Actuarial Statistics · Central limit theorem

Statement of the Central Limit Theorem and Its Conditions

Updated 11 October 2026 · Fact-checked

The central limit theorem says that if X₁, X₂, … are independent, identically distributed with finite mean μ and finite variance σ², then (X̄ₙ − μ) ÷ (σ/√n) converges in distribution to N(0,1) as n grows. So sums and means become approximately normal for large n, whatever the original distribution.

Understand Statement of the Central Limit Theorem

Start with a simple question. You add up many small, independent random amounts, such as claims on a large portfolio. What does the total look like? The central limit theorem (CLT) says it looks roughly normal, even if each single amount is skewed or discrete.

Here is the formal statement. Let X₁, X₂, …, Xₙ be independent and identically distributed (iid) with mean μ and variance σ², where σ² is finite and positive. Let Sₙ = X₁ + … + Xₙ and X̄ₙ = Sₙ ÷ n. Then the standardised versions (Sₙ − nμ) ÷ (σ√n) and (X̄ₙ − μ) ÷ (σ/√n) converge in distribution to the standard normal N(0,1) as n → ∞.

Convergence in distribution means the CDF of the standardised variable approaches the standard normal CDF Φ(z) at every z. It does not say that the variable becomes exactly normal for any finite n. It also does not say the individual Xᵢ become normal. Only the standardised sum or mean does.

Why standardise? Without it, Sₙ has mean nμ and variance nσ², which both grow with n, and X̄ₙ collapses to the point μ. Subtracting the mean and dividing by the standard deviation keeps a fixed mean 0 and variance 1, so a limit shape can appear.

In practice you use the result as an approximation: for large n, Sₙ is approximately N(nμ, nσ²) and X̄ₙ is approximately N(μ, σ²/n). The proof idea uses moment generating functions: the MGF of the standardised sum tends to exp(t²/2), the MGF of N(0,1), and a limit of MGFs identifies the limiting distribution. This needs the MGF to exist near 0 in this version. A general proof uses characteristic functions.

Key rules to remember

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.

How to solve Statement of the Central Limit Theorem questions

Use this method for any question that asks you to state the CLT, check its conditions, or use it to find a probability.

1Check the conditions: are the variables independent, identically distributed, with finite variance? Say so explicitly.
2Find μ = E[X] and σ² = Var(X) for one variable. Work these out from the given distribution.
3Decide whether the question is about the sum Sₙ or the mean X̄ₙ.
4Write the approximate distribution: Sₙ ≈ N(nμ, nσ²) or X̄ₙ ≈ N(μ, σ²/n).
5Standardise: Z = (value − mean) ÷ standard deviation, using √(nσ²) or σ/√n.
6Read the probability from the standard normal table and state the answer.
7State that the result is an approximation, and add a continuity correction if the variable is discrete.

Quickest way: Sum or mean, then standardise

When to use it: Use this for numerical MCQs on probabilities of totals or averages when n is large.

Write μ, σ² and n on one line.
For a sum, use mean nμ and sd σ√n. For a mean, use mean μ and sd σ/√n.
Compute z and look up Φ(z).
For integer-valued variables, move the boundary by 0.5 before computing z.

Common mistakes in Statement of the Central Limit Theorem

Saying the individual Xᵢ become normal as n grows.
The words 'tends to normal' are remembered without saying what tends to normal.
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.
Students confuse the spread of one observation with the spread of the average.
Fix: Always write Var(X̄ₙ) = σ²/n first, then take the square root.
Using variance as the second parameter incorrectly, for example N(μ, σ/√n).
Mixing up N(mean, variance) notation with standard deviation.
Fix: In IAI notation N(μ, σ²) takes the variance. Write the variance, then take its square root when standardising.
Leaving out the conditions (independence, identical distribution, finite variance).
The theorem is remembered as a slogan.
Fix: Write all three conditions in any statement. Mention that a distribution such as the Cauchy has no finite variance, so the CLT does not apply.
Claiming the CLT is exact for n = 30 or any fixed n.
The rule of thumb n ≥ 30 is taken as a law.
Fix: Say it is an approximation whose quality depends on the skewness of the original distribution. Skewed distributions need larger n.
Forgetting the continuity correction for discrete variables.
Students treat the normal approximation as exact for counts.
Fix: For integer-valued sums, use P(S ≤ k) ≈ Φ((k + 0.5 − nμ) ÷ (σ√n)).

Worked examples

Example 1

State the central limit theorem and use it to approximate P(X̄ > 5.2) where X̄ is the mean of 100 iid observations, each with mean 5 and variance 4.

Show the solution

Conditions: the observations are iid with finite mean μ = 5 and finite variance σ² = 4, so the CLT applies.
Statement: (X̄ − μ) ÷ (σ/√n) converges in distribution to N(0,1) as n → ∞.
For n = 100, X̄ ≈ N(5, 4/100) = N(5, 0.04). Standard deviation = √0.04 = 0.2.
z = (5.2 − 5) ÷ 0.2 = 1.
P(X̄ > 5.2) ≈ 1 − Φ(1) = 1 − 0.8413 = 0.1587.

Answer: Approximately 0.159.

Example 2

Claims on a policy are iid with mean ₹10,000 and standard deviation ₹6,000. Approximate the probability that the total of 400 claims exceeds ₹41,00,000.

Show the solution

Check conditions: iid, finite variance (σ² = 6,000² = 3,60,00,000), so use the CLT for the sum.
Mean of S = 400 × 10,000 = ₹40,00,000.
Standard deviation of S = 6,000 × √400 = 6,000 × 20 = ₹1,20,000.
S ≈ N(40,00,000, 1,20,000²).
z = (41,00,000 − 40,00,000) ÷ 1,20,000 = 1,00,000 ÷ 1,20,000 = 0.8333.
P(S > 41,00,000) ≈ 1 − Φ(0.8333). Φ(0.83) = 0.7967 and Φ(0.84) = 0.7995, so Φ(0.8333) ≈ 0.7976.
Probability ≈ 1 − 0.7976 = 0.2024.

Answer: Approximately 0.20 (about 20%).

Exam tips

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.
Always say 'approximately' and, for discrete variables, show the continuity correction.
Mention a case where the CLT fails, such as infinite variance, if the question asks about conditions.

Practice questions from Central limit theorem

Statement of the Central Limit Theorem in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

Statement of the Central Limit Theorem: frequently asked questions

What is the central limit theorem in simple words?

If you add or average many independent values from the same distribution, the result looks roughly like a normal curve. This holds even if the original values are not normal, provided the variance is finite. The more values you combine, the better the approximation.

What are the conditions for the central limit theorem?

The variables must be independent and identically distributed, with a finite mean and a finite, positive variance. Then the standardised sum or mean converges in distribution to N(0,1). Some versions relax the identical-distribution condition, but the basic IAI statement uses iid.

How does the MGF proof of the CLT work?

Standardise each variable to Y with mean 0 and variance 1. The MGF of the standardised sum is [M_Y(t/√n)]ⁿ. Expanding M_Y gives 1 + t²/(2n) + smaller terms, and the nth power tends to exp(t²/2), the MGF of N(0,1).

How large must n be for the CLT to work?

There is no single number. A common rule of thumb is about 30, but a strongly skewed distribution needs more, while a symmetric one needs fewer. Treat the rule as a guide, not a guarantee.