Economic Modelling · Role of insurance in reducing or removing risk

Pooling and the Law of Large Numbers in Insurance

Updated 11 October 2026 · Fact-checked

Pooling combines many independent risks in one portfolio. If each policy has mean μ and variance σ², the average claim per policy has mean μ and variance σ² ÷ n. The variance per policy falls as n grows, so total claims become more predictable relative to their mean. This is the law of large numbers at work.

Understand Pooling and the Law of Large Numbers

An insurer takes on many policies. Each policy's claim is uncertain. For one policyholder, the loss may be zero or very large. The insurer can reduce its uncertainty by holding many such policies together. This is pooling.

Suppose you hold n policies. Let the claim on policy i be Xᵢ. Assume the Xᵢ are independent and identically distributed, each with mean μ and variance σ². The total claim is S = X₁ + ... + Xₙ. Its mean is nμ and its variance is nσ². The average claim per policy is S ÷ n. Its mean is μ and its variance is σ² ÷ n.

The key point: the variance of the average shrinks as n grows. The standard deviation of the average is σ ÷ √n. It falls with the square root of n, not with n. To halve the standard deviation per policy you need four times as many policies.

The law of large numbers says that, for independent identically distributed claims with finite mean, the average claim gets closer to μ as n grows. So the insurer can charge a premium near μ plus a loading and expect results close to expectation. The total claim S does not become less variable in rupees. Its standard deviation √n σ actually rises. What falls is the risk relative to the mean. The coefficient of variation of S is σ ÷ (μ√n), which falls as n rises.

Pooling only removes diversifiable risk. If claims are positively correlated, for example flood or pandemic losses hitting many policies at once, the variance of the average does not fall to zero. It tends to a floor set by the covariance. Also, pooling does not remove parameter risk, such as having μ wrong for every policy.

Key rules to remember

Total claim: S = X₁ + X₂ + ... + Xₙ; Sum of the claims on n policies.
Mean and variance of total claim: E(S) = nμ; Var(S) = nσ²; Requires the Xᵢ to be independent (or uncorrelated) with common mean μ and variance σ².
Average claim per policy: E(S ÷ n) = μ; Var(S ÷ n) = σ² ÷ n; Variance per policy falls in proportion to 1 ÷ n.
Standard deviation of average claim: SD(S ÷ n) = σ ÷ √n; Falls with the square root of n.
Coefficient of variation of total claim: CV(S) = √(nσ²) ÷ (nμ) = σ ÷ (μ√n); Measures relative risk. It falls as n rises.
Variance with common correlation ρ: Var(S ÷ n) = σ² ÷ n + ((n − 1) ÷ n) ρσ²; As n → ∞ this tends to ρσ². Correlated risk cannot be pooled away.
Weak law of large numbers: P(|S ÷ n − μ| > ε) → 0 as n → ∞, for any ε > 0; Holds for independent identically distributed claims with finite mean.

How to solve Pooling and the Law of Large Numbers questions

Use this method for numerical and discussion questions on pooling.

1Define the claim on one policy: its mean μ and variance σ² (or standard deviation σ). State the assumptions: independence and identical distribution.
2Decide whether the question asks about the total claim S or the average claim S ÷ n. Mixing these up loses marks.
3Write the mean and variance: E(S) = nμ, Var(S) = nσ², and E(S ÷ n) = μ, Var(S ÷ n) = σ² ÷ n.
4Substitute the numbers. Take the square root only at the end if a standard deviation is needed.
5If the question gives a correlation, use the correlated variance formula instead and note that the risk has a floor.
6If a probability is asked, use the normal approximation (central limit theorem) for large n. Standardise using the mean and standard deviation of the quantity you chose.
7Comment on the result: which risk fell (per policy or relative), which did not (absolute total), and what limits the benefit, such as correlation.

Quickest way: Square root shortcut

When to use it: For multiple-choice questions comparing risk at different portfolio sizes.

Standard deviation per policy scales as 1 ÷ √n. Variance per policy scales as 1 ÷ n.
To go from n₁ to n₂ policies, multiply the per-policy standard deviation by √(n₁ ÷ n₂).
Total claim standard deviation scales as √n. Its coefficient of variation scales as 1 ÷ √n.
Check for correlation. If claims are correlated, the per-policy variance does not fall below ρσ².

