Question

Let \( X_1, X_2, \dots \) be a sequence of i.i.d. discrete random variables with the probability mass function \[ P(X_1 = m) = \frac{(\log 2)^m}{2(m!)} \quad \text{for} \quad m = 0, 1, 2, \dots \] If \( S_n = X_1 + X_2 + \dots + X_n \), then which one of the following sequences of random variables converges to 0 in probability?

• A. \( \frac{S_n}{n \log 2} \)
• B. \( \frac{S_n}{n} \log 2 \)
• C. \( S_n - \log 2 \)
• D. \( \frac{S_n - n}{\log 2} \)

Hint:

In probability theory, to check for convergence in probability, look at the behavior of the sample mean \( \frac{S_n}{n} \) and apply the Law of Large Numbers.

Answer 1

The Correct Option is B

Step 1: Understanding the distribution of \( X_1, X_2, \dots \).
The random variables \( X_1, X_2, \dots \) are i.i.d., and their distribution is given by the probability mass function \( P(X_1 = m) \). To find which sequence of random variables converges to 0, we analyze the expected value and variance of \( S_n \).
Step 2: Analyze the options.
Using the Law of Large Numbers, we know that \( \frac{S_n}{n} \) converges to the expected value of \( X_1 \), and the variance of \( S_n \) decreases with \( n \). This allows us to conclude that \( \frac{S_n}{n \log 2} \) converges to 0 in probability.
Step 3: Conclusion.
The correct answer is (A) \( \frac{S_n}{n \log 2} \).