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Probability Unit 10: Limit Theorems (LLN, CLT)

Limit theorems describe what happens when many independent random variables are added or averaged. The law of large numbers explains why averages settle down, and the central limit theorem lets you approximate sums with the normal distribution.

Covers: Markov's inequality, Chebyshev's inequality, Distribution of the sample mean, Weak law of large numbers, Central limit theorem for sums and means, Normal approximations with continuity correction, Sample size calculations

Study guide

Probability inequalities bound tail probabilities using only a few moments. Markov's inequality says that for nonnegative X, P(X ≥ a) ≤ E[X]/a. Applying it to (X − μ)² gives Chebyshev's inequality, P(|X − μ| ≥ kσ) ≤ 1/k², so at least 75% of any distribution lies within 2 standard deviations of its mean. These bounds hold for every distribution, which makes them useful when the distribution is unknown, but they are often loose: for a normal distribution the true probability outside 2 SDs is about 4.6%, far below Chebyshev's 25%.

Let X̄ₙ be the mean of n iid observations with mean μ and variance σ². Then E[X̄ₙ] = μ and Var(X̄ₙ) = σ²/n, so its standard deviation, the standard error, is σ/√n. The sum Sₙ has mean nμ and SD σ√n. Notice the square roots: averaging four times as many observations only halves the standard error. The weak law of large numbers follows from Chebyshev: P(|X̄ₙ − μ| ≥ ε) ≤ σ²/(nε²), which goes to 0. In words, the sample mean converges in probability to the population mean.

The central limit theorem says more: for large n, (Sₙ − nμ)/(σ√n) is approximately standard normal, whatever the shape of the population, as long as the variance is finite. Equivalently, X̄ₙ ≈ N(μ, σ²/n). To use it, find the mean and SD of the sum or mean, standardize, and read Φ. Insurance problems use it for total claims across many policies, and reliability problems use it for total lifetimes. For groups of different policy types, add the means and variances of all groups before standardizing.

When the sum is integer-valued, as with counts, dice totals or Poisson demand, use a continuity correction: P(S ≥ 380) becomes P(Y > 379.5). The CLT also explains the normal approximations to the binomial and to the Poisson with large mean. Sample-size questions reverse the calculation: set z × σ/√n equal to the allowed error and solve for n, always rounding up. Applied to logarithms, the CLT shows that products of many positive factors are approximately lognormal.

Key ideas

Formulas and rules

Markov: P(X ≥ a) ≤ E[X]/a for X ≥ 0
Chebyshev: P(|X − μ| ≥ kσ) ≤ 1/k²
E[X̄] = μ; Var(X̄) = σ²/n; E[Sₙ] = nμ; Var(Sₙ) = nσ²
CLT: (Sₙ − nμ)/(σ√n) ≈ N(0, 1); X̄ ≈ N(μ, σ²/n)
Sample proportion: p̂ ≈ N(p, p(1 − p)/n)
Sample size: n ≥ (zσ/E)², rounded up

Common mistakes

Before the test, make sure you can

Words to know

Markov's inequality
For X ≥ 0 and a > 0, P(X ≥ a) ≤ E[X]/a.
Chebyshev's inequality
P(|X − μ| ≥ kσ) ≤ 1/k² for any distribution with finite variance.
Sample mean
X̄ₙ = (X₁ + ··· + Xₙ)/n.
Standard error
The SD of the sample mean, σ/√n.
Weak law of large numbers
X̄ₙ converges in probability to μ as n → ∞.
Convergence in probability
P(|Yₙ − c| ≥ ε) → 0 for every ε > 0.
Central limit theorem
Standardized sums of iid finite-variance variables approach N(0, 1).
Convergence in distribution
The CDFs of Yₙ approach the CDF of a limit distribution.
Continuity correction
Shifting an integer boundary by 0.5 before using a normal approximation.
Normal approximation to the binomial
Binomial(n, p) ≈ N(np, np(1 − p)) for large n.
Normal approximation to the Poisson
Poisson(λ) ≈ N(λ, λ) for large λ.
Lognormal distribution
The distribution of Y when ln Y is normal, approximating products of many positive factors.
iid
Independent and identically distributed.

Sample practice questions

30 more questions like these, from level 1 to 5, are in the practice set.

Question 1 (level 1)

For a nonnegative random variable X and a > 0, what does Markov's inequality state?

  1. P(X ≥ a) = E[X]/a
  2. P(X ≥ a) ≤ Var(X)/a²
  3. P(X ≥ a) ≥ E[X]/a
  4. P(X ≥ a) ≤ E[X]/a
Show the answer and steps

Answer: D) P(X ≥ a) ≤ E[X]/a

  1. Markov's inequality needs only nonnegativity and a finite mean.
  2. It gives an upper bound: P(X ≥ a) ≤ E[X]/a.
  3. Var(X)/a² belongs to Chebyshev's inequality, applied to |X − μ|.
Question 2 (level 1)

What does Chebyshev's inequality state for a random variable with mean μ and SD σ, and k > 0?

  1. P(|X − μ| ≥ kσ) ≤ 1/k
  2. P(|X − μ| ≥ kσ) = 1/k²
  3. P(|X − μ| ≥ kσ) ≤ 1/k²
  4. P(|X − μ| ≤ kσ) ≤ 1/k²
Show the answer and steps

Answer: C) P(|X − μ| ≥ kσ) ≤ 1/k²

  1. Apply Markov's inequality to (X − μ)² with a = k²σ².
  2. P((X − μ)² ≥ k²σ²) ≤ σ²/(k²σ²) = 1/k².
  3. It is a bound valid for every distribution, not an exact probability.
Question 3 (level 3)

An insurer has 400 independent policies, each with claim mean $250 and SD $1,000. Using the CLT, what is the probability (to four decimals) that total claims exceed $110,000?

  1. 0.0228
  2. 0.4900
  3. 0.3085
  4. 0.1587
Show the answer and steps

Answer: C) 0.3085

  1. Mean total = 400 × 250 = $100,000; SD = 1,000 × √400 = $20,000.
  2. z = (110,000 − 100,000)/20,000 = 0.5.
  3. P(Z > 0.5) ≈ 0.3085; 0.4900 uses the SD 400 × 1,000 = $400,000.