Consider a sequence of random variables $(X_n)_{n \in \mathbb{N}}$ wich are independently normal distributed $N(0,\sigma^2)$.
Set $M_0$=1 and $$ M_n =\exp \left( \sum_{i=1}^n X_i - \frac{n\sigma^2}{2} \right) = \prod_{n=1}^n \exp \left(X_i - \frac{\sigma^2}{2} \right)$$
I already showed that $(M_n)_{n \in \mathbb{N}}$ is a martingale w.r.t. $\mathscr{F}_n=\sigma(X_m, m\leq n)$. Therefore, by the martingale convergence theorem, $M_n$ converges to some $M>0$ almost surely and $E[M] \leq E[M_0]$.
My task is to determine the limit random variable $M$. I'm not sure how to do that.
What I did so far. Since $X_i \sim N(0,\sigma^2)$, we have $\sum_i^n X_i \sim N(0,n\sigma^2)$. Therefore, $$Z := \exp \left( \sum_i^n X_i \right)$$ is log-normal distributed $\mathcal{L}N(0, n \sigma^2)$ with $$E[Z]=\exp(n \sigma/2)$$ and $$\operatorname{var}(Z)=\exp(n\sigma^2)(\exp(n \sigma^2)-1).$$ Therefore, it hold that $$E[M_n]=1$$ and $$\operatorname{var}(M_n)=\exp(n \sigma^2)-1$$ for all $n \in \mathbb{N}$.
This seems to suggest that $\operatorname{var} (M)=\infty$. Is this true? Does in this case the a.s.-convergence imply the convergence of moments? Still, this does not help me to determine $M$.
My Question. In which sense can I "determine" $M$? Its distribution? Is it a constant? Can anyone give me a hint on how to do that?
Any help is much appreciated!
Hint: By the strong law of large numbers,
$$\frac{1}{n} \sum_{i=1}^n X_i \to \mathbb{E}X_1 = 0$$
almost surely. Write
$$\exp \left( \sum_{i=}^n X_i - \frac{n \sigma^2}{2} \right) = \exp \left( n \left[ \frac{1}{n} \sum_{i=1}^n X_i - \frac{\sigma^2}{2} \right] \right)$$
in order to deduce that $M=0$ a.s.