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Gumbel exceedance probability and 0.1 flood

Short answer

P(X>15000) =1 - exp[-e^{-α(15000-β)}] with α=1.28255/3000, β=10000-0.48×3000, giving 0.0617. Dischrge with exceedance probability 0.1: Q_0.1 = β - (1/α)ln(-ln0.9) = 13824 m³/s.

The annual peak discharge of a river follows the Gumbel's extreme value distribution with a mean of 10000 m3^3/s and a standard deviation of 3000 m3^3/s. What is the probability that the annual peak discharge is more than 15000 m3^3/s? What is the magnitude of the peak discharge with an exceedance probability of 0.1? [Hint: α=1.28255σ\alpha = \frac{1.28255}{\sigma}; β=μ−0.48σ\beta = \mu - 0.48\sigma]

Gumbel's cumulative distribution (probability of non-exceedance):

F(x)=P(X≤x)=exp⁡[−e−α(x−β)]F(x)=P(X\le x)=\exp\left[-e^{-\alpha(x-\beta)}\right]

Using the hint:

α=1.282553000=4.2752×10−4 per m3/s,β=10000−0.48×3000=8560 m3/s\alpha=\frac{1.28255}{3000}=4.2752\times10^{-4}\ \text{per m}^3/\text{s},\qquad \beta=10000-0.48\times3000=8560 \text{ m}^3/\text{s}

Probability that the annual peak exceeds 15000 m³/s

α(x−β)=4.2752×10−4×(15000−8560)=2.753\alpha(x-\beta)=4.2752\times10^{-4}\times(15000-8560)=2.753 F(15000)=exp⁡(−e−2.753)=0.9383F(15000)=\exp(-e^{-2.753})=0.9383 P(X>15000)=1−F=0.0617P(X>15000)=1-F=0.0617

Answer: probability ≈ 0.0617 (about 6.17%), a return period of ≈ 16.2 years.

Discharge with exceedance probability 0.1

Exceedance 0.1 means F=0.9F=0.9:

−e−α(x−β)=ln⁡0.9 ⇒ x=β−1αln⁡(−ln⁡0.9)-e^{-\alpha(x-\beta)}=\ln 0.9\ \Rightarrow\ x=\beta-\frac{1}{\alpha}\ln(-\ln0.9) x=8560−ln⁡(−ln⁡0.9)4.2752×10−4=8560−(−2.2504)4.2752×10−4=13824 m3/sx=8560-\frac{\ln(-\ln0.9)}{4.2752\times10^{-4}}=8560-\frac{(-2.2504)}{4.2752\times10^{-4}}=13824 \text{ m}^3/\text{s}

Answer: Q0.1Q_{0.1} ≈ 13824 m³/s (the 10-year flood).

✓ Answer checked by Claude Sonnet 5.5

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