A casino game has a house edge of μ = $0.05 per play. The Weak Law of Large Numbers guarantees that for large n, the average profit per game is close to $0.05 with high probability. What does the Strong Law add that the Weak Law does not?
AThe Strong Law guarantees the total profit grows without bound
BThe Strong Law guarantees that with probability 1, the running average of profit per game converges permanently to $0.05, not just that it is close at each fixed large n
CThe Strong Law guarantees convergence for dependent random variables, while the Weak Law requires independence
DThe Strong Law gives a faster rate of convergence than the Weak Law
The Weak Law says P(|Sₙ/n − μ| > ε) → 0 for any fixed ε — a statement about snapshots at individual n. It does not rule out the average escaping ε infinitely often, as long as those excursions become progressively rarer. The Strong Law closes this gap: P(limₙ Sₙ/n = μ) = 1. This is almost sure convergence — the set of sample paths that fail to converge to μ has probability zero. The trajectory itself settles down permanently.
Question 2 Multiple Choice
A textbook states: 'Since P(|Sₙ/n − μ| > ε) → 0 for all ε, the sample average must eventually stay within ε of μ for all sufficiently large n.' Is this a valid conclusion from the Weak Law alone?
AYes — this follows immediately from the definition of convergence in probability
BNo — convergence in probability only controls probabilities at each fixed n; it does not prevent the average from returning outside ε infinitely often as n grows
CYes, but only when the random variables are bounded
DNo — this would require the Central Limit Theorem, not just the Weak Law
This is the key misconception separating the Weak and Strong Laws. Convergence in probability says: for any fixed ε, the probability of exceeding it at step n goes to zero. But it allows the running average to wander outside ε on a sparse-but-infinite set of times — the 'occasionally bad' behavior can persist forever while still having vanishingly small probability at each individual n. Almost sure convergence (the Strong Law) rules this out by asserting that with probability 1, there exists N such that for all n ≥ N, |Sₙ/n − μ| < ε.
Question 3 True / False
Almost sure convergence implies convergence in probability.
TTrue
FFalse
Answer: True
Almost sure convergence is strictly stronger. If Sₙ/n → μ almost surely, then for any ε > 0, the set of ω where Sₙ(ω)/n fails to be close to μ for large n has probability zero — and in particular, the probability of the event {|Sₙ/n − μ| > ε} must go to zero. The reverse is not true: there exist sequences that converge in probability but not almost surely (the 'sliding bump' example in probability textbooks).
Question 4 True / False
The classical Strong Law of Large Numbers requires that the random variables have finite second moment (finite variance).
TTrue
FFalse
Answer: False
The SLLN holds under the weaker condition that the Xᵢ are i.i.d. with finite first moment E[|X₁|] < ∞. Finite variance is sufficient for the Weak Law via Chebyshev's inequality, but it is not necessary for the Strong Law. The proof for unbounded variables uses a truncation argument — approximate Xᵢ by truncated versions with bounded support, prove the SLLN for those, then show the error from truncating is negligible almost surely.
Question 5 Short Answer
Explain why the Weak Law of Large Numbers does not guarantee that the sample average 'eventually stays close' to μ, and what the Strong Law adds to give this guarantee.
Think about your answer, then reveal below.
Model answer: The Weak Law says P(|Sₙ/n − μ| > ε) → 0 for each fixed ε — a pointwise statement about individual time steps. This is compatible with the average oscillating away from μ infinitely often, as long as those excursions become increasingly rare. The Strong Law asserts P(lim_{n→∞} Sₙ/n = μ) = 1 — a statement about the whole trajectory. Almost sure convergence means: except for a set of sample paths of measure zero, the running average eventually settles within ε of μ and stays there. The Borel-Cantelli approach makes this precise: by showing the sum of probabilities P(|Sₙ/n − μ| > ε) converges, one concludes (via first Borel-Cantelli) that the bad events occur only finitely many times almost surely.