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Find the asymptotic distribution of the MME and MLE.

For the top answer there, my lecturer showed us a similar question and I recall him saying that we do not have asymptotics depending on 'n'. As in, the part where it says:
$$g(\bar X)\sim N\left(g(E(X)), \frac{\mathrm{Var}(X)(g'(E(X))^2}{n}\right)$$

Should we have an $n$ in the right hand side? My lecturer said that it should really be on the left hand side so it should read:
$$\sqrt{n}g(\bar X)\sim N\left(g(E(X)), \mathrm{Var}(X)(g'(E(X))^2\right).$$

I just accepted it (I'm not sure why) but would like some clarification on this. Should we never have $n$ on the RHS for asymptotic distributions, and if not, why?

Natash1
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    Think of it as a limit. You can write $n x_n \xrightarrow[n\to\infty]{} 2$, this is perfectly valid. But you cannot write $x_n \xrightarrow[n\to\infty]{} \frac{2}{n}$... why? Because on the RHS, you "have taken $n\to \infty$: there is no such thing as the variable $n$ anymore. (Now, what is confusing is the use of the word "asymptotic distribution" and of the symbol $\sim$, since that symbol has a different meaning for asymptotics of sequences, namely "equivalents"; but in your context, that boils down to the above). – Clement C. Jun 12 '17 at 12:43
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    Note also that instead of $\sqrt{n}g(\bar X)\sim N\left(g(E(X)), \mathrm{Var}(X)(g'(E(X))^2\right)$ it should be $$\sqrt{n}\left(g(\bar X)-g(E(X))\right) \xrightarrow{D} N\left(0, \mathrm{Var}(X)(g'(E(X))^2\right)$$ – NCh Jun 12 '17 at 17:13

1 Answers1

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A sequence of random variables $X_1,\cdots,X_n,\cdots$ is said to converge in distribution to another random variable $Y$ if $\lim_{n\to\infty} F_n(x)=F(x)$ for all $x\in\mathbb R$, where $F_n$ is the CDF of $X_n$ and $F$ is the CDF of $X$.

Now let's come to your question: if you have $n$ on the RHS then it's no longer a single distribution to converge to. Whenever you write $A_n \overset{d}{\to} B$ you must make sure that $B$ is a single random variable, with a well-defined distribution.

Yining Wang
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