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Let $\{X_i\}_{i=1}^n$ be independent zero mean random variables with finite variance and $\{r_n\},\{d_n\}$ positive monotone increasing real sequences. Assume that $$\frac{\sum_{i=1}^{r_n} X_i}{Var(\sum_{i=1}^{r_n} X_i)^{1/2}}\overset{d}{\to} N(0,1), \text{ as } n\to\infty,$$ and that $$\frac{Var(\sum_{i=1}^{r_n}X_i)}{d_n}\to \sigma^2.$$ Can you tell me the convergence results needed to obtain that: $$\frac{\sum_{i=1}^{r_n} X_i}{d_n^{1/2}}\overset{d}{\to} N(0,\sigma^2) ?$$

My View

As far as I understand $\frac{Var(\sum_{i=1}^{r_n}X_i)}{d_n}$ is a real sequence, and so for the continous function $f(x)=\sqrt{x}$, we have $f(\frac{Var(\sum_{i=1}^{r_n}X_i)}{d_n})\to f(\sigma^2)=\sigma $ as $n\to\infty$. Next, Slutsky's Theorem states that if $Y_n\overset{p}{\to}c, Z_n\overset{d}{\to}Z$, then $Y_nZ_n\overset{d}{\to}cZ$. In my case I have $Y_n(w)=\sqrt{\frac{Var(\sum_{i=1}^{r_n}X_i)}{d_n}}$ for all $w\in\Omega$, and so $Y_n\to\sigma$ pointwise/uniformly (almost everywhere). I remember that when I studied measure theory, $Y_n\to\sigma$ pointwise implies $Y_n\overset{p}{\to}\sigma$. If these steps are correct, then we are done: by Slutsky's theorem

$$\frac{\sum_{i=1}^{r_n} X_i}{Var(\sum_{i=1}^{r_n} X_i)^{1/2}}\sqrt{\frac{Var(\sum_{i=1}^{r_n}X_i)}{d_n}}=\frac{\sum_{i=1}^{r_n} X_i}{d_n^{1/2}}\overset{d}{\to} \sigma N(0,1)=N(0,\sigma^2).$$

However, I think I used too much theory for the required result. Do you have any recommendations for more efficient ways to show this?

Thanks in advance!

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  • $\begingroup$ This looks like overkill, because your $Y_n$ is not a sequence of random variables: it's a sequence of non-negative real numbers. You don't need any theorems: a two-epsilon argument will do nicely. In carrying it out you might discover you need to assume $\sigma \gt 0.$ $\endgroup$ – whuber Oct 15 '19 at 19:25
  • $\begingroup$ @whuber If $X_n\to X$ in distribution and $b_n\to b$ as $n\to\infty$, how to show that $b_nX_n\to bX$ in distribution? I tried to use $\epsilon$'s definition but didn't suceed. $\endgroup$ – Celine Harumi Oct 15 '19 at 20:16
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    $\begingroup$ Employ the definition. That is, let $x$ be a fixed value, let $F$ be the limiting distribution function, and consider the sequence $\Pr(b_nX_n\le x).$ Assuming the limit of the $b_n$ is nonzero, eventually all $b_n$ are nonzero and you can express this as $\Pr(X_n \le x/b_n)=F_n(x/b_n).$ Because $F$ is continuous, your assumptions imply $\lim_{n\to\infty}F_n(z)=F(z)$ for all $z.$ Now you're at the point of analyzing sequences of real numbers. One epsilon relates $F_n(x/b_n)$ to $F_n(x/b)$ and the other epsilon relates $F_n(x/b)$ to $F(x/b).$ $\endgroup$ – whuber Oct 15 '19 at 20:26
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    $\begingroup$ The Normal$(0,1)$ distribution is continuous. $\endgroup$ – whuber Oct 15 '19 at 21:45
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    $\begingroup$ Thanks whuber. Its all clear now. :) $\endgroup$ – Celine Harumi Oct 15 '19 at 22:13

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