\sqrt{1\pm\varepsilon}x\pm\delta_{1} \biggr) \\ &\quad \leq\lim_{n\rightarrow\infty}\frac{1}{D_{n}}\sum _{k=1}^{n}d_{k}\mathrm{I} \biggl( \frac{{T}_{k,i}-E{T}_{k,i}}{\beta\delta_{k}\sqrt{k}}\leq\sqrt {1\pm\varepsilon} x\pm\delta_{1}+\alpha \biggr) \\ &\quad \rightarrow \Phi(\sqrt{1\pm\varepsilon}x\pm\delta_{1}+\alpha)\quad \mbox{a.s.}, \end{aligned} \end{aligned}$$
(21)
letting
\(\alpha\rightarrow0\) in (
20) and (
21), (
16) holds.
Now, we prove (
17); by
\(\mathrm{E}(Y^{2})<\infty\), we know
\(\lim_{x\rightarrow\infty}x^{2}P(|Y|>x)=0\), such that
$$\mathrm{E}\mathrm{I}\Biggl(\bigcup_{i=1}^{k} \bigl( \vert Y_{i} \vert >\sqrt{k}\bigr)\Biggr)\leq\sum _{i=1}^{k}P\bigl( \vert Y_{i} \vert > \sqrt{k}\bigr)\leq kP\bigl( \vert Y \vert >\sqrt{k}\bigr)\rightarrow 0,\quad k \rightarrow\infty, $$
by the Toeplitz lemma, we get
$$ \lim_{n\rightarrow\infty}\frac{1}{D_{n}}\sum_{k=1}^{n}d_{k} \mathrm{E}\mathrm{I}\Biggl(\bigcup_{i=1}^{k} \bigl( \vert Y_{i} \vert >\sqrt{k}\bigr)\Biggr)\rightarrow0\quad \mbox{a.s.}, $$
(22)
hence, to prove (
17), it suffices to prove
$$ \lim_{n\rightarrow\infty}\frac{1}{D_{n}}\sum_{k=1}^{n}d_{k} \Biggl(\mathrm{I}\Biggl(\bigcup_{i=1}^{k} \bigl( \vert Y_{i} \vert >\sqrt{k}\bigr)\Biggr)-\mathrm{E} \Biggl[ \mathrm{I}\Biggl(\bigcup_{i=1}^{k}\bigl( \vert Y_{i} \vert >\sqrt{k}\bigr)\Biggr) \Biggr] \Biggr) \rightarrow0\quad \mbox{a.s.}, $$
(23)
writing
$$\mathscr {Z}_{k}= \mathrm{I}\Biggl(\bigcup _{i=1}^{k}\bigl( \vert Y_{i} \vert > \sqrt{k}\bigr)\Biggr)-\mathrm{E} \Biggl[\mathrm{I}\Biggl(\bigcup _{i=1}^{k}\bigl( \vert Y_{i} \vert > \sqrt{k}\bigr)\Biggr) \Biggr], $$
for every
\(0\leq2k< l\), so by the definition of
\(\rho^{-}\)-mixing sequence, we have
$$\begin{aligned} \mathrm{E} \vert \mathscr {Z}_{k}\mathscr {Z}_{l} \vert =& \Biggl\vert \operatorname{Cov}\Biggl(\mathrm{I}\Biggl(\bigcup _{i=1}^{k}\bigl( \vert Y_{i} \vert > \sqrt{k}\bigr)\Biggr),\mathrm{I}\Biggl(\bigcup_{i=1}^{l} \bigl( \vert Y_{i} \vert >\sqrt{l}\bigr)\Biggr) \Biggr) \Biggr\vert \\ \leq& \Biggl\vert \operatorname{Cov}\Biggl(\mathrm{I}\Biggl(\bigcup _{i=1}^{k}\bigl( \vert Y_{i} \vert > \sqrt {k}\bigr)\Biggr),\mathrm{I}\Biggl(\bigcup_{i=1}^{l} \bigl( \vert Y_{i} \vert >\sqrt{l}\bigr)\Biggr)-\mathrm{I}\Biggl( \bigcup_{i=2k+1}^{l}\bigl( \vert Y_{i} \vert >\sqrt{l}\bigr)\Biggr) \Biggr) \Biggr\vert \\ &{}+ \Biggl\vert \operatorname{Cov}\Biggl(\mathrm{I}\Biggl(\bigcup _{i=1}^{k}\bigl( \vert Y_{i} \vert > \sqrt {k}\bigr)\Biggr),\mathrm{I}\Biggl(\bigcup_{i=2k+1}^{l} \bigl( \vert Y_{i} \vert >\sqrt{l}\bigr)\Biggr) \Biggr) \Biggr\vert \\ \leq&\mathrm{E} \Biggl\vert \mathrm{I}\Biggl(\bigcup _{i=1}^{l}\bigl( \vert