Let $\eta \sim \mathcal{U}(0, 1)$, and let $\theta$ be an independent Bernoulli random variable such that $P(\theta = 1) = P(\theta = -1) = 1/2$.
Determine the distribution of $\xi := \theta/\sqrt{\eta}$.
Let $\{\xi_n\}$ be an i.i.d. sequence distributed as $\xi$. Show that
$$\lim_{n\to\infty} \frac{\xi_1 + \cdots + \xi_n}{\sqrt{n\log(n)}} = N(0,1) $$
I am not so sure how to approach this problem as I've never had to find a distribution in terms of more than one variable before. Typically, I can just work backwards and use the pdf/cdf of a single random variable, but when I try that for $\xi$, I get:
$$P(\xi \leq x) = P(\theta/\sqrt{\eta} \leq x) \ldots $$
and you can't really move on from here. I guess you can write it as $P(\theta \leq x\sqrt{\eta})$ so that it's the cdf of $\theta$, but I can't do much more from here.
I tried all sorts of things like conditioning on $\eta$, but I guess there must be some easier way to do this.
Your initial approach is the right idea, and you can get more leverage out of it than you think.
First, note that $\xi$ is supported on $[-1, 1]$. We'll begin by evaluating $\mathbb P(\xi \leq x)$ for $x \in [-1, 0)$. In order to have $\xi \leq x$, we first need to have $\theta = -1$, so $$\mathbb P(\xi \leq x) = \mathbb P(\theta= -1, -1/\sqrt \eta \leq x) = \mathbb P \left(\eta \geq \frac 1 {x^2} \right) \mathbb P(\theta = -1).$$ The left term gives you a foothold to use the CDF of $\eta$, and the right term is $1/2$.
If $x \in (0, 1]$, then we begin with $\mathbb P(\xi \leq x) = 1 - \mathbb P(\xi > x)$ and note that the event $\{\xi > x\}$ requires $\theta = 1$ to proceed similarly.
Can you take it from here?