Central limit theorem: formula sheet

Key formulas

Expectation of a sum

E[X₁ + … + Xₙ] = E[X₁] + … + E[Xₙ]

Always true. No independence needed.

Linear combination

E[aX + bY] = aE[X] + bE[Y]; Var(aX + bY) = a²Var(X) + b²Var(Y) + 2ab Cov(X, Y)

The covariance term is zero if X and Y are independent (or uncorrelated).

Variance of a sum (independent)

Var(X₁ + … + Xₙ) = Var(X₁) + … + Var(Xₙ)

Needs independence, or at least zero covariances between all pairs.

Sum of n iid variables

E[S] = nμ; Var(S) = nσ²; SD(S) = σ√n

Mean and variance of each variable are μ and σ².

Sample mean

E[X̄] = μ; Var(X̄) = σ² ÷ n; SD(X̄) = σ ÷ √n

SD of X̄ is the standard error. Needs independent, identically distributed variables.

MGF of a sum

M_{X+Y}(t) = M_X(t) × M_Y(t)

For independent X and Y. For n iid variables, M_S(t) = [M_X(t)]ⁿ.

MGF of a linear function

M_{aX+b}(t) = e^{bt} × M_X(at)

Use this for the sample mean: M_X̄(t) = [M_X(t ÷ n)]ⁿ.

CLT for the sum

(Sₙ − nμ) ÷ (σ√n) → N(0,1) in distribution as n → ∞

Needs X₁, …, Xₙ iid with finite variance σ² > 0 and mean μ.

CLT for the sample mean

(X̄ₙ − μ) ÷ (σ/√n) → N(0,1) in distribution

Same conditions. Standard deviation of X̄ₙ is σ/√n.

Approximate distributions for large n

Sₙ ≈ N(nμ, nσ²); X̄ₙ ≈ N(μ, σ²/n)

Approximation, not exact. The second parameter is the variance, not the standard deviation.

MGF of standardised sum

M_Zₙ(t) = [M_Y(t ÷ √n)]ⁿ → exp(t²/2), where Y = (X − μ) ÷ σ and Zₙ = (Sₙ − nμ) ÷ (σ√n)

Basis of the MGF proof: M_Y(s) = 1 + s²/2 + o(s²) since E[Y] = 0 and Var(Y) = 1.

Mean and variance of Sₙ and X̄ₙ

E[Sₙ] = nμ; Var(Sₙ) = nσ²; E[X̄ₙ] = μ; Var(X̄ₙ) = σ²/n

Uses independence for the variance.

Binomial approximation

X ~ Bin(n, p) ≈ N(np, np(1 − p))

Works well for large n with p not near 0 or 1. A common rule of thumb is np and n(1 − p) both above about 5 to 10; it is a guide, not a law.

Poisson approximation

X ~ Poisson(λ) ≈ N(λ, λ)

Mean and variance are both λ. Better for larger λ, for example above about 10 to 20.

Gamma approximation

X ~ Gamma(α, λ) ≈ N(α ÷ λ, α ÷ λ²)

Uses the rate parameter λ. If the question gives a scale parameter, convert first. Improves as α grows.

Chi-squared approximation

X ~ χ²(k) ≈ N(k, 2k)

Chi-squared is gamma with α = k/2 and λ = 1/2. Mean k, variance 2k.

Continuity correction

P(X ≤ r) ≈ P(Z ≤ (r + 0.5 − μ) ÷ σ); P(X ≥ r) ≈ P(Z ≥ (r − 0.5 − μ) ÷ σ)

For integer-valued X. P(X = r) ≈ P(r − 0.5 < Y < r + 0.5). Not used for gamma or chi-squared.

Standardisation

Z = (Y − μ) ÷ σ, where σ = √variance

Take the square root of the variance before dividing.

Normal approximation to the binomial

X ~ Bin(n, p) ⇒ Y ~ N(np, np(1 − p))

A common guide is that both np and n(1 − p) should be reasonably large, for example above 5. Check what the question says.

Normal approximation to the Poisson

X ~ Poisson(λ) ⇒ Y ~ N(λ, λ)

Works better for larger λ. The standard deviation is √λ, not λ.

Single value

P(X = k) ≈ P(k − 0.5 < Y < k + 0.5)

Valid for integer-valued X with spacing 1.

Lower tail, inclusive

P(X ≤ k) ≈ P(Y < k + 0.5)

Add 0.5 to the upper limit.

Lower tail, strict

P(X < k) = P(X ≤ k − 1) ≈ P(Y < k − 0.5)

Convert to ≤ first, then correct.

Upper tail, inclusive

P(X ≥ k) ≈ P(Y > k − 0.5)

Subtract 0.5 from the lower limit.

Upper tail, strict

P(X > k) = P(X ≥ k + 1) ≈ P(Y > k + 0.5)

Convert to ≥ first, then correct.

Range

P(a ≤ X ≤ b) ≈ P(a − 0.5 < Y < b + 0.5)

Widen the interval by 0.5 at each end. For strict inequalities, convert to inclusive integer limits first.

Standardising

Z = (Y − μ) ÷ σ

Apply it to the corrected limit, not the original integer.

Sum of n iid claims

S = X₁ + ... + Xₙ ≈ N(nμ, nσ²)

Needs independence, identical distribution, finite variance and large n.

