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Let \(\{X_{n}, n \ge 1\}\) be independent and identically distributed random variables with mean 0 and variance 1. Which of the following is true as \(n \to \infty\) ?

  • A.\(\sum_{i=1}^{n} X_{i} / \sqrt{n}\) converges in distribution to a degenerate random variable
  • B.\(\sum_{i=1}^{n} X_{i} / \sum_{i=1}^{n} X_{i}^{2}\) converges in probability to 0
  • C.\(\sqrt{n} \sum_{i=1}^{n} X_{i}\) converges in distribution to a non-zero random variable
  • D.\(\dfrac{\sum_{i=1}^{n} X_{i}^{2}}{\sqrt{n} \sum_{i=1}^{n} X_{i}}\) converges in probability to 1

Answer: B

\(\sum X_{i} / \sum X_{i}^{2}\) converges in probability to 0.

Divide numerator and denominator by \(n\). By the weak law of large numbers, \(\frac{1}{n}\sum X_{i} \to E X = 0\) and \(\frac{1}{n}\sum X_{i}^{2} \to E X^{2} = 1\) in probability. The ratio of the two converges to \(0/1 = 0\).

The others fail. By the central limit theorem \(\sum X_{i}/\sqrt{n} \to N(0, 1)\), a non-degenerate law. \(\sqrt{n}\sum X_{i} = n \cdot (\sum X_{i}/\sqrt{n})\) blows up rather than converging. And dividing the last ratio through by \(n\) gives \(\dfrac{\frac{1}{n}\sum X_{i}^{2}}{\sum X_{i}/\sqrt{n}}\), which behaves like \(1/Z\) with Z standard normal, which is random, not 1.

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