Suppose that $X_n$ are iid and for $\theta > 0$ we have the martingale: $$M_n = \frac{\prod_{i=1}^n e^{\theta X_i}}{ (E[e^{\theta X_i}])^n}$$ I know by the martingale convergence theorem that $M_n \rightarrow M_\infty$ almost surely as $n\rightarrow\infty$, but how can I find what the actual limit $M_\infty$ is? I think it's zero but I am having trouble formalising the proof (I was thinking of using Jensen's inequality and the squeeze law but this feels wrong to me).
I also have another related question that asks me to prove that $(M_n)^{\frac{1}{n}} \rightarrow a$ for some $a \geq 0$ (which I need to explicitly identify) but if I think that $M_\infty = 0$ then I'm not sure I understand how to find this limit either. Could anyone provide any tips?
Take the logarithm of $\sqrt[n]{M_n}$, i.e. $$ \log \sqrt[n]{M_n} = \frac1n \log M_n = \theta \frac1n \sum_{i=1}^n X_i - \log \mathbb E[e^{\theta X_1}] .$$ If $\exists \theta>0$ s.t. $\mathbb E[e^{\theta X_1}]<\infty$, then all the moments are finite, hence we have SLLN, and so $$ \lim_n \log \sqrt[n]{M_n} \stackrel{a.s.}{=} \theta \mathbb E[X_1]- \log \mathbb E[e^{\theta X_1}], $$ meaning that $$ \lim_n \sqrt[n]{M_n} \stackrel{a.s}{=} \frac{e^{\theta \mathbb E[X_1]}}{\mathbb E[e^{\theta X_1}]}. $$