---
title: "Gumbel's distribution and frequency factor derivation"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:gumbel-distribution-frequency-factor"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-distribution-frequency-factor
site: BCT Engineering
---

# Gumbel's distribution and frequency factor derivation

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2072 Kartik, 7 marks
- Video: [Flood discharge at various return periods using Gumbel's extreme value distribution | Hydrology](https://www.youtube.com/watch?v=RE7Q4NDVPQQ) — The Study Sphere
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-distribution-frequency-factor
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

Explain Gumbel's Distribution function. Derive frequency factor (k) using Gumbel's distribution.

## Short answer

Gumbel's extreme value distribution models annual maxima with probability P=1-e^{-e^{-y}} and reduced variate y_T=-ln[ln(T/(T-1))]. The frequency factor K for Chow's equation x_T=mean+K*std is derived as K=(y_T - y_bar_n)/S_n, where y_bar_n and S_n are reduced mean and standard deviation for sample size N.

## Answer

#### Gumbel's distribution (extreme value type I)
Gumbel considered a series of annual maxima, each the largest of many daily values, and showed that for a large sample the probability that a value $x$ is equalled or exceeded in a year is

$$
P(X\ge x)=1-e^{-e^{-y}}
$$

where $y$ is the **reduced variate**, a linear function of $x$:

$$
y=a(x-u), \qquad a=\frac{1.2825}{\sigma},\qquad u=\bar x-0.45\sigma
$$

($\bar x$ and $\sigma$ = mean and standard deviation of the annual maxima.) The non-exceedance probability is $F(x)=e^{-e^{-y}}$. It has a long right tail, so it suits floods and maximum rainfall.

#### Return period and reduced variate
Since $P=1/T$:

$$
\frac1T=1-e^{-e^{-y_T}}\ \Rightarrow\ e^{-e^{-y_T}}=\frac{T-1}{T}
$$

Taking natural logs twice:

$$
y_T=-\ln\left[\ln\frac{T}{T-1}\right]
$$

For large T, $y_T\approx\ln T$.

#### Derivation of the frequency factor
Chow's general equation: $x_T=\bar x+K\sigma$.

From $y=a(x-u)$, the value at return period T is $x_T=u+\dfrac{y_T}{a}$. For the infinite sample the mean of the reduced variate is $\bar y=0.5772$ (Euler's constant) and its standard deviation is $\sigma_y=1.2825$. Since $\bar x=u+\bar y/a$ and $\sigma=\sigma_y/a$:

$$
x_T-\bar x=\frac{y_T-\bar y}{a}=\sigma\,\frac{y_T-\bar y}{\sigma_y}
$$

Comparing with $x_T=\bar x+K\sigma$:

$$
K=\frac{y_T-0.5772}{1.2825}
$$

For a finite sample of N years the theoretical mean and standard deviation of $y$ are replaced by the **reduced mean $\bar y_n$ and reduced standard deviation $S_n$** (tabulated against N):

$$
\boxed{K=\frac{y_T-\bar y_n}{S_n}}\qquad\text{so}\qquad x_T=\bar x+K\sigma
$$

with $\bar y_n\to0.5772$ and $S_n\to1.2825$ as $N\to\infty$. For example, T = 100 years gives $y_T=4.600$ and $K=(4.600-0.5772)/1.2825=3.14$ for a large sample.

## Check yourself

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

1. In Gumbel's distribution the probability of exceedance is:
   A. P = e^(-e^(-y))
   B. P = 1 - e^(-y)
   C. P = 1 - e^(-e^(-y))
   D. P = y/T
   Answer: C. Non-exceedance is exp(-e^-y). (Key point: Reduced variate)

2. y_T in terms of T is:
   A. ln T
   B. -ln(1/T)
   C. ln[ln T]
   D. -ln[ln(T/(T-1))]
   Answer: D. Taking natural logs twice of (T-1)/T. (Key point: y_T)

3. The frequency factor is:
   A. K = (y_T - y_n)/S_n
   B. K = y_T S_n
   C. K = y_T - sigma
   D. K = (x - y_n)/T
   Answer: A. From x_T - mean = sigma (y_T - y_n)/S_n. (Key point: Frequency factor)

4. For large N, y_n and S_n tend to:
   A. 0 and 1
   B. 0.5772 and 1.2825
   C. 0.45 and 1.0
   D. 1.28 and 0.577
   Answer: B. Euler constant and sigma_y. (Key point: Infinite sample)

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