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Let $N_n$ - random variable with integer values and $\frac{N_n}{n} \xrightarrow[]{P} a$. Let $S_n = X_1 + \cdots + X_n$, where $X_i$ - iid with $EX_i=0$, $Var X_i = 1$. $N_n$ and $\{X_i\}$ are independent. Prove that $$\frac{S_{N_n}}{\sqrt{n}} \xrightarrow[]{d} U \sim \mathcal{N}(0,a).$$

I tried to repeat the last part of proof of CLT which is using characteristic function:

$$\psi_{\frac{X_1 +\cdots + X_{N_n}}{\sqrt n}} (t) = \psi_{\frac{X_1}{\sqrt n}} (t) \cdots \psi_{\frac{X_{N_n}}{\sqrt n}} (t) = \left(\psi_{\frac{X_1}{\sqrt n}} (t) \right)^{N_n} = \\ = \left( 1 - \frac{t^2}{2n} + O\left(\frac{1}{n}\right) \right)^{n \cdot\frac{N_n}{n}} \rightarrow \left(e^{\frac{t^2}{2}}\right)^a = e^{-\frac{at^2}{2}}.$$ Of course, the limit is not correct because $N_n$ - random. But this is kind of idea. And I have no ideas how to use this convergence in probability $\frac{N_n}{n} \xrightarrow[]{P} a$.

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    $\begingroup$ Are $N_n$ and $\{X_n\}$ independent? $\endgroup$ Commented Feb 18 at 22:42
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    $\begingroup$ @Zhanxiong yes, thanks for the question $\endgroup$ Commented Feb 19 at 7:34
  • $\begingroup$ I wonder whether the result is even true without some uniform integrability assumption. Even almost sure convergence of $N_n/n$ to 1 does not guarantee that variance converges to 1. We can just take $N_n = n$ with probability $1-1/n$ and $N_n$ is astronomically large with probability $1/n$. This would give a very high variance for $S_Nn$. $\endgroup$ Commented Feb 19 at 9:28
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    $\begingroup$ @JayanthRVarma No. Variance of limiting distribution and limit of variance are completely different things. Just think about your example, $N_{n}/n$ converges almost surely (hence in probability, hence weakly) to $1$. Does the large variance of $N_{n}/n$ block it converging to a constant, something with zero variance? $\endgroup$ Commented Feb 19 at 11:53
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    $\begingroup$ @Q9y5 You are right. $\endgroup$ Commented Feb 21 at 4:47

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After a second thought, the conclusion can hold without the independence assumption between $N_n$ and $\{X_i\}$, and the proof goes as follows (the key is to apply Kolmogorov's inequality).

Without loss of generality, we can assume $a = 1$. Rewrite $\frac{S_{N_n}}{\sqrt{n}}$ as \begin{align*} \frac{S_{N_n}}{\sqrt{n}} = \frac{S_n}{\sqrt{n}} + \frac{S_{N_n} - S_n}{\sqrt{n}}. \tag{1}\label{1} \end{align*} By the classical CLT, the first term in the right hand side of $\eqref{1}$ converges in distribution to $N(0, 1)$. Therefore it suffices to show the remainder term (by the Slutsky's theorem) \begin{align*} \frac{S_{N_n} - S_n}{\sqrt{n}} \Rightarrow 0. \tag{2}\label{2} \end{align*} To prove $\eqref{2}$, notice that for any $\epsilon > 0$, \begin{align*} & P[|S_{N_n} - S_n| \geq \sqrt{n}\epsilon] \\ =& P[|S_{N_n} - S_n| \geq \sqrt{n}\epsilon, |N_n - n| \geq \epsilon^3n] + P[|S_{N_n} - S_n| \geq \sqrt{n}\epsilon, |N_n - n| < \epsilon^3n] \\ \leq & P[|N_n - n| \geq \epsilon^3n] + P\left[\max_{|k - n| < \epsilon^3n}|S_k - S_n| \geq \sqrt{n}\epsilon\right]. \tag{3}\label{3} \end{align*} The first term in the right hand side of $\eqref{3}$ converges to $0$ as $n \to \infty$ because $\frac{N_n}{n} \to_P 1$. To bound the second term, by Kolmogorov's inequality (or more precisely, repeat the proof of it), \begin{align*} P\left[\max_{|k - n| < \epsilon^3n}|S_k - S_n| \geq \sqrt{n}\epsilon\right] \leq \frac{1}{n\epsilon^2}\sum_{k: |k - n| < \epsilon^3n}\operatorname{Var}(X_k) \leq \frac{2n\epsilon^3}{n\epsilon^2} = 2\epsilon. \end{align*} This completes the proof.

