---
title: "Prove mean annual flood has 2.33-year return period"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:gumbel-mean-flood-2-33yr"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-mean-flood-2-33yr
site: BCT Engineering
---

# Prove mean annual flood has 2.33-year return period

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2081 Baisakh, 4 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-mean-flood-2-33yr
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

Prove that for a large sample as per Gumbel's distribution, the mean annual flood will have a return period of 2.33 years.

## Short answer

For large N, reduced mean yn=0.5772 (Euler's constant). Exceedance probability P=1-e^{-e^{-0.5772}}=0.4296, so return period T=1/P≈2.33 years.

## Answer

**Proof.** In Gumbel's method $x_T=\bar x+K\sigma$, so the mean flood $x=\bar x$ corresponds to $K=0$:

$$
K=\frac{y_T-\bar y_n}{S_n}=0\ \Rightarrow\ y_T=\bar y_n
$$

For a large sample ($N\to\infty$), the reduced mean is $\bar y_n=0.5772$ (Euler's constant). The exceedance probability of this value is

$$
P(X\ge\bar x)=1-e^{-e^{-y_T}}=1-e^{-e^{-0.5772}}
$$

$$
e^{-0.5772}=0.5615,\qquad e^{-0.5615}=0.5704,\qquad P=1-0.5704=0.4296
$$

The return period is

$$
T=\frac1P=\frac{1}{0.4296}\approx2.33 \text{ years}
$$

Hence the **mean annual flood has a return period of about 2.33 years** for Gumbel's distribution (large sample). ∎

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What condition characterises the mean flood in Gumbel's method?
   A. K = 0, so y_T = y_n
   B. K = 1
   C. y_T = 0
   D. T = 1
   Answer: A. x_T = mean + K sigma equals the mean when K = 0, so y_T = y_n-bar. (Key point: Mean corresponds to K = 0)

2. What is y_n for a large sample?
   A. 1.2825
   B. 1
   C. 0.5772 (Euler's constant)
   D. 0
   Answer: C. As N tends to infinity y_n tends to 0.5772. (Key point: Large-sample value)

3. What is P(X >= mean)?
   A. 0.5
   B. exp(-0.5772)
   C. 0.5772
   D. 1 - exp(-exp(-0.5772)) = 0.4296
   Answer: D. Gumbel is skewed, so the mean is exceeded 43% of the time, not 50%. (Key point: Exceedance probability)

4. How is T obtained?
   A. T = 1/0.5772
   B. T = 1/P = 1/0.4296 = 2.33 years
   C. T = 2 years
   D. T = P
   Answer: B. Return period is the reciprocal of exceedance probability. (Key point: Return period)

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