Convergence in Probability and in the Mean, Part 2

What this lecture covers

This short problem continues the previous part's comparison of random variable sequences by introducing a new, stronger notion of convergence: convergence in mean square. It proves that mean-square convergence implies convergence in probability, then uses the earlier sequence Y_n to show the converse is false.

It follows directly from Part 1's analysis of X_n and Y_n and their convergence in probability. After watching, you should be able to state the definition of mean-square convergence, prove it implies convergence in probability using the Markov inequality, and explain why the reverse implication fails.

Key ideas

• Mean-square convergence: X_n converges to c in mean square if E[(X_n - c)^2] -> 0 as n -> infinity; when c equals a constant mean, this is equivalent to the variance of X_n going to 0.
• Mean-square implies convergence in probability: squaring the inequality |X_n - c| > epsilon and applying the Markov inequality to (X_n - c)^2 shows P(|X_n - c| > epsilon) -> 0 whenever E[(X_n - c)^2] -> 0.
• The converse is false: convergence in probability does not imply mean-square convergence.
• Counterexample: Y_n (value n with probability 1/n, else 0) converges to 0 in probability, but E[Y_n^2] = n, which diverges, so Y_n does not converge to 0 in mean square.
• Hierarchy of convergence notions: mean-square convergence is strictly stronger than convergence in probability.

Before you watch

• Watch Part 1 of this problem first, which defines X_n and Y_n and establishes their convergence in probability.
• Review the Markov inequality and the definition of variance.

Check your understanding

1. How does squaring the inequality |X_n - c| > epsilon let you apply the Markov inequality to prove convergence in probability?
2. Why does Y_n's convergence in probability to 0 not carry over to mean-square convergence?
3. If a sequence has a constant mean, how does mean-square convergence relate to its variance?