Question:
Let \( \{X_n\}_{n \geq 1} \) be a sequence of i.i.d. random variables having common probability density function
Let \( \bar{X}_n = \frac{1}{n} \sum_{i=1}^{n} X_i \), \( n = 1, 2, \dots \). Then \[ \lim_{n \to \infty} P(\bar{X}_n = 2) \] equals
Let \( \{X_n\}_{n \geq 1} \) be a sequence of i.i.d. random variables having common probability density function
Let \( \bar{X}_n = \frac{1}{n} \sum_{i=1}^{n} X_i \), \( n = 1, 2, \dots \). Then \[ \lim_{n \to \infty} P(\bar{X}_n = 2) \] equals
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The strong law of large numbers tells us that the sample mean converges almost surely to the expected value, meaning the probability of the sample mean equaling the expected value exactly is 0.
Updated On: Dec 15, 2025
- 0
- \( \frac{1}{4} \)
- \( \frac{1}{2} \)
- 1
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the distribution.
The random variables \( X_n \) are i.i.d. with a probability density function given by \( f(x) = x e^{-x} \), which is the gamma distribution with shape parameter 2 and rate parameter 1. The mean \( \mathbb{E}[X_n] \) is 2.
Step 2: Applying the law of large numbers.
By the strong law of large numbers, \( \bar{X}_n \) converges almost surely to the expected value \( \mathbb{E}[X_n] = 2 \). Therefore, the probability that \( \bar{X}_n \) equals exactly 2 becomes 0 as \( n \to \infty \).
Step 3: Conclusion.
Thus, the correct answer is (A) 0.
The random variables \( X_n \) are i.i.d. with a probability density function given by \( f(x) = x e^{-x} \), which is the gamma distribution with shape parameter 2 and rate parameter 1. The mean \( \mathbb{E}[X_n] \) is 2.
Step 2: Applying the law of large numbers.
By the strong law of large numbers, \( \bar{X}_n \) converges almost surely to the expected value \( \mathbb{E}[X_n] = 2 \). Therefore, the probability that \( \bar{X}_n \) equals exactly 2 becomes 0 as \( n \to \infty \).
Step 3: Conclusion.
Thus, the correct answer is (A) 0.
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