Let $\eta_k$ be i.i.d. random variables having an exponential distribution, $$F_\lambda(x) = P(\eta_k \leq x) = 1-e^{-\lambda x}$$ for $x \geq 0$. Consider a sequence $\xi_k = F_{\lambda}(\eta_k)$. Define $$X_n = \left(\prod_{k=1}^{n} \xi_k\right)^{1/n}.$$ Find $\lim_{n \to \infty} X_n$. Is $X_n$ a martingale, submartingale, or super martingale ?
My attempts/remarks:
$lim_{n \to \infty} X_n = lim_{n \to \infty} (\prod_{k=1}^{n} \xi_k)^{1/n} =lim_{n \to \infty} (\prod_{k=1}^{n} 1-e^{-\lambda \eta_k})^{1/n}$
How can I continue from here? I also have problems with the definition of $\xi_k := F_{\lambda}(\eta_k)$, since $F_{\lambda}(\eta_k) = P(\eta_k \leq \eta_k) = P(0 \leq 0)=1$. Huh? Could the author have meant something else? What am I missing here?
My work from some hints in the comments: We can easily compute that $f_{\xi_k}(x)=1$ for $x \in (0,1)$. Then, we compute:
$$\lim_{n \to \infty} \log{X_n} = \lim_{n \to \infty} \dfrac{\sum_{k=1}^{n}\log{\xi_k}}{n}=E[log(\xi_k)] = \int_{0}^{1}1*log(\tau) d\tau = -1.$$ It follows by monotone convergence theorem that $$\lim_{n \to \infty} X_n = \lim_{n \to \infty} e^{\log{X_n}} = e^{-1}$$
Note that for every $n\ge2$, $$E[X_n|F_{n-1}] = E[(\prod_{k=1}^{n} \xi_k)^{1/n}|F_{n-1}]= E[(\xi_n)^{1/n}{(\prod_{k=1}^{n-1} \xi_k)^{1/(n-1)}}^{(n-1)/n}|F_{n-1}] = (X_{n-1})^{(n-1)/n}E[(\xi_n)^{1/n}] = (X_{n-1})^{(n-1)/n} \int_{0}^{1}(\tau)^{1/n} d \tau = (X_{n-1})^{(n-1)/n} \dfrac{n}{n+1} =??$$ I am stuck here.