Discussion
Let \(X_{1}, X_{2}, \ldots\) be a sequence of independent random variables with
\[ \begin{aligned} P(X_{n} = n) &= \frac{1}{n^{2}} = 1 - P(X_{n} = 0), \\ &\text{for } n = 1, 2, \ldots. \end{aligned} \]
Consider the following statements.
P: \(\{X_{n}\}\) converges in mean-square to \(0\).
Q: \(\{X_{n}\}\) converges in probability to \(0\).
Then
Answer: B
P is NOT correct and Q is correct.
Mean square: \(E(X_{n}^{2}) = n^{2} \cdot \frac{1}{n^{2}} + 0 = 1\) for every \(n\). It does not tend to 0, so \(X_{n}\) does not converge to 0 in mean square.
Probability: for any \(\varepsilon\) with \(0 \lt \varepsilon \lt 1\),
\[ \begin{aligned} P(|X_{n}| \gt \varepsilon) &= P(X_{n} = n) \\ &= \frac{1}{n^{2}} \to 0 \end{aligned} \]
so \(X_{n} \to 0\) in probability. The example shows that convergence in probability does not imply convergence in mean square: the rare value \(n\) is large enough to keep the second moment at 1. (Since \(\sum \frac{1}{n^{2}}\) converges, the Borel-Cantelli lemma even gives almost sure convergence.)
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