Risk Modelling and Survival Analysis · Introduction to extreme value theory

Distribution of the Maximum and the Block Maxima Method

Q: What is the distribution of the maximum of n iid random variables?

Its CDF is P(Mₙ ≤ x) = [F(x)]ⁿ, where F is the common CDF. This works because the maximum is at most x only when every variable is at most x, and independence lets you multiply.

Q: What is the block maxima method?

You split a data series into equal blocks, such as years, and take the maximum from each block. You then fit a GEV distribution to those maxima. It rests on the limit result for normalised maxima.

Q: How do I choose the block size?

There is a trade-off. Large blocks make the GEV approximation more reliable but leave few maxima, so parameter estimates are uncertain. Small blocks give more maxima but the approximation may be poor, which causes bias.

Q: Why do we need normalising constants for the maximum?

As n increases, the maximum drifts towards the upper end of the distribution, so its distribution degenerates. Subtracting bₙ and dividing by aₙ gives a non-degenerate limit, which is a GEV distribution under suitable conditions.

Updated 11 October 2026 · Fact-checked

For n independent identically distributed variables with CDF F, the maximum Mₙ has CDF P(Mₙ ≤ x) = [F(x)]ⁿ. The block maxima method splits data into equal blocks, takes the maximum of each block, and fits an extreme value distribution to those maxima.

Understand Distribution of the Maximum and Block Maxima

Insurers care about the largest claim, the highest flood level or the worst yearly loss. These are maxima, not averages. Extreme value theory studies how maxima behave.

Start with the exact result. Let X₁, X₂, ..., Xₙ be independent and identically distributed with CDF F(x). The maximum Mₙ = max(X₁, ..., Xₙ) is at most x only if every Xᵢ is at most x. Independence lets you multiply the probabilities. So P(Mₙ ≤ x) = [F(x)]ⁿ.

The density follows by differentiating: fₘ(x) = n [F(x)]ⁿ⁻¹ f(x). The survival function of the maximum is P(Mₙ > x) = 1 − [F(x)]ⁿ. Compare this with the minimum, which is below x unless all variables exceed x: P(min ≤ x) = 1 − [1 − F(x)]ⁿ.

There is a practical problem. The exact formula needs F, and F is rarely known well in the tail, where data is scarce. Also, as n grows, [F(x)]ⁿ tends to 0 for any x where F(x) < 1, so the distribution of Mₙ collapses to the upper endpoint. To get a useful limit you need to normalise: look at (Mₙ − bₙ) ÷ aₙ for suitable constants aₙ > 0 and bₙ. Under suitable conditions the limit is a member of the generalised extreme value (GEV) family.

The block maxima method uses this result in practice. Divide the data into blocks of equal size, such as years or months. Take the maximum in each block. Treat those maxima as approximately independent draws from a GEV distribution, and fit its parameters. Block size is a trade-off: large blocks make the GEV approximation better but leave you with few maxima, so estimates have more variance. Small blocks give more data but a poorer approximation, which adds bias. The method also wastes data, since only one value per block is used.

Key rules to remember

CDF of the maximum: Fₘ(x) = P(Mₙ ≤ x) = [F(x)]ⁿ; Needs independence and identical distribution. Without independence you cannot multiply.
Density of the maximum: fₘ(x) = n [F(x)]ⁿ⁻¹ f(x); Differentiate [F(x)]ⁿ using the chain rule.
Survival function of the maximum: P(Mₙ > x) = 1 − [F(x)]ⁿ; Use this for the chance that at least one value exceeds x.
CDF of the minimum: P(min ≤ x) = 1 − [1 − F(x)]ⁿ; Condition for the minimum to exceed x: all variables exceed x.
Normalised maximum: (Mₙ − bₙ) ÷ aₙ → GEV in distribution, with aₙ > 0; The limit exists only under suitable conditions on F. The constants depend on F.
Exceedance in n blocks from one block: If a block maximum has P(M ≤ x) = p, then P(M ≤ x in all k independent blocks) = pᵏ; Follows from the same product rule applied to block maxima.

How to solve Distribution of the Maximum and Block Maxima questions

Use this method for any question on the distribution of a maximum or on block maxima.

1Identify the variables and check that they are independent and identically distributed. State this assumption.
2Write the CDF F(x) of a single variable. If only the survival function is given, convert it using F(x) = 1 − S(x).
3Write P(Mₙ ≤ x) = [F(x)]ⁿ, with the correct n (the block size).
4If the question asks for a density, differentiate: n [F(x)]ⁿ⁻¹ f(x). If it asks for an exceedance probability, use 1 − [F(x)]ⁿ.
5For a quantile, set [F(x)]ⁿ equal to the required probability p, take the nth root to get F(x) = p^(1/n), then invert F.
6For block maxima, define the block (for example one year), find the number of observations per block, and say how the maxima will be fitted to a GEV distribution.
7Check the answer: it must lie between 0 and 1 and increase with x. The maximum should be stochastically larger than a single variable.
8State the limitations when asked: loss of data, block size trade-off, and the independence assumption.

Quickest way: Maximum in three lines

When to use it: Use when the MCQ or short question gives F(x) and n and asks for a probability or quantile of the maximum.