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  • $\begingroup$ Then we can conclude that if we have independence between $N_n$ and $X_i$ the statement can be proved as @Q9y5 suggested without Kolmogorov's inequality, BUT the statement is also true without assumption about independence, just need to use Kolmogorov's inequality. Am I right? $\endgroup$ Commented Feb 19 at 14:41
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    $\begingroup$ Nice, $S_{n}$ is a martingale, therefore $S_{n}/\sqrt{n}$ indeed implies a uniform continuity required for the convergence, upvoted. $\endgroup$ Commented Feb 19 at 14:49
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    $\begingroup$ @Alex Yes, but I cannot endorse that Q9y5's proof is rigorous enough. $\endgroup$ Commented Feb 19 at 14:50
  • $\begingroup$ @Zhanxiong one more question. It seems that it's not so obvious moment with reducing to the case $a=1$. I needed to represent $\frac{S_{N_n}}{\sqrt{n}}$ as $\sqrt{a} \frac{S_{N_n}}{\sqrt{\lfloor na\rfloor}} ( 1 + O(\frac{1}{\sqrt{n}} ))$. And then we can continue with $\frac{S_{N_n}}{\sqrt{\lfloor na\rfloor}}$. Does it looks like true? $\endgroup$ Commented Feb 20 at 21:30
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    $\begingroup$ @Alex Good question. A more straightforward decomposition is rewriting $\frac{S_{N_n}}{\sqrt{n}}$ as $\frac{S_{N_n}}{\sqrt{\lfloor na \rfloor}} \times \frac{\sqrt{\lfloor na \rfloor}}{\sqrt{n}}$ then use the $a = 1$ case result and the Slutsky's theorem (the second term converges to $\sqrt{a}$ as $n \to \infty$). $\endgroup$ Commented Feb 20 at 22:22
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Update: Since $S_{n}$ is a partial sum process, the uniform continuity condition required in Aldous (1978) naturally holds using Doob's martingale inequality. Therefore, the result remains true even if $N_{n}$ and $S_{n}$ are not independent. See @Zhanxiong's post for a detailed proof.

However, independence is not completely useless: the convergence $S_{N_{n}}/\sqrt{N_{n}}\to_{\mathcal{L}|N_{n}}\mathcal{N}(0,1)$ together with $N_{n}/n\to_{\mathbb{P}}a$ yields a joint stable convergence in law of $(S_{N_{n}}/\sqrt{N_{n}},N_{n}/n)$. As a result, with independence, the desired convergence still holds even if $a$ is a random variable with $0<a<\infty$ a.s., and the limit would be $\mathcal{MN}(0,a)$, a mixed normal distribution.

For independent $N_{n}$, this is straightforward: $N_{n}\to\infty$ with probability approaching 1, conditional on this event, $S_{N_{n}}/\sqrt{N_{n}}\to_{\mathcal{L}|N_{n}}\mathcal{N}(0,1)$, since the limit is independent of $N_{n}$, it holds unconditionally. Applying Slutsky lemma together with $N_{n}/n\to_{\mathbb{P}} a$ gives desired result.

For general $N_{n}$ that is not independent of $S_{n}$, an additional (also necessary) uniform continuity condition is needed, see Aldous, David J (1978) "Weak convergence of randomly indexed sequences of random variables," Math. Proc. Camb. Phil. Soc, Vol. 83(1), 117-126.

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    $\begingroup$ could you please explain place where we claim that $S_{N_n}/\sqrt{N_n}$ converges to $\mathcal{N}(0,1)$ in distribution? The initial problem is that $N_n$ converges to $\infty$ in probability but in CLT we have ordinary convergence. I cannot understand that hack with conditional convergence. $\endgroup$ Commented Feb 19 at 21:32
  • $\begingroup$ Condition on $N_{n}$ means treating $N_{n}$ as nonrandom, hence your characteristic function method follows. By definition of convergence in dist, for any $\epsilon>0$, there exists a $N_{\epsilon}$ such that $|\mathbb{P}(S_{N_{n}}/\sqrt{N_{n}}\leq x\mid N_{n})-\Phi(x)|<\epsilon$ for all $N_{n}\geq N_{\epsilon}$. Law of iterated expectation $\mathbb{E}\bigl[\mathbb{E}[1_{A}\mid \mathcal{G}]\bigr]=\mathbb{E}[1_{A}]$ shows this inequality holds without conditioning as long as $N_{n}\geq N_{\epsilon}$, and you know by convergence in prob that $\mathbb{P}(N_{n}\geq N_{\epsilon})\to 1$. $\endgroup$ Commented Feb 20 at 5:20
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    $\begingroup$ Ok, put it this way: $A$ is the event $S_{N_{n}}/\sqrt{N_{n}}\leq x$, $B$ is the event $N_{n}>N_{\epsilon}$. You can have $|\mathbb{P}(A\mid B)-\Phi(x)|<\epsilon/2$, want to show $|\mathbb{P}(A)-\Phi(x)|<\epsilon$ for all $n$ greater than some $n_{\epsilon}$. By law of total probability $\mathbb{P}(A)-\Phi(x)=\mathbb{P}(A\mid B)\mathbb{P}(B)-\Phi(x)\mathbb{P}(B)+\mathbb{P}(A\mid B^{\complement})\mathbb{P}(B^{\complement})-\Phi(x)\mathbb{P}(B^{\complement})<\epsilon/2+\mathbb{P}(B^{\complement})$. $\endgroup$ Commented Feb 21 at 1:26
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    $\begingroup$ The probability of $B^{\complement}=\{N_{n}<N_{\epsilon}\}$ can be bounded by $\epsilon/2$ for $n$ sufficiently large. Similar for the other side. $\endgroup$ Commented Feb 21 at 1:29
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    $\begingroup$ Seems like much more clear, thank you! I would add indices $A_n$ and $B_n$, but it's not essential thing obviously. So we have convergence in distribution for $S_{N_n}/\sqrt{N_n}$ to $\mathcal{N}(0,1)$ (without any condition) and then it remains to apply Slutsky's lemma. $\endgroup$ Commented Feb 21 at 11:28

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