Gumbel's distribution and frequency factor derivation

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.