If I have $\sqrt{n} (X_n - c) \xrightarrow[]{d} N(0,v) $ does it make any sense at all to say this implies that $X_n \xrightarrow[]{d} N(c, \frac{v}{n})$.
If not, what is the accurate way/notation that allows me to express the above?
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1$\begingroup$ $v$/n goes to 0 as n approaches infinity. So $X_n$ can only converge to c in some sense. but not to a non-degenerate normal distribution. $\endgroup$– Michael R. ChernickCommented Feb 2, 2019 at 18:40
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1$\begingroup$ You cannot use $n$ in the limit since $n$ goes to $\infty$. $\endgroup$– Xi'anCommented Feb 3, 2019 at 11:02
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1 Answer
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No, it doesn't make much sense to say that $$X_n \overset{d}{\to} N\left(c, \frac{v}{n}\right)\,.$$ The main reason is because the $\overset{d}{\to}$ is as $n\to \infty$, and if $n\to \infty$ in the limit, then on the right hand side $v/n$ makes no sense. Instead, an appropriate consequence for the CLT is the use of the "approximately distributed" notation which is usually "$\approx$". So,
$$X_n \approx N\left(c, \frac{v}{n}\right)\,. $$
answered Feb 2, 2019 at 16:02
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1$\begingroup$ +1 though some people object to $\approx$ for "approximately distributed" as they think it is too close to using $=$ for "exactly distributed" $\endgroup$– HenryCommented Feb 2, 2019 at 18:39
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$\begingroup$ @Henry hmm yes. What do they recommend to use for "approximately distributed" then? $\endgroup$ Commented Feb 2, 2019 at 18:42
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1$\begingroup$ Somewhat like $\stackrel{D}{\approx}$? $\endgroup$ Commented Feb 2, 2019 at 18:50
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$\begingroup$ Writing "approximately distributed as" can work. Wikipedia's article on the central limit theorem tends to spell it out except for a couple of cases of $\approx$ preceded by a suggestion such usage is informal $\endgroup$– HenryCommented Feb 2, 2019 at 18:50
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$\begingroup$ And would it be ok to say that $X_n$ has asymptotic variance equal to $\frac{v}{n}$ if $X_n$ is approximately distributed as in your answer? $\endgroup$– nsimplexCommented Feb 2, 2019 at 18:52