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I've needed to brush up on asymptotics recently, and I came across this interesting problem.

Let $(Z_{1}, Z_{2},...,Z_{N})$ be a random sample from a distribution from mean $\mu$ and variance $\sigma^2$. What is the asymptotic distribution of $Y=$ exp($\bar{Z}$)?

Here's what I've got so far. Just from the information of $Z$ alone, coupled with the fact that it's a random sample from some distribution, I can say $$E(\bar Z) = E(Z) = \mu$$ and $$Var(\bar Z ) = \frac{Var(Z)}{N} = \frac{\sigma^2}{N}$$.

I was wondering what $Y$ would look like. Intuitively, it looks like an exponential random variable, but I'm at a struggle as to why we would need an asymptotic distribution in that case. I would appreciate a hint or guidance as to what to do from here. I'm not very good with asymptotics, so if you could also give theorems or references, I would appreciate it (but it's not necessary).

igiari21
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    Note that asymptotically, $\bar{Z}$ converges to $\mu$. – Glen_b Nov 16 '17 at 21:08
  • Use the delta method, as in @Alecos's answer. [For intuition](https://stats.stackexchange.com/questions/243510/how-to-interpret-the-delta-method/243525#243525), observe that in any tiny ball $(c - \epsilon, c + \epsilon)$, the function $f(x) = e^x$ is just a line, an affine function $f(x) \approx e^c + e^c (x - c)$. So as $n$ gets larger, you're getting closer to an affine function of a normally distributed random variable (since the sample mean converges on the normal distribution). An affine function of a normally distributed random variable also follows the normal distribution. – Matthew Gunn Dec 05 '17 at 21:12

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By the Delta theorem, if

$${\sqrt{n}[\bar Z_n-\mu]\,\xrightarrow{D}\,\mathcal{N}(0,\sigma^2)}$$

then

$${\sqrt{n}[g(\bar Z_n)-g(\mu)]\,\xrightarrow{D}\,\mathcal{N}(0,\sigma^2\cdot[g'(\mu)]^2)}$$

I guess you can do the rest.

Alecos Papadopoulos
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