Y_{i} \vert > \sqrt{l}\bigr)\Biggr)-\mathrm{I}\Biggl(\bigcup_{i=2k+1}^{l} \bigl( \vert Y_{i} \vert >\sqrt{l}\bigr)\Biggr) \Biggr\vert \\ &{}+ \rho^{-}(k)\sqrt{\operatorname{Var}\Biggl(\mathrm{I} \Biggl(\bigcup_{i=1}^{k}\bigl( \vert Y_{i} \vert >\sqrt {k}\bigr)\Biggr)\Biggr)\operatorname{Var}\Biggl( \mathrm{I}\Biggl(\bigcup_{i=2k+1}^{l}\bigl( \vert Y_{i} \vert >\sqrt{l}\bigr)\Biggr)\Biggr)} \\ \leq&\mathrm{E}\Biggl[\mathrm{I}\Biggl(\bigcup_{i=1}^{2k} \bigl( \vert Y_{i} \vert >\sqrt{l}\bigr)\Biggr) \Biggr]+C \rho^{-}(k) \\ \leq&\sum_{i=1}^{k}P\bigl( \vert Y_{i} \vert >\sqrt{l}\bigr)+C\rho^{-}(k) \\ \leq& kP\bigl( \vert Y \vert >\sqrt{l}\bigr)+C\rho^{-}(k) \\ \leq& C\biggl(\frac{k}{l}+\rho^{-}(k)\biggr), \end{aligned}$$
so by Lemma
2.4, (
23) holds. And combining with (
22), we know that (
17) holds.
Next, we prove (
18); by
\(\mathrm{E}(\bar{V}_{k}^{2})=k\delta_{k}^{2}\),
\(\bar{V}_{k}^{2}=\bar{V}_{k,1}^{2}+\bar{V}_{k,2}^{2}\),
\(\mathrm{E}(\bar {V}_{k,l}^{2})=k\delta_{k,l}^{2}\), and
\(\delta_{k,1}^{2}\leq\delta_{k}^{2}\),
\(l=1,2\), we have
$$\begin{aligned} \mathrm{I}\bigl(\bar{V}_{k}^{2}>(1+\varepsilon)k \delta_{k}^{2}\bigr) =&\mathrm{I}\bigl(\bar {V}_{k}^{2}-\mathrm{E}\bigl(\bar{V}_{k}^{2} \bigr)>\varepsilon k\delta_{k}^{2}\bigr) \\ \leq& \mathrm{I}\bigl(\bar{V}_{k,1}^{2}-\mathrm{E}\bigl(\bar {V}_{k,1}^{2}\bigr)>\varepsilon k\delta_{k}^{2}/2 \bigr)+\mathrm{I}\bigl(\bar {V}_{k,2}^{2}-\mathrm{E}\bigl( \bar{V}_{k,2}^{2}\bigr)>\varepsilon k\delta _{k}^{2}/2\bigr) \\ \leq& \mathrm{I}\biggl(\bar{V}_{k,1}^{2}> \biggl(1+ \frac{\varepsilon}{2}\biggr)k\delta _{k,1}^{2} \biggr)+ \mathrm{I} \biggl(\bar{V}_{k,2}^{2}> \biggl(1+\frac{\varepsilon}{2}\biggr)k \delta_{k,2}^{2}\biggr), \end{aligned}$$
therefore, by the arbitrariness of
\(\varepsilon>0\), to prove (
18), it suffices to prove
$$ \lim_{n\rightarrow\infty}\frac{1}{D_{n}}\sum_{k=1}^{n}d_{k} \mathrm{I} \biggl(\bar{V}_{k,l}^{2}>\biggl(1+\frac{\varepsilon}{2} \biggr)k\delta_{k,l}^{2}\biggr)=0\quad \mbox{a.s. } l=1,2, $$
(24)
when
\(l=1\), for given
\(\varepsilon>0\), let
f be a bounded function with bounded continuous derivative such that
$$ \mathrm{I}(x>1+\varepsilon)\leq f(x)\leq\mathrm{I}\biggl(x>1+\frac{\varepsilon }{2} \biggr), $$
(25)
under the condition
$$\mathrm{E}\bigl(\bar{V}_{k,1}^{2}\bigr)=k \delta_{k,1}^{2},\qquad \mathrm{E}\bigl(Y^{2}\bigr)< \infty,\qquad \mathrm{E}\bigl(Y^{2}\mathrm{I}(Y\geq0)\bigr)>0, $$
by the Markov inequality, and Lemma
2.2, we get
