Events A_n are independent and satisfy P(A_n) = 1/n for all n ≥ 1. What does the Borel-Cantelli theory predict about how often these events occur?
AP(lim sup A_n) = 0, because each individual probability P(A_n) → 0
BP(lim sup A_n) = 1/2, by the law of large numbers
CP(lim sup A_n) = 1, because the A_n are independent and Σ P(A_n) = Σ 1/n diverges
DP(lim sup A_n) cannot be determined without knowing the sample space
This is precisely the setup for the second Borel-Cantelli lemma: independence plus divergent sum. Even though P(A_n) → 0, the sum Σ 1/n is the harmonic series and diverges. With independence, the second lemma guarantees P(lim sup A_n) = 1 — the events occur infinitely often with probability 1. The common mistake is thinking that because individual probabilities go to zero, the events must eventually stop. The divergent sum overrides that intuition.
Question 2 Multiple Choice
Events A_n are defined such that P(A_n) = 1/n, but they are NOT independent. What can be concluded about P(lim sup A_n)?
AP(lim sup A_n) = 1, because the sum Σ P(A_n) diverges
BP(lim sup A_n) = 0, because independence fails
CThe second Borel-Cantelli lemma does not apply; P(lim sup A_n) could be anywhere in [0, 1]
DP(lim sup A_n) = 1/2, by symmetry of the non-independence
Independence is genuinely required by the second Borel-Cantelli lemma — it is not just a technical nicety. Without independence, a divergent sum does not guarantee P(lim sup A_n) = 1. For example, if all A_n are the same event (say, a coin lands heads), then P(A_n) = 1/2, the sum diverges, but P(lim sup A_n) = 1/2, not 1. The proof of the second lemma relies on the product structure that independence provides.
Question 3 True / False
If Σ P(A_n) diverges and the events A_n are independent, then P(lim sup A_n) = 1, regardless of any other properties of the sequence.
TTrue
FFalse
Answer: True
This is exactly the second Borel-Cantelli lemma. Independence plus a divergent sum is sufficient to guarantee that the lim sup has probability 1 — meaning the events recur infinitely often almost surely. The key conditions are independence and divergence of the sum; no other assumptions about the events are needed.
Question 4 True / False
If Σ P(A_n) diverges, then P(lim sup A_n) = 1 regardless of whether the events are independent.
TTrue
FFalse
Answer: False
This is the most common mistake when applying Borel-Cantelli. The second lemma requires independence — a divergent sum alone is not enough. Without independence, you cannot use the exponential product bound that drives the proof. There exist explicit counterexamples where Σ P(A_n) = ∞ but P(lim sup A_n) = 0 (e.g., if A_n = A for all n with P(A) > 0 but less than 1, the sum diverges yet the lim sup equals A, not the whole space). The first lemma (convergent sum → probability 0) needs no independence; the second does.
Question 5 Short Answer
Explain why the independence assumption is essential for the second Borel-Cantelli lemma but is not needed for the first.
Think about your answer, then reveal below.
Model answer: The first lemma follows purely from the union bound (subadditivity of probability) and the assumption that series tail sums go to zero — no independence is needed. The second lemma requires independence because its proof multiplies individual non-occurrence probabilities across trials: P(none of A_{n+1},...,A_{n+k} occur) = ∏(1 − P(A_i)), which requires independence. This product is then bounded by exp(−Σ P(A_i)) → 0 when the sum diverges. Without independence, you cannot factor the joint probability this way, and the argument collapses.
Independence is a statement about the joint probability structure: it allows you to factor P(∩ A_i^c) = ∏ P(A_i^c). The exponential bound 1 − x ≤ e^{−x} then lets you show the probability of avoiding all events in a tail goes to zero. The first lemma is purely about marginal probabilities and their sum — subadditivity works event-by-event without needing to know how events relate to each other. This asymmetry is fundamental: ruling something out (first lemma) is easier than guaranteeing recurrence (second lemma).