Computing the distribution of a random variable that involves two other random variables

Let $$\eta \sim \mathcal{U}(0, 1)$$, and let $$\theta$$ be an independent Bernoulli random variable such that $$P(\theta = 1) = P(\theta = -1) = 1/2$$.

1. Determine the distribution of $$\xi := \theta/\sqrt{\eta}$$.

2. 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.

• Are $\theta$ and $\eta$ independent? – sorva Dec 11 '19 at 10:59
• Yes, they are independent – hom Dec 11 '19 at 11:00
• Does $\mathcal{U}(0,1)$ here mean uniform distribution? – Ajay Kumar Nair Dec 11 '19 at 14:02

$$\mathbf{Hint}$$ for (1): $$P(\frac{\theta}{\sqrt{\eta}} \leq t) = P(\frac{1}{\sqrt{\eta}} \leq t, \theta =1) + P(\frac{-1}{\sqrt{\eta}} \leq t, \theta =-1).$$ Now use independence and properties of uniform distribution.

$$\mathbf{Hint}$$ for (2): Use Lindeberg-Feller C.L.T

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?

• Isn't the left term $P(\eta \geq \frac{1}{x^2})$? – Ajay Kumar Nair Dec 11 '19 at 14:43
• @AjayKumarNair Ah, yep -- thanks for the catch! – Aaron Montgomery Dec 11 '19 at 14:45