Question

Let \( \{ X_n \}_{n \geq 1} \) be a sequence of independent random variables with \[ \Pr(X_n = -\frac{1}{2^n}) = \Pr(X_n = \frac{1}{2^n}) = \frac{1}{2}, \quad \forall n \in \mathbb{N}. \] Suppose that \( \sum_{i=1}^{n} X_i \) converges to \( U \) as \( n \to \infty \). Then \( 6 \Pr(U \leq \frac{2}{3}) \) equals ___________ (answer in integer).

Hint:

When dealing with sums of independent random variables with symmetric distributions, their limit often results in a uniform distribution if the individual steps decrease exponentially.

Answer 1

Correct Answer: 5

Step 1: Understanding the distribution of \( X_n \). 
Each \( X_n \) is a random variable with two possible outcomes, \( -\frac{1}{2^n} \) and \( \frac{1}{2^n} \), each with probability \( \frac{1}{2} \). 
Step 2: Analyze the sum \( S_n = \sum_{i=1}^{n} X_i \).
The random variables \( X_n \) are independent and symmetric, meaning that each step contributes positively or negatively by a decreasing amount, \( \pm \frac{1}{2^n} \). As \( n \to \infty \), the sum \( S_n = \sum_{i=1}^{n} X_i \) converges to the limit \( U \), which is a random variable. The limit \( U \) is the sum of the infinite series: \[ U = \sum_{n=1}^{\infty} X_n. \] This sum converges because the magnitude of each term decreases exponentially. 
Step 3: Distribution of \( U \). 
The limiting distribution of \( U \) is a uniform distribution on the interval \( [-1, 1] \), since the steps \( X_n \) contribute to the value of \( U \) in a balanced way and the sum converges. Therefore, \( U \) is uniformly distributed on \( [-1, 1] \). 
Step 4: Calculate \( \Pr(U \leq \frac{2}{3}) \). 
For a uniform random variable \( U \) on \( [-1, 1] \), the probability that \( U \leq \frac{2}{3} \) is simply the proportion of the interval \( [-1, 1] \) that is less than or equal to \( \frac{2}{3} \). The length of the interval from \( -1 \) to \( \frac{2}{3} \) is: \[ \frac{2}{3} - (-1) = \frac{5}{3}. \] Since the total length of the interval \( [-1, 1] \) is 2, the probability is: \[ \Pr(U \leq \frac{2}{3}) = \frac{\frac{5}{3}}{2} = \frac{5}{6}. \] Step 5: Multiply by 6. 
We are asked to find \( 6 \Pr(U \leq \frac{2}{3}) \), so: \[ 6 \times \frac{5}{6} = 5. \] Thus, the correct answer is \( \boxed{5} \).