Central limit theorem in CS1, with a worked check

CS1 · Actuarial Statistics · 9 min read

What the theorem actually claims

Individual claim amounts are not normally distributed. They cannot be: they are positive, and they have a long right tail, and a normal distribution has neither property. Yet an actuary reserving for a book of ten thousand policies will reach for a normal distribution without apology, and be right to.

The central limit theorem is what licenses that. Its claim is narrow and precise, and the whole topic turns on reading it correctly. It says nothing about the distribution of the data. It is a statement about the distribution of a sample mean, or equivalently of a total: as the number of observations grows, the standardised version of that mean behaves more and more like a standard normal, whatever the shape of the thing being averaged.

So the data stay as skewed as they ever were. What becomes normal is the average of them.

The statement

Take X₁, …, X_n independent, from the same distribution, with mean μ and finite variance σ². Let X̄ be their average. Then as n grows,

Z_n = (X̄ − μ) / (σ/√n) → N(0, 1)

In words: subtract what the average should be, divide by how much the average typically wobbles, and the result settles down to the standard normal bell.

In practice the theorem is used in one of two equivalent forms, and being fluent in both saves time:

X̄ ≈ N(μ, σ²/n) for the average
S_n = ∑X_i ≈ N(nμ, nσ²) for the total

They contain the same information, because the total is just n times the average. Which one to use is decided by the question: if it asks about total claims, work with the total and avoid dividing anything.

Where the square root of n comes from

This is the part worth understanding rather than memorising, because it is the source of the most common error in the topic.

Variances add; standard deviations do not. For independent variables, Var(X₁ + … + X_n) = nσ². Take the square root and the standard deviation of the total is σ√n — it grows, but more slowly than n. Divide the total by n to get the average and the standard deviation divides by n as well:

sd(X̄) = σ√n / n = σ/√n

A warm-up with round numbers. Four independent claims, each with standard deviation 10. The variance of the total is 4 × 100 = 400, so the standard deviation of the total is 20. The standard deviation of the average is 20/4 = 5, and 10/√4 = 5 agrees. Averaging four observations has halved the uncertainty, not quartered it.

That quantity, σ/√n, is the standard error of the mean, and it reappears in every confidence interval and every test in the rest of the paper. It is the same object each time.

What has to be true for it to work

Independence. If the observations are correlated, the variance of the total is not nσ² and the theorem in this form does not apply. A portfolio where one storm damages every roof is not a portfolio of independent claims, and this is where naive normal approximations in insurance go badly wrong.

Finite variance. The theorem needs σ² to exist. Some distributions used for very large claims have infinite variance, and for those the sample mean does not settle down to a normal at all.

An n that is large enough for the skewness you actually have. There is no universal threshold, and "n > 30" is folklore rather than mathematics. The honest version: the error in the approximation is governed mainly by the skewness of one observation divided by √n. A symmetric distribution needs very little n. A claim-size distribution with a skewness of 4 needs far more than thirty observations before the normal is trustworthy in the tail, which is precisely where a reserving question wants it.

The practical reading: the approximation is always best in the middle of the distribution and worst in the tails. If you are approximating a 99.5th percentile, be much more careful than if you are approximating a median.

The continuity correction

When the variable being approximated takes whole-number values — a count of claims, a number of deaths — and the approximating normal is continuous, the two do not line up at the edges. A count of exactly 110 occupies no width at all for the normal, so the convention is to give each integer the interval that surrounds it:

P(N ≤ 110) ≈ P(Z ≤ (110.5 − μ)/σ)
P(N ≥ 110) ≈ P(Z ≥ (109.5 − μ)/σ)

Half a unit, in the direction that makes the interval wider. The correction is small and it is routinely dropped, which is a mistake: the worked example below shows it cutting the error by a factor of about forty.

Where marks are lost

Dividing by σ instead of σ/√n. The commonest error in the topic. The test of whether you have the right denominator is to ask what the question is about: one observation uses σ, an average of n uses σ/√n, a total of n uses σ√n. Three different denominators, and the words of the question decide.

Dividing by n rather than √n. This makes the standard error far too small and the resulting probability far too extreme. A sanity check: the standard error can never be larger than σ and it can never be zero, so an answer that collapses the spread to almost nothing is wrong.

Dropping the continuity correction on a count. Free marks, lost for a line of arithmetic.

Claiming the data are now normal. A question that asks for the probability that one claim exceeds some amount is not a central limit theorem question at all, however large the portfolio is.

Forgetting that the theorem is about convergence, not equality. Writing an exact equals sign where an approximation is meant will usually cost nothing, but it signals the misunderstanding that costs marks in the written parts.

Questions students ask

How large does n have to be?

