Consider $X_i\sim Exp(\lambda=1)$, i.e. $f_X(x)=e^{-x}$. Is it true that for a large $N$ the CLT can be used for the average $\frac{1}{N}\sum\limits_{i=1}^{N}ln(X_i)$ can be approached by the normal distribution? I am aware that it is sufficient to prove that $V[ln(X_i)]<\infty$ and $E[ln(X_i)]<\infty$, but does that apply in this case?
My first result was that $E[ln(X_i)]=\infty$, but I am doubting that is correct.
If the pdf of $X$ is given by: $$ f_X(x) = e^{-x}\cdot\mathbb{1}_{x>0}$$ then $Y=\log(X)$ is supported on $\mathbb{R}$ and: $$\mathbb{P}[Y<c]=\mathbb{P}[X<e^c]=1-e^{-e^c}$$ so: $$ f_Y(x) = \exp(x-e^x)$$ and: $$ \mathbb{E}[Y]=-\gamma,\qquad \mathbb{E}[Y^2]=\gamma^2+\zeta(2), $$ hence $\frac{1}{N}\sum_{i=1}^{N}\log(X_i)$ is approximated by a normal variable with mean $-\gamma$ and variance $\zeta(2)$.
Here $\gamma$ is the Euler-Mascheroni constant and $\zeta(2)=\sum_{n\geq 1}\frac{1}{n^2}=\frac{\pi^2}{6}.$