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Tully ERE 323
HW # 10
Problem d: Gumbel Distribution
1/5
Problem Statement
Assume that the mean and variance of a random variable are µ = 6.267 and σ 2 = 10.73 .
Assuming the random variable has a Gumbel distribution,
i. what are the distribution’s parameters?
ii. what are the 1st, 10th, 50th, 90th, and 99th percentiles?
iii. graph the distribution.
Background
The Gumbel distribution is good for modeling extreme values of a random variable.
Specifically, it is useful when looking at the maximum value of a set of random variables. For
example, one might be interested in the probability of a flood occurring. Knowing that a flood
will ensue when a certain amount of rainfall has been received, it would be useful to look at the
maximum amount of rainfall. If the random variable AR is a set of rainfall values, the Gumbel
could help to model the maximum amount of rainfall by looking at X, the maximum of AR. From
the Gumbel distribution of that maximum, one could infer the likelihood of a seeing a flood.
The following formulas describe the parameters, probability density function, and
cumulative distribution function of the Gumbel distribution.
π2
E(X) =u + 0.5772156649/α
V(X) =
6α 2
f(x) = α exp [-α(x-u) -exp[-α(x- u)]] for - ∞ < x < ∞
F(x) = exp[- exp[-α(x- u)]]
Solution
Define the random variable. Here, we are told to assume that the random variable has a
Gumbel distribution.
X ∼ Gumbel(α,u)
Tully ERE 323
HW # 10
Problem d: Gumbel Distribution
2/5
i. what are the distribution’s parameters?
First find the scale parameter α. In the problem statement, we are given the variance or
V(X). Use the following formula to find α:
π2
V(X) =
6α 2
π2
10.73 =
6α 2
2
2
64.38 α = π
π2
α =
64.38
2
α = 0.392
Next find the location parameter u.
E(X) = u + 0.5772156649/α
6.267 = u + 0.5772156649/0.392
u = 6.267 - 1.472
u = 4.795
Now we have the following:
X ∼ Gumbel(α=0.392, u =4.795)
ii. what are the 1st, 10th, 50th, 90th, and 99th percentiles?
There are many ways to solve this problem. One way is to take the negative of the random
variable values and then solve by using the Weibull distribution. Since the Weibull
distribution is used in the next problem in this homework set, this method will not be
explored further here.
Tully ERE 323
HW # 10
Problem d: Gumbel Distribution
3/5
A second way to solve this problem is to use the formula for the CDF or cumulative
distribution function of the Gumbel. To get the 1st percentile, set F(x) = .01 and solve to
find x. The x value obtained ( x1 ) will be the value at which 1% of all the data lies at or
below. The other percentile values, x P , are found in a similar manner. See below for an
example solution.
The general formula to use:
1st percentile:
F( x P ) = exp[- exp[-α( x P -u)]] = percentile/100
F( x1 ) = exp[- exp[-α( x1 -u)]] = .01
Ln[exp[- exp[-α( x1 -u)]]] = Ln[.01]
exp[-α( x1 -u)]= - Ln[.01]
Ln[exp[-α( x1 -u)]]= Ln[- Ln[.01]]
-α( x1 - u)= Ln[- Ln[.01]]
x1 = [-Ln[- Ln[.01]] / α ] + u
x1 = [-Ln[- Ln[.01]] / 0.392 ] + 4.795
x1 = 0.89913
Note that the next to the last line above can be restated to form a shortcut formula:
x P = [-Ln[- Ln[(percentile/100)]] / 0.392 ] + 4.795
The remaining percentiles can be found using the shortcut formula. The percentiles are
summarized below.
x1 = 0.89913
x10 = 2.66737
x50 = 5.72998
x90 = 10.53573
x99 = 16.53007
Tully ERE 323
HW # 10
4/5
Problem d: Gumbel Distribution
Another way to solve this problem that’s faster is to use Microsoft Excel. It is based on
the same principles as above, but letting the computer solve the CDF for x P using the goal
seek function. The CDF was defined as follows:
$C11=F($B10)=EXP( -EXP(-$B$1*($B10-$B$2)))
$B10 = the x P value
$B$1 = α
$B$2 = u
The values of the percentiles are listed below.
x 1 = 0.90514
x 1 0 = 2.66508
x 5 0 = 5.72947
x 9 0 = 10.51996
x 9 9 = 16.46885
iii. graph the distribution
For our problem, α = 0.392 and u = 4.795.
f(x)
PDF for a Gumbel distribution
0.16
0.14
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f(x)
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Tully ERE 323
HW # 10
5/5
Problem d: Gumbel Distribution
CDF for a Gumbel distribution
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F(x)
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F(x)
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