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It is known that from the CLT, if $X_i \stackrel{\text{iid}}{\sim} F$ for some distribution $F$ with finite variance, then $$\frac{1}{\sqrt{n}} \sum_{i=1}^n (X_i - \text{E}[X]) \stackrel{d}{\to} N(0,\sigma^2)$$ for some $\sigma^2$. Now, define $n$ different sequences of random variables of the form $\{A^i_k\}_{k=1}^\infty$ such that $A^i_k \stackrel{p}{\to} 1$ as $k\to\infty$ for all $i=1,2,\ldots,n$.

Here is my question. Is the following statement true? I really want to apply Slutsky's theorem in some form, but I can't seem to do it because there are different variables being multiplied at each term in the sum. $$\frac{1}{\sqrt{n}} \sum_{i=1}^n (X_i A^i_n - \text{E}[X]) \stackrel{d}{\to} N(0,\sigma^2) \quad \text{as } n\to\infty$$

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    $\begingroup$ I haven't written out a proof, but intuitively this looks hopeless without more assumptions. The Lindeberg condition is sufficient and almost-necessary for a CLT to hold in these settings, which suggests that you would need some moment condition on the $A_n^i$'s; the convergence in probability doesn't seem strong enough. $\endgroup$
    – guy
    Oct 20, 2020 at 20:23
  • $\begingroup$ There's no assurance that $X_iA^i_n$ even has a finite expectation, much less a finite variance. $\endgroup$
    – whuber ♦
    Oct 20, 2020 at 20:54

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