Convergent hyperbolic cosine martingale $X_n = \frac{\text{exp} \big (\sum_{j=1}^n a_j Y_j \big )}{\prod_{j=1}^n \text{cosh}(a_j)}$, $Y_n$ Rademacher

Consider a sequence of i.i.d. random variables $\{ Y_n : n \in \mathbb{N} \}$ where $P(Y_n = 1) = P(Y_n = -1) = \frac{1}{2}$. Consider the following sequence

$$X_0 \equiv 1, \ X_n := \frac{\text{exp} \big (\sum_{j=1}^n a_j Y_j \big )}{\prod_{j=1}^n \text{cosh}(a_j)} \ \forall n \geq 1. $$

Moreover, suppose the sequence $\{a_n : n \in \mathbb{N} \} \subset \mathbb{R}_+$ satisfies $a_n > a_{n+1} \ \forall n \geq 1$ and $\sum_{n=1}^\infty a_n^2 < \infty$.

Show that there exists $X \in L^2$ such that $X_n \to X$ almost surely and in $L^2$

Based on the question, I want to appeal to the martingale convergence theorems. In particular, I want to show that $\text{sup}_n E[X_n^2] < \infty$.

I will skip the proof that $X_n$ is a martingale as this is pretty straightforward. As for bounding $E[X_n^2]$, I so far have worked out the following

$\begin{align} E[X_n^2] &= E \bigg [ \frac{\text{exp} \big (\sum_{j=1}^n 2 a_j Y_j \big )}{\prod_{j=1}^n \text{cosh}^2(a_j)} \bigg ] \\ &= \frac{1}{\prod_{j=1}^n \text{cosh}^2(a_j)} \prod_{j=1}^n E [ \text{exp} (2 a_j Y_j)] \\ &= \frac{1}{\prod_{j=1}^n \text{cosh}^2(a_j)} \prod_{j=1}^n \text{cosh}(2 a_j) \end{align}$

Basically, I am stuck at this last line. Not sure how I am supposed to use the facts that $a_n > a_{n+1} \ \forall n \geq 1$ and $\sum_{n=1}^\infty a_n^2 < \infty$. I tried to use some hyperbolic cosine identities to simplify the above expression but I didn't see that help too much.

If you have any hints/ideas I would appreciate hearing them. Many thanks in advance for taking the time to help out.

First, observe that $\cosh(2x) = \cosh^2(x) + \sinh^2(x)$. Therefore $\mathbb{E}[X_n^2] = \prod_{i \le n} (1+ \tanh^2(a_i)).$

Now, notice that $\mathbb{E}[X_n^2]$ converges if and only if $\log \mathbb{E}[X_n^2]$ converges. Further, by using that $\log(1+z) \le z,$ we can observe that it suffices to show that $\sum \tanh^2(a_i)$ is convergent. Finally, we observe that for $x \ge 0,$ $\tanh(x) \le x.$ Indeed, the function $x - \tanh(x)$ is $0$ at $x = 0$ and has the non-negative derivative $\tanh^2(x)$.

But then, it follows that $$ \sum \tanh^2(a_n) \le \sum a_n^2 < \infty,$$ and we're done. In particular, note that the monotonicity of $a_n$ is not necessary in the above.