$$\begin{aligned} &P\biggl(\bar{V}_{k,1}^{2}>\biggl(1+\frac{\varepsilon}{2}\biggr)k \delta _{k,1}^{2}\biggr) \\ &\quad =P\biggl(\bar{V}_{k,1}^{2}- \mathrm{E}\bigl(\bar{V}_{k,1}^{2}\bigr)>\frac {\varepsilon}{2} k \delta_{k,1}^{2}\biggr) \\ &\quad \leq C\frac{\mathrm{E}(\bar{V}_{k,1}^{2}-\mathrm{E}(\bar {V}_{k,1}^{2}))^{2}}{k^{2}} \leq C\frac{\sum_{i=1}^{k}\mathrm{E}(\bar{Y}_{ki}^{2}\mathrm{I}(\bar {Y}_{ki}\geq0))^{2}}{k^{2}} \\ &\quad \leq C\frac{\mathrm{E}\bar{Y}_{k1}^{4}\mathrm{I}(\bar{Y}_{k1}\geq0)}{k} \leq C\frac{\mathrm{E}Y^{4}\mathrm{I}(0\leq Y\leq\sqrt{k})+k^{2}P(Y>\sqrt{k})}{k}, \end{aligned}$$
(26)
because
\(\mathrm{E}(Y^{2})<\infty\) implies
\(\lim_{x\rightarrow\infty }x^{2}P(|Y|>x)=0\), we have
$$\begin{aligned} \mathrm{E}Y^{4}\mathrm{I}(0\leq Y\leq\sqrt{k}) =& \int _{0}^{\infty}P\bigl( \vert Y \vert \mathrm{I}(0 \leq Y\leq\sqrt{k})\geq t \bigr)4t^{3}\,dt \\ \leq&C \int_{0}^{\sqrt{k}}P\bigl( \vert Y \vert \geq t \bigr)t^{3}\,dt \\ =& \int_{0}^{\sqrt{k}}o(1)t\, dt =o(1)k, \end{aligned}$$
thus, combining with (
26),
$$P\biggl(\bar{V}_{k,1}^{2}>\biggl(1+\frac{\varepsilon}{2}\biggr)k \delta _{k,1}^{2}\biggr)\rightarrow0 ,\quad k\rightarrow\infty. $$
Therefore, from (
5), (
25) and the Toeplitz lemma
$$\begin{aligned} 0 \leq& \frac{1}{D_{n}}\sum_{k=1}^{n}d_{k} \mathrm{I} \biggl(\bar{V}_{k,1}^{2}>\biggl(1+\frac{\varepsilon}{2} \biggr)k\delta_{k,1}^{2}\biggr) \\ \leq&\frac{1}{D_{n}}\sum_{k=1}^{n}d_{k}f \biggl(\frac{\bar {V}_{k,1}^{2}}{k\delta_{k,1}^{2}}\biggr) \\ =&\frac{1}{D_{n}}\sum_{k=1}^{n}d_{k} \mathrm{E} \biggl(f\biggl(\frac{\bar {V}_{k,1}^{2}}{k\delta_{k,1}^{2}}\biggr) \biggr) +\frac{1}{D_{n}} \sum_{k=1}^{n}d_{k} \biggl(f \biggl(\frac{\bar {V}_{k,1}^{2}}{k\delta_{k,1}^{2}}\biggr)-\mathrm{E}\biggl(f\biggl(\frac{\bar {V}_{k,1}^{2}}{k\delta_{k,1}^{2}} \biggr)\biggr) \biggr) \\ \leq&\frac{1}{D_{n}}\sum_{k=1}^{n}d_{k} \mathrm{E}\biggl(\mathrm{I} \biggl(\bar {V}_{k,1}^{2}>\biggl(1+ \frac{\varepsilon}{2}\biggr)k\delta_{k,1}^{2}\biggr)\biggr) + \frac{1}{D_{n}}\sum_{k=1}^{n}d_{k} \biggl(f\biggl(\frac{\bar {V}_{k,1}^{2}}{k\delta_{k,1}^{2}}\biggr)-\mathrm{E}\biggl(f\biggl( \frac{\bar {V}_{k,1}^{2}}{k\delta_{k,1}^{2}}\biggr)\biggr) \biggr) \\ =&\frac{1}{D_{n}}\sum_{k=1}^{n}d_{k}P \biggl(\bar{V}_{k,1}^{2}>\biggl(1+\frac {\varepsilon}{2}\biggr)k \delta_{k,1}^{2}\biggr) +\frac{1}{D_{n}}\sum _{k=1}^{n}d_{k} \biggl(f\biggl( \frac{\bar {V}_{k,1}^{2}}{k\delta_{k,1}^{2}}\biggr)-\mathrm{E}\biggl(f\biggl(\frac{\bar {V}_{k,1}^{2}}{k\delta_{k,1}^{2}}\biggr) \biggr) \biggr) \\ \rightarrow&0 \quad \mbox{a.s.}, k\rightarrow\infty, \end{aligned}$$
hence, (
24) holds for
\(l=1\). Similarly, we can prove (
24) for
\(l=2\), so (
18) is true. By similar methods used to prove (
18), we can prove (
19), this completes the proof of Theorem
1.