Common mistakes in Pooling and the Law of Large Numbers

Saying pooling reduces the total variance of claims.
Students confuse variance per policy with variance of the total.
Fix: Var(S) = nσ² rises with n. Only Var(S ÷ n) and the coefficient of variation fall.
Writing Var(S ÷ n) = σ² ÷ √n or SD = σ ÷ n.
The square root is applied to the wrong quantity.
Fix: Variance is σ² ÷ n. Standard deviation is σ ÷ √n. Check which one you are writing.
Applying the formulas without independence.
The condition is easy to forget.
Fix: State independence explicitly. With correlation ρ, use σ² ÷ n + ((n − 1) ÷ n) ρσ².
Claiming pooling removes all risk.
The phrase law of large numbers sounds absolute.
Fix: Pooling removes only diversifiable risk. Systematic risk, correlated claims and parameter uncertainty remain.
Assuming that doubling policies halves the standard deviation per policy.
Students think the effect is linear.
Fix: Doubling n divides the standard deviation by √2. You need four times n to halve it.

Worked examples

Example 1

An insurer writes 2,500 independent policies. Each policy has mean annual claim ₹4,000 and standard deviation ₹12,000. Find the mean and standard deviation of (a) the total claim and (b) the average claim per policy.

Show the solution

μ = ₹4,000, σ = ₹12,000, n = 2,500. σ² = 14,40,00,000.
(a) E(S) = 2,500 × 4,000 = ₹1,00,00,000.
Var(S) = 2,500 × 14,40,00,000. SD(S) = σ√n = 12,000 × 50 = ₹6,00,000.
(b) E(S ÷ n) = ₹4,000.
SD(S ÷ n) = σ ÷ √n = 12,000 ÷ 50 = ₹240.
Check: 6,00,000 ÷ 2,500 = ₹240.

Answer: Total claim: mean ₹1,00,00,000, standard deviation ₹6,00,000. Average claim: mean ₹4,000, standard deviation ₹240.

Example 2

Using the portfolio in the previous example, with the same assumptions, find the approximate probability that the total claim exceeds ₹1,01,20,000 (use the normal approximation). Then state how many policies would be needed to cut the coefficient of variation of the total claim to one third of its current value.

Show the solution

Total claim mean ₹1,00,00,000, SD ₹6,00,000.
Excess over mean = 1,01,20,000 − 1,00,00,000 = ₹1,20,000.
z = 1,20,000 ÷ 6,00,000 = 0.2.
P(S > 1,01,20,000) ≈ 1 − Φ(0.2) = 1 − 0.5793 = 0.4207.
CV(S) = σ ÷ (μ√n), so it is proportional to 1 ÷ √n.
To divide the CV by 3, √n must be multiplied by 3, so n is multiplied by 9.
New n = 2,500 × 9 = 22,500.

Answer: Probability ≈ 0.42. About 22,500 policies are needed, assuming independence still holds.

Exam tips

Always state independence and identical distribution before using σ² ÷ n. Examiners give marks for assumptions.
Read whether the question is about the total claim or the average per policy. Write the answer for that quantity.
In discussion parts, name what pooling cannot remove: correlated risks, catastrophe exposure and parameter risk.
For written answers, show the formula in notation, the substitution and the result with units in rupees.
Link to insurance demand: pooling lets insurers offer cover near expected cost, which is why risk-averse people buy it.

Practice questions from Role of insurance in reducing or removing risk

Pooling and the Law of Large Numbers: frequently asked questions

How does pooling reduce risk in insurance?

Pooling combines many independent claims. The average claim per policy has variance σ² ÷ n, which falls as n grows. The insurer's result becomes more predictable relative to expected claims.

Does the law of large numbers mean total claims become less variable?

No. The standard deviation of total claims is σ√n, which rises with n. The relative variability, measured by the coefficient of variation σ ÷ (μ√n), falls.

What happens to pooling if claims are correlated?

The variance of the average claim does not fall to zero. It tends to ρσ² for common correlation ρ. This is why catastrophe and pandemic risks are hard to pool and often need reinsurance.

Why does the standard deviation fall with √n and not n?

Variances of independent claims add, giving nσ² for the total. Dividing by n squares to a factor of n², so the average has variance σ² ÷ n. Taking the square root gives σ ÷ √n.