Write Fₘ(x) = F(x)ⁿ immediately.
For 'at least one exceeds x', compute 1 − F(x)ⁿ.
For a quantile at level p, solve F(x) = p^(1/n), then invert F.

Common mistakes in Distribution of the Maximum and Block Maxima

Writing P(Mₙ ≤ x) = n F(x) or F(x) ÷ n.
Students confuse the maximum with a sum or mean, or try to add probabilities.
Fix: Remember that the maximum is at most x only if all are at most x, so you multiply: F(x)ⁿ.
Using 1 − F(x)ⁿ for P(Mₙ ≤ x).
This is the formula for the minimum, or for an exceedance, and it gets mixed up.
Fix: Ask whether the event is 'all below x' (F(x)ⁿ) or 'at least one above x' (1 − F(x)ⁿ).
Forgetting the factor n and the power n − 1 in the density.
Students differentiate F(x)ⁿ as if F were a constant.
Fix: Use the chain rule: derivative is n F(x)ⁿ⁻¹ times f(x).
Applying the formula to dependent or non-identical variables.
The formula looks general, so the conditions are skipped.
Fix: State independence and identical distribution. If variables are dependent, say the product rule does not hold.
Taking the nth power instead of the nth root when finding a quantile.
Students solve in the wrong direction.
Fix: Set F(x)ⁿ = p, so F(x) = p^(1/n). Then invert F.
Choosing block size without discussing the trade-off.
Students treat blocks as a fixed convention.
Fix: Mention that bigger blocks improve the GEV approximation but give fewer maxima, and smaller blocks do the reverse.

Worked examples

Example 1

Claim sizes X are independent and exponential with mean ₹50,000. A portfolio has 10 claims in a year. Find the probability that the largest claim is at most ₹1,00,000, and the probability that at least one claim exceeds ₹1,00,000. Give answers to 4 decimal places.

Show the solution

The CDF of one claim is F(x) = 1 − e^(−x ÷ 50,000). Claims are assumed iid.
At x = 1,00,000: F = 1 − e^(−2). Now e^(−2) = 0.135335, so F = 0.864665.
P(M₁₀ ≤ 1,00,000) = 0.864665¹⁰. Take logs: ln(0.864665) = −0.145413 (this equals ln(1 − e^(−2))). Multiply by 10: −1.45413.
Then e^(−1.45413) = 0.2336 (e^(−1.45413) ≈ 0.23356).
P(at least one exceeds) = 1 − 0.2336 = 0.7664.

Answer: P(largest ≤ ₹1,00,000) ≈ 0.2336. P(at least one claim exceeds ₹1,00,000) ≈ 0.7664.

Example 2

Daily losses are modelled as iid with CDF F(x) = 1 − x⁻² for x ≥ 1 (units of ₹ lakh). A block is 5 days. (a) Find the CDF and density of the block maximum M. (b) Find the median of M.

Show the solution

(a) For x ≥ 1, F(x) = 1 − x⁻². Then P(M ≤ x) = (1 − x⁻²)⁵. For x < 1 the CDF is 0.
The density of one day's loss is f(x) = 2x⁻³.
The density of M is fₘ(x) = 5 (1 − x⁻²)⁴ × 2x⁻³ = 10 x⁻³ (1 − x⁻²)⁴, for x ≥ 1.
(b) Median m satisfies (1 − m⁻²)⁵ = 0.5. So 1 − m⁻² = 0.5^(0.2).
Compute 0.5^(0.2) = e^(−0.693147 × 0.2) = e^(−0.138629) = 0.870551.
Then m⁻² = 1 − 0.870551 = 0.129449. So m² = 7.7250 and m = 2.779.

Answer: CDF: (1 − x⁻²)⁵ for x ≥ 1. Density: 10 x⁻³ (1 − x⁻²)⁴ for x ≥ 1. Median of M ≈ 2.78 (₹ lakh).

Exam tips

In MCQs, first decide whether the event is 'all below x' or 'at least one above x'. This picks the formula in seconds.
In written answers, always state the iid assumption before using F(x)ⁿ. Markers look for it.
When asked to derive the density, show the differentiation step and keep the factor n and the power n − 1.
If asked to discuss block maxima, give a clear point on each side of the block-size trade-off and mention that only one value per block is used.
In computer-based questions, show the formula and then compute with a vectorised call, for example F(x)^n, and check that the result is between 0 and 1.

Practice questions from Introduction to extreme value theory

Distribution of the Maximum and Block Maxima: frequently asked questions

What is the distribution of the maximum of n iid random variables?

Its CDF is P(Mₙ ≤ x) = [F(x)]ⁿ, where F is the common CDF. This works because the maximum is at most x only when every variable is at most x, and independence lets you multiply.

What is the block maxima method?

You split a data series into equal blocks, such as years, and take the maximum from each block. You then fit a GEV distribution to those maxima. It rests on the limit result for normalised maxima.

How do I choose the block size?

There is a trade-off. Large blocks make the GEV approximation more reliable but leave few maxima, so parameter estimates are uncertain. Small blocks give more maxima but the approximation may be poor, which causes bias.

Why do we need normalising constants for the maximum?

As n increases, the maximum drifts towards the upper end of the distribution, so its distribution degenerates. Subtracting bₙ and dividing by aₙ gives a non-degenerate limit, which is a GEV distribution under suitable conditions.