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$X_1,X_2,...,X_n,...$ are independent random variables such that $X_k \sim N(\mu,\sigma^2)$ for all $k$. I want to calculate the asymptotic distribution of $$ \left(\sqrt{n}(e^{T_n} - e^{\mu})\right)_{n \geq 1}, $$ where $T_n = \frac{1}{n}\sum\limits_{k=1}^n X_k$, as $n \to \infty$.

Any suggestions for how this can be done?

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Write $\tilde{T}_n = \frac{1}{n} \sum_{k=1}^{n} (X_k - \mu)$ and define $f : \Bbb{R} \to \Bbb{R}$ by $f(x) = \sum_{n=0}^{\infty} \frac{x^n}{(n+1)!} = \frac{\mathrm{e}^x - 1}{x}$. Then we can write

$$ \sqrt{n} (\mathrm{e}^{T_n} - \mathrm{e}^{\mu}) = \sqrt{n} \tilde{T}_n \mathrm{e}^{\mu} f(\tilde{T}_n). $$

Here, SLLN shows that $f(\tilde{T}_n) \to f(0) = 1$ a.s. and the classical CLT shows that $\sqrt{n} \tilde{T}_n \Rightarrow \sigma Z$ for a standard normal variable $Z \sim \mathcal{N}(0, 1)$. (Here, $\Rightarrow$ denotes convergence in distribution.) Thus by the Slutsky's theorem,

$$ \sqrt{n} (\mathrm{e}^{T_n} - \mathrm{e}^{\mu}) \quad \Rightarrow \quad \sigma \mathrm{e}^{\mu} Z. $$

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    $\begingroup$ Great answer (+1)...would you mind pointing out where I went wrong my my answer? $\endgroup$ – user237392 Sep 26 '16 at 17:50

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