It depends entirely on the shape of what you are averaging, and anyone who gives a single number is guessing. For a distribution that is already symmetric, n of 5 or 10 is often ample. For a right-skewed claim-size distribution, hundreds may be needed before the upper tail is reliable. The useful habit is to compare: if the exact answer is available for a similar case, compute both and see how far apart they are. The worked example below does exactly that, and finds the answer trustworthy at n = 100 in the body of the distribution.

Does it apply if the observations have different distributions?

Yes, under conditions, and this is the version that matters in insurance, where no two policies are identical. Provided no single observation dominates the total variance, a sum of independent but differently distributed variables also tends to normality. The identically-distributed version in the syllabus is the simplest case rather than the only one.

Why the normal distribution and not some other?

Because of a property no other distribution with finite variance has: add two independent normals and you get a normal again. A limit of sums has to be stable under adding, and the normal is where that requirement leads. The proof, through cumulant generating functions, makes this visible — everything beyond the second cumulant dies away as n grows, and a distribution described by its first two cumulants alone is a normal.

A worked example

Two parts. The first applies the theorem to a portfolio total; the second tests the approximation against an answer that can be computed exactly, which is the only honest way to find out whether it can be trusted. All figures below were invented for this example.

Part one: the total cost of 400 claims

Suppose individual claim amounts on a motor book have mean ₹28,000 and standard deviation ₹42,000. Note the shape that implies: the standard deviation is one and a half times the mean, so the distribution is strongly right-skewed and certainly not normal. Four hundred claims are expected in the year, treated as independent. What is the probability the total exceeds ₹1.20 crore?

The total, not one claim, is what the question asks about, so use the total form:

E[S] = 400 × 28,000 = ₹1,12,00,000
sd(S) = 42,000 × √400 = 42,000 × 20 = ₹8,40,000

Standardise the threshold:

z = (1,20,00,000 − 1,12,00,000) / 8,40,000 = 8,00,000 / 8,40,000 = 0.9523810

P(S > ₹1.20 crore) ≈ 1 − Φ(0.9523810) = 0.1705

About a 17% chance, or rather better than one year in six.

Check it through the average instead. A total of ₹1.20 crore over 400 claims is an average of 1,20,00,000/400 = ₹30,000 a claim. The standard error of the average is 42,000/√400 = ₹2,100. So

z = (30,000 − 28,000) / 2,100 = 2,000 / 2,100 = 0.9523810

The same z to seven decimal places, as it must be — the two forms of the theorem are the same statement. If the two routes had disagreed, one of the two denominators would have been wrong, and that is the error this check is designed to catch.

How much should this answer be trusted? Four hundred observations is a reasonable number, and 0.95 standard deviations from the mean is well inside the body of the distribution rather than out in the tail. Both facts favour the approximation. Had the question asked for the probability of exceeding ₹1.6 crore — nearly six standard errors out — the skewness of the underlying claims would matter a great deal and the normal answer would understate the risk.

Part two: testing the approximation where the exact answer is known

Claim numbers on a different book are modelled as Poisson with mean 100. For a Poisson, both the mean and the variance are 100, so the standard deviation is 10. What is the probability of 110 claims or fewer?

A Poisson probability can be summed exactly, so there is a correct answer to compare against. Summing e⁻¹⁰⁰ 100^k/k! for k = 0 to 110 gives

P(N ≤ 110) = 0.8528627

(Computed twice, once by direct summation of all 111 terms and once through the incomplete gamma function; the two agree to twelve decimal places.)

Now the normal approximation, first without the continuity correction:

z = (110 − 100)/10 = 1.00 → Φ(1.00) = 0.8413447

And with it:

z = (110.5 − 100)/10 = 1.05 → Φ(1.05) = 0.8531409

Method	Answer	Error
Exact Poisson sum	0.8528627	—
Normal, no correction	0.8413447	0.0115179
Normal, with correction	0.8531409	0.0002783

The uncorrected approximation is out by 1.15 percentage points. The corrected one is out by 0.03 of a percentage point — an error about forty-one times smaller, for the cost of adding a half. That is the whole argument for the continuity correction, and it is why dropping it is not a rounding matter.

It is also a fair verdict on the theorem itself: at n = 100, in the body of the distribution, the approximation is good to three decimal places. The honest caveat is that this was a Poisson, whose skewness is only 1/√100 = 0.1. A claim-size distribution with the skewness of 1.15 seen on the generating-functions page would not do nearly this well at the same n.

Where this fits in CS1

This page explains one idea from CS1 Actuarial Statistics. The full CS1 syllabus, with what each part is examined on, sits on the course page, and past-paper style mock exams with full solutions are where you find out whether the idea has stuck under time pressure.

The central limit theorem, and when the normal approximation is safe