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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.

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

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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?

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

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