IAI Actuarial Core Principles · Actuarial Statistics

Central Limit Theorem for IAI Actuarial Statistics

The central limit theorem says that the sum or mean of many independent, identically distributed random variables with finite variance is approximately normal, whatever the original distribution. To solve a question, find the mean and variance of the sum or mean, standardise, apply a continuity correction if the variable is discrete, and read the normal table.

What this chapter covers

This chapter answers one question: when can you use the normal distribution for a sum or an average of random variables? You start with the mean and variance of sums and means of independent variables. Then you learn the theorem itself, and finally you use it to approximate probabilities.

The working is short. For independent, identically distributed X₁, ..., Xₙ with mean μ and variance σ², the sum Sₙ has mean nμ and variance nσ². The sample mean X̄ has mean μ and variance σ²/n. For large n, Sₙ is approximately N(nμ, nσ²) and X̄ is approximately N(μ, σ²/n). Most marks come from getting these parameters right.

The chapter links to many other parts of the paper. Sampling distributions and confidence intervals rely on the sample mean being approximately normal. Hypothesis tests for means and proportions use the same idea. Normal approximations to the binomial and Poisson depend on it. In risk modelling, aggregate claims over many policies are treated as approximately normal. Paper B questions in R often check the theorem by simulation.

The central limit theorem is a short chapter, but it supports large parts of statistical inference, so errors here carry over into confidence intervals, tests and aggregate claims questions. Multiple-choice questions often ask for a quick normal approximation, and written questions ask you to justify the approximation and compute a probability. The method is the same each time, so once you have it you can score reliably in both papers.

Central limit theorem: topics in the order to study them

1Sums and Means of Independent Random VariablesYou need the mean and variance of Sₙ and X̄ before the theorem means anything, and these formulas are used in every later topic.
2Statement of the Central Limit TheoremOnce you can find the parameters, you learn the conditions and the standardised form Z = (X̄ − μ) ÷ (σ ÷ √n) that the theorem gives.
3Normal Approximations to Standard DistributionsThis applies the theorem to binomial, Poisson, gamma and similar distributions, where you see how to match mean and variance.
4Continuity CorrectionIt comes after the approximations because it only matters when you approximate a discrete variable with a continuous one.
5Applications to Aggregate Claims and Sample SizesThis is the exam-style use, combining all earlier steps to find probabilities for total claims or the n needed for a target probability.

How to prepare Central limit theorem

Treat this chapter as one procedure that you practise until it is automatic, rather than a set of facts to memorise.

Derive the mean and variance of a sum and of a mean yourself, using independence. Check that you can explain why the variance of X̄ is σ²/n and not σ²/n².
Write the theorem in your own words, including the conditions: independent, identically distributed, finite variance, n large.
For each standard distribution, write down the normal approximation. For example, Bin(n, p) is approximately N(np, np(1 − p)) and Poisson(λ) is approximately N(λ, λ) for large λ.
Practise a fixed routine: define the variable, find the mean and variance, apply continuity correction if discrete, standardise, then use the table. Write every step in written answers.
Solve sample size questions by setting up the standardised inequality and solving for n. Round up to the next whole number.
Do aggregate claims questions with both the number of policies fixed and a given claim distribution. State the assumptions of independence and identical distribution.
Finish with past-paper MCQs under time pressure, then repeat one question in R by simulation to see the approximation for yourself.

Common mistakes in Central limit theorem

Using the variance where the standard deviation is needed when standardising.
Fix: Always write the standard deviation as a separate line, such as √(nσ²) or σ ÷ √n, before you standardise.
Writing Var(X̄) as σ²/n² or σ/n.
Fix: Derive it each time: Var(X̄) = Var(Sₙ) ÷ n² = nσ² ÷ n² = σ²/n.
Forgetting the continuity correction, or applying it in the wrong direction.
Fix: Treat each integer k as the interval from k − 0.5 to k + 0.5. Decide which integers are included, then write the boundary.
Applying the continuity correction to a continuous variable such as a sum of gamma or exponential variables.
Fix: Check whether the original variable is discrete. Only then use ±0.5.
Using the theorem without stating its conditions or assuming independence without comment.
Fix: Write one line stating independence, identical distribution, finite variance and that n is large enough.
Not rounding up in sample size questions, or solving the inequality in the wrong direction.
Fix: Solve for √n first, then square it, and round up to the next whole number so the probability requirement is met.

Last-day revision: Central limit theorem

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.

Central limit theorem practice questions

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.

Central limit theorem: frequently asked questions

What is the central limit theorem in simple words?

It says that if you add up many independent variables from the same distribution with finite variance, the total looks approximately normal. The same holds for the average. The original distribution does not need to be normal.

When do I need a continuity correction?

You need it when you approximate a discrete distribution, such as binomial or Poisson, by the normal distribution. You do not need it when the original variable is already continuous. Adjust the boundary by 0.5 according to which integers the event includes.

How large must n be for the normal approximation to work?

There is no single value that works for every case. A skewed distribution needs a larger n than a symmetric one. Use the sizes stated in the question or study material, and say that the approximation is reasonable rather than exact.

How does this chapter appear in aggregate claims questions?

You are given the number of independent policies and the mean and variance of each claim. You find the mean and variance of total claims as nμ and nσ², then use a normal approximation to find the probability that claims exceed a given amount.

Is this chapter tested in the computer-based paper?

It can be. You may be asked to simulate sample means in R and compare their histogram with the normal curve. Know the theory so you can explain what the simulation shows.