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Suppose $\{X_n,n\ge1\}$ are independent r.v., $E(X_n)=0$, $\operatorname{Var} \left(X_n\right)=\sigma_n^2<\infty$. Set $S_n=\sum_{i=1}^nX_i$ and $s_n^2=\sum_{i=1}^n\sigma_i^2$, assume $$\frac{S_n}{s_n}\overset{d}\to N(0,1)~~~ \text{and}~~~ \frac{\sigma_n}{s_n}\to \rho.$$ Show that $$\frac{X_n}{s_n}\overset{d}\to N\left(0,\rho^2\right).$$

Can we say that since $S_n/s_n=(S_{n-1}+X_n)/s_n\overset{d}\to N(0,1)$ and $S_{n-1}$ is independent with $X_n$, we can conclude that $X_n/s_n$ approaches a normal distribution?

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Notice that $$\frac{S_{n-1}}{s_n}=\frac{S_{n-1}} {s_{n-1} }\frac{s_{n-1}} {s_n} $$ and by assumption, $\lim_{n\to +\infty}s_{n-1} /s_n=\sqrt{1-\rho^2} $, hence $S_{n-1} / s_n\to\mathcal N\left(0, 1-\rho^2 \right)$ in distribution as $n$ goes to infinity. Now, denoting by $\varphi_Y$ the characteristic function of a random variable $Y$, the equality $$\varphi_{\dfrac{S_n}{s_{n}}}(t) = \varphi_{\dfrac{X_n}{s_n}}(t) \varphi_{\dfrac{S_{n-1} }{s_n}}(t) $$
is valid for any real number $t$ (since $S_{n-1}$ is independent of $X_n$). Since the sequence $\left(\varphi_{\dfrac{S_{n-1} }{s_n}}(t) \right)_{n\geqslant 2}$ converges to a non-zero number, we get that for any $t$, for $n$ large enough, $$\frac{\varphi_{\dfrac{S_n}{s_{n}}}(t)}{\varphi_{\dfrac{S_{n-1} }{s_n}}(t)  } = \varphi_{\dfrac{X_n}{s_n}}(t) $$
and taking the limit as $n$ goes to infinity, we derive that for any $t$, $$\lim_{n\to +\infty}\varphi_{\dfrac{X_n}{s_n}}(t)=\frac{\exp\left(-\frac{t^2}2\right)} {\exp\left(-\left(1-\rho^2\right) \frac{t^2}2\right)}, $$ giving the wanted result.

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This problem is a tailor-cut application of Slutsky's Lemma.

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Doesn't it follow from the Lévy's continuity theorem? I mean, consider the characteristic functions of $S_{n-1}/s_n$ and $X_n/s_n$, the product of them converges to $e^{-t^2/2}$, so the characteristic function of $X_n/s_n$ converges to what it should, so $X_n/s_n$ converges.

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