Question

Let \( \{X_n\}_{n \geq 1} \) be a sequence of independent and identically distributed random variables with mean 0 and variance 1, all of them defined on the same probability space. For \( n = 1, 2, 3, \dots \), let \[ Y_n = \frac{1}{n} \left( X_1 X_2 + X_3 X_4 + \dots + X_{2n-1} X_{2n} \right). \] Then which one of the following statements is/are true?

• A. $\{ \sqrt{n} Y_n \}_{n \geq 1}$ converges in distribution to a standard normal random variable.
• B. $\{ Y_n \}_{n \geq 1}$ converges in 2nd mean to 0.
• C. $\{ Y_n + \frac{1}{n} \}_{n \geq 1}$ converges in probability to 0.
• D. $\{ X_n \}_{n \geq 1}$ converges almost surely to 0.

Hint:

- The Central Limit Theorem helps to understand convergence to the normal distribution for sums of independent random variables.
- Convergence in second mean (mean square) implies that the sequence converges to zero in expectation and variance.
- Convergence in probability means that the sequence will get arbitrarily close to the limiting value with increasing probability as $n$ increases.

Answer 1

The Correct Option is A

1) Analyzing statement (A):
$\{ \sqrt{n} Y_n \}_{n \geq 1}$ involves scaling the sequence $Y_n$ by $\sqrt{n}$. Since $Y_n$ is the sum of independent random variables with mean 0 and variance 1, by the Central Limit Theorem, this sequence converges in distribution to a standard normal random variable.

2) Analyzing statement (B):
Since $Y_n$ is the average of independent identically distributed random variables, we expect that $Y_n$ converges in 2nd mean (or mean square) to 0, as the variance of $Y_n$ tends to 0 with increasing $n$.

3) Analyzing statement (C):
$\{ Y_n + \frac{1}{n} \}_{n \geq 1}$ converges in probability to 0 because the addition of $\frac{1}{n}$ does not affect the convergence of $Y_n$ to 0, and $Y_n$ converges to 0 in probability.

Thus, the correct answer is (A), (B), and (C).