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Let $X_1, X_2,...$ be i.i.d. random variables with finite mean $\mu$ and finite variance $\sigma^2$. From the Central Limit Theorem, we know that $\sqrt{n}(\bar{X_n}-\mu)$ tends in distribution to $N(0,\sigma^2)$ as $n\rightarrow\infty$, where $\bar{X_n}=\frac{1}{n}\sum_{i=1}^nX_i$.

I'm interested in the distribution of $f(X^n)$, where $X^n$ is the sum $X^n=n\bar{X_n}$ and $f$ is continuous and differentiable. This question is relevant and does this for $f(x)=\sqrt{x}$, but I'd like a more general result. Is the following reasoning correct?

We apply the delta method with $g(x)=f(nx)$ to see that $\frac{(f(X^n)-f(n\mu))}{\sqrt{n}f'(n\mu)}$ converges in distribution to $N(0,\sigma^2)$. (Is $g$ allowed to depend on $n$ like this?)

For sufficiently large $n$ then, $f(X^n)$ approximately follows a normal distribution $N(f(n\mu),nf'(n\mu)^2\sigma^2)$.

On the one hand, I get the same result as the question linked above for $f(x)=\sqrt{x}$, which is promising. But something slightly weird happens for $f(x)=\log(x)$. Then $g'(\mu)=\frac{n}{n\mu}=\frac{1}{\mu}$ and so $\sqrt{n}(\log(X^n)-\log(n\mu))$ converges in distribution to $N(0,\frac{\sigma^2}{\mu^2})$. But then if I want to look at the approximation of $\log(X^n)$ for large $n$, I get $N(\log(n\mu),\frac{\sigma^2}{n\mu^2})$, which has variance that decreases as $n$ increases. That seems weird but I guess it makes sense. It definitely makes sense when $f$ is bounded, as $f(X^n)$ will just approach that bound and the variance should drop to $0$. For $\log(X^n)$, the mean is still moving up but the distribution just get more narrowly concentrated around it, since $\log(X^n)$ grows so slowly for large $n$. (I'm assuming for these examples that the $X_i$ are positive.)

My main question is: am I allowed to use the delta method like this?

If so, I'd also be interested to hear if my interpretation of the $\log(X^n)$ approximation is correct.

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    $\begingroup$ Yes, this looks correct to me. Interestingly, I also get $\frac{\sigma^2}{n\mu^2}$ as the asymptotic variance for $\text{log}(\bar{X})$. I suppose the best way to double check all of this is through simulation. I'll give it a try... $\endgroup$ Aug 16 at 23:31
  • $\begingroup$ Thanks. Yeah, it seems surprising at first that $\log(\bar{X_n})$ and $\log(X^n)$ have the same variance, but I guess that actually makes a lot of sense, since $\log(\bar{X_n})=\log(\frac{1}{n}X^n)=\log(X^n)-\log(n)$. $\endgroup$
    – DM-97
    Aug 18 at 10:58
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Here is a simulation that confirms your suspicions. The variance decreases with increasing $n$.

data norm;
do n=20, 50, 100;
do sim=1 to 100000;
    do i=1 to n;
        x=rand('normal',2,3);
        output;
    end;
end;
end;
run;

proc means data=norm noprint;
by n sim;
var x;
output out=out sum(x)=sum;
run;

data out;
set out;
log_sum=log(sum);
run;

proc means data=out var;
class n;
var log_sum;
run;
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  • $\begingroup$ Thanks. I ran some more simulations to test a few different distributions for $X_i$ and functions $f$, and found approximately the right mean, variance, skew (0) and kurtosis (3) most of the time. The approximation is poor for lognormal $X_i$ unless (presumably) $n$ is larger than I'm patient enough to simulate. But that's not too surprising as it's heavy-tailed. $\endgroup$
    – DM-97
    Aug 18 at 11:32

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