Let \( X_1, X_2, \dots \) be a sequence of i.i.d. continuous random variables with the probability density function
If \( S_n = X_1 + X_2 + \dots + X_n \) and \( \bar{X}_n = \frac{S_n}{n} \), then the distributions of which of the following sequences of random variables converge(s) to a normal distribution with mean 0 and a finite variance?
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The Central Limit Theorem applies to sums of i.i.d. random variables. In this case, the normalized sum \( \sqrt{n} \left( \bar{X}_n - \mu \right) \) converges to a normal distribution.
Step 1: Understand the setup. We are given that \( X_1, X_2, \dots \) are i.i.d. continuous random variables. The Central Limit Theorem (CLT) states that the distribution of \( \frac{S_n - n\mu}{\sigma \sqrt{n}} \) will approach a normal distribution as \( n \to \infty \), where \( \mu \) and \( \sigma^2 \) are the mean and variance of the individual random variables. In this case, the mean of \( X_i \) is \( \mu = 1 \), and the variance is \( \sigma^2 = \frac{1}{4} \). Therefore, CLT applies to \( \sqrt{n} \left( \bar{X}_n - 1 \right) \). Step 2: Analyzing the options. (A) \( \frac{S_n - n}{\sqrt{n}} \): This is incorrect. The expected value is \( n \), not 0. (B) \( \frac{S_n}{\sqrt{n}} \): This does not give a distribution with mean 0. The sum \( S_n \) grows linearly, so this option is not correct. (C) \( \sqrt{n} \left( \bar{X}_n - 1 \right) \): This is correct. By the Central Limit Theorem, \( \sqrt{n} \left( \bar{X}_n - 1 \right) \) will converge to a normal distribution with mean 0 and finite variance. (D) \( \sqrt{n} \left( \bar{X}_n - 1 \right)/2 \): This is incorrect. While it is related to (C), the factor of 2 in the denominator changes the distribution. Step 3: Conclusion. The correct answer is C, as it represents the normalized form of the sample mean converging to a normal distribution.
