Find the MGF of $\frac{1}{n} \sum\limits_{i=1}^{n}\ln X_i$ for $(X_i)$ i.i.d. gamma random variables

Question

Let $X_1, X_2, X_3, ..., X_n \stackrel{iid}{\sim}$ Gamma ($\alpha, \lambda$).

Define $\overline{lnX} = \frac{1}{n} \sum_{i=1}^{n}ln X_i$.

Find the MGF of $\overline{lnX}$.

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So far I've got $E(e^{tX}) = E(e^{t\overline{lnX}})$

But i'm stuck here because I'm not sure if I can do $= E({\bar{X}}^t) = E([\frac{1}{n}\sum_{i=1}^{n} X_i]^t)$ or if that is not correct I'd have to do $E(e^{\frac{1}{n} \sum_{i=1}^{n}ln X_i})$ and that seems much more complicated and I am unsure how to approach the problem.

Answer 1

Suppose $\ln X_1, \dots, \ln X_n$ have MGF $M$.

Suppose $Y = \dfrac{1}{n}\sum_{i=1}^{n}\ln X_i$.

Then $$\begin{align} M_{Y}(t) &= \mathbb{E}[e^{(t/n)\sum_{i=1}^{n}\ln X_i}] \\ &= \mathbb{E}[e^{(t/n)\ln X_1 + (t/n)\ln X_2 + \cdots+(t/n)\ln X_n}] \\ &= \mathbb{E}[e^{(t/n)\ln X_1}e^{(t/n)\ln X_2} \cdots e^{(t/n)\ln X_n}] \\ &= \mathbb{E}[e^{(t/n)\ln X_1}]\cdot\mathbb{E}[e^{(t/n)\ln X_2}] \cdots \mathbb{E}[e^{(t/n)\ln X_n}]\text{ by independence} \\ &= \underbrace{M(t/n) \cdot M(t/n) \cdots M(t/n)}_{n\text{ times}} \\ &= [M(t/n)]^{n}\text{.} \end{align}$$