Sample mean

X̄ ≈ N(μ, σ² ÷ n)

The standard deviation of X̄ is σ ÷ √n.

Standardisation

Z = (S − nμ) ÷ (σ√n) ≈ N(0, 1)

Use Z to read Φ from the normal table.

Compound distribution moments

E[S] = E[N] E[X]; Var(S) = E[N] Var(X) + Var(N) (E[X])²

Use when the number of claims N is random and independent of the claim amounts.

Compound Poisson with mean λ

E[S] = λ E[X]; Var(S) = λ E[X²]

Use E[X²], not Var(X). The CLT applies when λ is large.

Sample size for a mean

P(|X̄ − μ| ≤ ε) ≈ 2Φ(ε√n ÷ σ) − 1, so n ≥ (zσ ÷ ε)²

z is the standard normal value with P(|Z| ≤ z) equal to the required probability. For 95%, z = 1.96. Round n up.

Sample size for a proportion

n ≥ z² p(1 − p) ÷ ε²

If p is unknown, p = 0.5 gives the largest n and is the safe choice.

Continuity correction

For integer-valued S: P(S ≤ k) ≈ Φ((k + 0.5 − mean) ÷ sd)

Use it when S is discrete and takes integer values, for example a count of claims.

Quick revision

• For independent variables, E(ΣXᵢ) = ΣE(Xᵢ) and Var(ΣXᵢ) = ΣVar(Xᵢ).
• For iid variables, E(Sₙ) = nμ and Var(Sₙ) = nσ².
• E(X̄) = μ and Var(X̄) = σ²/n, so the standard deviation of X̄ is σ ÷ √n.
• CLT: for large n, Sₙ is approximately N(nμ, nσ²) and X̄ is approximately N(μ, σ²/n).
• The theorem needs independence, identical distribution and finite variance.
• Bin(n, p) is approximately N(np, np(1 − p)); Poisson(λ) is approximately N(λ, λ) for large λ.
• Continuity correction applies only when approximating a discrete variable. Use ±0.5 for integer values.
• P(X ≤ k) becomes P(Y < k + 0.5) and P(X ≥ k) becomes P(Y > k − 0.5).
• P(X < k) becomes P(Y < k − 0.5) and P(X > k) becomes P(Y > k + 0.5).
• Standardise with Z = (value − mean) ÷ standard deviation, using the standard deviation, not the variance.
• For sample size questions, solve for n and round up.
• A larger n gives a better approximation, and a very skewed distribution needs a larger n.

Common mistakes

• Adding variances for dependent variables without checking covariance. Fix: Always state independence (or zero covariance) before adding. Otherwise include 2Cov terms.
• Writing Var(X − Y) = Var(X) − Var(Y). Fix: Var(X − Y) = Var(X) + Var(Y) for independent variables. Variances never subtract.
• Saying the individual Xᵢ become normal as n grows. Fix: Say that the standardised sum or mean converges in distribution to N(0,1). The distribution of each Xᵢ does not change.
• Dividing by σ instead of σ/√n when standardising a mean. Fix: Always write Var(X̄ₙ) = σ²/n first, then take the square root.
• Forgetting the continuity correction for binomial or Poisson. Fix: Whenever the original variable is discrete, write the integer as an interval r ± 0.5 before standardising.
• Applying the correction in the wrong direction, for example using P(X ≥ 30) ≈ P(Y > 30.5). Fix: For X ≥ r, the value r must be included, so use Y > r − 0.5. For X > r, use Y > r + 0.5. Sketch the integers if unsure.
• Moving the boundary in the wrong direction, for example using 29.5 for P(X ≤ 30). Fix: Draw the integer bars. For ≤ 30, the bar for 30 must be included, so go up to 30.5.
• Treating X < k as X ≤ k and correcting to k + 0.5. Fix: Convert to integers first: X < k is X ≤ k − 1, so the cut-off is k − 0.5.
• Using variance nσ as the variance of the sum, or taking SD as nσ. Fix: Variance of the sum is nσ². SD is σ√n. Always compute SD last by taking a square root.
• Using σ² instead of σ in the sample size formula, or forgetting to square at the end. Fix: Solve ε√n ÷ σ = z step by step. First √n = zσ ÷ ε, then square to get n.

Exam tips

• Always write down the independence assumption when you add variances or multiply MGFs. Examiners give marks for stating it.
• In MCQs, check whether the question asks for variance or standard deviation, and for the sum or the mean. Wrong options are often the right answer to a nearby question.
• For MGF questions, write the product first, then simplify the exponents. Compare with the Tables to name the distribution.
• Be careful with differences and scaled variables, such as 2X − 3Y. Square the constants and add the variances.
• Use the check E[S] = nμ and Var(S) = nσ² on your final answer. It catches slips quickly.
• When asked to state the CLT, write four items: the iid assumption, finite mean and variance, the standardised expression, and convergence in distribution to N(0,1).
• For MGF proof questions, set Y = (X − μ)/σ, expand M_Y(s) = 1 + s²/2 + o(s²), raise to the power n with s = t/√n, and use (1 + a/n)ⁿ → eᵃ.
• Decide at once whether the question is about a sum or a mean. Marks are lost most often on the standard deviation.