After 1000 fair coin flips, the sample proportion of heads is 0.52. A student claims the WLLN guarantees it will eventually equal exactly 0.5. What does the WLLN actually guarantee?
AThat the sample proportion must equal 0.5 exactly once n is large enough
BThat for any fixed ε > 0, the probability that the sample proportion differs from 0.5 by more than ε approaches 0 as n → ∞ — but individual sequences need not converge
CThat the sample proportion will converge to 0.5 at every possible infinite sequence of coin flips
DThat the sample proportion will decrease monotonically toward 0.5 after 1000 flips
The WLLN guarantees convergence in probability: P(|Sₙ/n − μ| > ε) → 0. This means the probability of being far from μ vanishes, but it does not mean any particular sample path converges. Individual sequences can fluctuate forever. The statement in option C describes almost sure convergence — the conclusion of the Strong Law, which is a strictly stronger result requiring a harder proof.
Question 2 Multiple Choice
The standard proof of the WLLN under finite variance σ² applies Chebyshev's inequality to Sₙ/n and obtains P(|Sₙ/n − μ| > ε) ≤ σ²/(nε²). What property of the random variables is essential for computing Var(Sₙ/n) = σ²/n?
AThat the variables have finite mean μ, which allows the expectation to be computed
BThat the variables are identically distributed, so each has the same variance σ²
CThat the variables are independent, which makes their variances additive: Var(Sₙ) = nσ²
DThat the variables take values in a bounded interval, ensuring Chebyshev applies
Independence is the key structural property that makes variances additive: for independent random variables, Var(X₁ + ··· + Xₙ) = Var(X₁) + ··· + Var(Xₙ) = nσ². Without independence, covariance terms appear and Var(Sₙ) could be much larger than nσ², preventing the bound σ²/(nε²) from going to zero. Identical distribution ensures each term contributes the same σ², but it is independence that allows addition of variances.
Question 3 True / False
The Weak Law of Large Numbers implies that individual sample paths of Sₙ/n is expected to converge to μ at nearly every outcome ω in the sample space.
TTrue
FFalse
Answer: False
This is the crucial distinction between the WLLN and the Strong Law. Convergence in probability means P(|Sₙ/n − μ| > ε) → 0: the probability mass outside any ε-neighborhood of μ goes to zero. But specific sample paths can still fluctuate outside that neighborhood forever, as long as the set of such paths has probability approaching zero. Almost sure convergence (P(Sₙ/n → μ) = 1), guaranteed by the Strong LLN, is the statement that essentially every sample path converges.
Question 4 True / False
The proof of the WLLN relies on the fact that for i.i.d. random variables with finite variance, the variance of the sample mean Sₙ/n tends to zero as n → ∞.
TTrue
FFalse
Answer: True
Var(Sₙ/n) = σ²/n → 0 as n → ∞. Chebyshev's inequality converts this vanishing variance into a vanishing probability: P(|Sₙ/n − μ| > ε) ≤ Var(Sₙ/n)/ε² = σ²/(nε²) → 0. The entire proof is a two-step argument: independence makes variance additive, causing Var(Sₙ/n) to shrink; Chebyshev converts that shrinking variance into a probability bound.
Question 5 Short Answer
Explain the difference between convergence in probability and almost sure convergence in the context of sample means. Why can the WLLN hold even while some individual sample paths continue to fluctuate away from μ?
Think about your answer, then reveal below.
Model answer: Convergence in probability means the probability of the sample mean being far from μ goes to zero: for any ε > 0, P(|Sₙ/n − μ| > ε) → 0. Almost sure convergence means the sample mean converges to μ at every outcome except a set of probability zero: P(lim Sₙ/n = μ) = 1. The WLLN guarantees the former. Individual sample paths can still fluctuate indefinitely away from μ — as long as the set of such paths has probability shrinking to zero. Convergence in probability is a statement about the distribution of the sample mean, not about every individual trajectory.
This distinction matters for understanding the limits of the WLLN. It justifies using sample means to estimate population means in practice (the probability of a bad estimate vanishes), without requiring that every conceivable sequence of observations converges — which would be a far stronger guarantee.