Limit value of the product martingale $\exp(uX_n - nu^2 \sigma^2 / 2)$ This question came from a problem I was solving for self-study.
Statement of the problem
Let $Y_n \sim \mathcal N(0,\sigma^2)$ be independent normally distributed variables, $X_n = Y_1+\cdots+Y_n$ their $n$-th partial sum and, for any real $u$, define $Z_n^u=\exp(uX_n-nu^2\sigma^2/2)$.
It is straightforward to prove that $Z^u$ is a positive martingale under the natural filtration $\mathscr F_n = \sigma(Y_1,\dots,Y_n)$.
Too see this, it suffices to know that  $Z_n^u$ is a product of $n$ independent variables of the form $\exp(uY_j-u^2\sigma^2/2)$, and that $E(e^{uY_j})$, the MGF of $Y_j$, is precisely $e^{u^2\sigma^2/2}$, so that $Z^u$ is in $L_1$ and $E(Z_n^u) = 1$ for all $n$. The martingale property follows from
$$E[Z_{n+1}^u/Z_n^u|\mathscr F_n] = E[\exp(uY_{n+1}-u^2\sigma^2/2)|\mathscr F_n] = E(e^{uY_{n+1}-u^2\sigma^2/2})=1.$$
It is known, by the martingale convergence theorem, that $Z_n^u$ converges almost surely to $Z_\infty^u$.
The problem asks to compute this limit and to say when it is true that $Z_n^u=E[Z_\infty^u|\mathscr F_n]$.
Some ideas
Of course, this last relation would hold if $Z^u$ was
a UI martingale. However, my intuition told me that this was not the case, as the partial sums, which have $\mathcal N(0,n\sigma^2)$ distributions, "concentrate their measure" in increasingly larger intervals $[-K, K]$ as $n$ increases (I haven't tried to prove this).
I started looking around on the internet and found the law of the iterated logarithm, which I used to (correctly, I hope) compute this limit value to be $0$ a.s. when $u$ is not zero and $1$ if it is. Moreover, I found that there is a product martingale theorem by Kakutani which mentions that $P (Z_\infty^u = 0) = 1$ if and only if $Z^u$ is not a UI family.
Main question
Is there a more elementary way of computing $Z_\infty^u$, which doesn't use the aforementioned law or Kakutani's theorem? Maybe using the fact that $X_n/n$ converges to $0$ a.s. and in $L_2$ (by the strong law and by a simple calculation of $V (X_n/n)$, respectively)?
Other questions
What am I dealing with here? Is it in some way natural to consider these exponential martingales? Searching on the internet I found the Wikipedia page for the Doléans exponential which mentions the Girsanov theorem for continuous time as an application, but it didn't give me much insight.
 A: Let $(Y_i)$ be i.i.d. random variables with $\mathbb{P}(Y\geq 0)=1$ and $\mathbb{E}(Y)\leq 1$.
The sequence of partial products $M_n:=\prod_{i=1}^n Y_i$ is a non-negative supermartingale
and so converges almost surely to a limit random variable $M_\infty$.
Martingale Lemma.  Either $\mathbb{P}(Y=1)=1$ or $\mathbb{P}(M_\infty=0)=1$.
Proof.  We have
$$\lim_{\varepsilon\downarrow 0} \mathbb{P}(|Y-1|>\varepsilon)=\mathbb{P}(|Y-1|>0)=\mathbb{P}(Y\neq 1).$$
If this probability is zero, we are in case 1.
Otherwise, we can choose $\varepsilon>0$ so that  $\mathbb{P}(|Y-1|>\varepsilon)>0$. Then the Borel-Cantelli
lemma shows that $\mathbb{P}(|Y_i-1|>\varepsilon \mbox{ infinitely often})=1$. This implies that $Y_i$ fails to
converge to 1 pointwise, so $\mathbb{P}(Y_i\not\to 1)=1$.
If an infinite product converges to a non-zero limit, the factors must converge to 1.
Therefore we finally  get
$$1=\mathbb{P}(M_n\to M_\infty)=\mathbb{P}(M_n\to M_\infty, Y_i\not\to 1)=\mathbb{P}(M_\infty=0),$$
which is case 2.
A: Here is an elementary computation of $Z^u_\infty$. Start with the fact that $X_n/n\to0$ almost surely by the law of large numbers hence, almost surely, $ uX_n\leqslant nu^2/4$ for every $n$ large enough. Assume that $u\ne0$. The preceding remark shows that $uX_n-nu^2/2\to-\infty$ almost surely when $n\to\infty$, thus $P(Z^u_\infty=0)=1$ and $E(Z^u_\infty\mid\mathscr F_n)=0$ almost surely, in particular $E(Z^u_\infty\mid\mathscr F_n)\ne Z^u_n$ almost surely.
