Case Study – Gumbel
Distribution
Rhone River – Maximum Daily
Discharge (Annual) – 1826-1936
Data Source: E. Gumbel (1941). “The Return Period of Flood Flows,”
The Annals of Mathematical Statistics, Vol. 12,#2, pp.163-190.
Data
• Years 1826-1936: maximum daily discharge of river obtained.
order
discharge
order
discharge
order
discharge
order
discharge
order
discharge
1
899
23
1992
45
2240
67
2586
89
3067
2
1172
24
1992
46
2258
68
2594
90
3126
3
1231
25
2006
47
2281
69
2594
91
3179
4
1272
26
2006
48
2296
70
2594
92
3214
5
1272
27
2013
49
2327
71
2602
93
3250
6
1432
28
2050
50
2342
72
2626
94
3266
7
1432
29
2050
51
2358
73
2627
95
3293
8
1439
30
2072
52
2381
74
2643
96
3310
9
1444
31
2094
53
2420
75
2675
97
3310
10
1502
32
2101
54
2444
76
2675
98
3354
11
1541
33
2115
55
2452
77
2773
99
3426
12
1560
34
2145
56
2467
78
2773
100
3444
13
1639
35
2145
57
2475
79
2773
101
3444
14
1706
36
2153
58
2475
80
2839
102
3480
15
1780
37
2160
59
2475
81
2856
103
3606
16
1829
38
2168
60
2491
82
2881
104
3625
17
1850
39
2175
61
2514
83
2881
105
3708
18
1857
40
2206
62
2514
84
2965
106
3801
19
1913
41
2206
63
2514
85
3007
107
3810
20
1913
42
2206
64
2514
86
3050
108
3905
21
1934
43
2221
65
2538
87
3058
109
4096
22
1955
44
2236
66
2554
88
3067
110
4105
111
4390
Gumbel Distribution

  ( y   ) 
F ( y )  1  exp  exp 
    y  ,      ,   0





 ( y  )
  ( y   ) 
exp 
exp

exp



    y  ,      ,   0








E (Y )    0.577216 
(0.577216  Euler' s Constant)
f ( y) 
1
( ) 2
V (Y ) 
6
Method of Moments Estimators :
 2 2
6
   
2
2
6 2

2
 
6

~
 0.7797    0.7797 S
~
~
  0.577216         0.577216     Y  0.577216   Y  0.4501S
n
where : Y 
Y
i 1
n
i
 Y  Y 
n
and S 
i 1
2
i
n 1
Maximum Likelihood Estimation (I)
n
n
i 1
i 1
L( ,  )  f ( y1 ,..., yn |  ,  )   f ( yi |  ,  )  

 ( y  )
 ( yi   )  
exp  i
exp

exp



 








1
 n  ( yi   )
1
 ( y   )  
   exp   
 exp  i
 
  
 i 1  
 

n
( y  )
 ( y   ) 
l  ln( L)   n ln(  )    i
 exp  i


 
i 1 

n
n 
 ( y   )  n 1
l
1 1

    exp  i
  


   
i 1  

n
 ( yi   ) 




 exp 
i 1
 ( yi   ) 
 ( yi   ) 
l
n n  ( y i   ) ( yi   )
n n ( yi   ) n ( yi   )

    

exp





exp






2
2

 i 1 
2
2




 
i 1
i 1



 l 
^ ^
   0
Goal, choose  ,  such that      using Newton - Raphson Algorithm
l
  0 
  
Maximum Likelihood Estimation (II)
 l 
^ ^
  0 
Goal, choose  ,  such that       using Newton - Raphson Algorithm :
l
  0 
  
 2l
1


 2

1
 ( y  )
1
  exp  i
 2

 

i 1   

n
 2l
n
1
 2  2



 (y ) 1
exp  i


  
i 1

n
n
 ( yi   ) 

 

 exp 
i 1
n

i 1
( yi   )
2
 ( y  )
exp  i

 

 ( yi   )  1 n
 ( yi   ) 
  2  2  exp 
  3  ( yi   ) exp 


 i 1
   i 1
 


 ( yi   )  n ( yi   ) ( yi   )
 ( yi   ) 
 2l
n
2 n
2 n


(
y


)

(
y


)
exp


exp







i
i
 2  2  3 i 1
 3 i 1
  i 1  2
2
 


n
n
1
 ( y ) 1
 2  3  ( yi   )  3  ( yi   ) exp  i


 i 1
 i 1
  4

n
 l 
 
f    
l
 
  
2
n
  2l

 2

H
2
 l
 

2
n
 2l 

 
 2l 
 2 
n
(y
i 1
 ( yi   ) 
2


)
exp


i
 

 ( new)   ( old ) 
1
( old )
,  ( old ) f  ( old ) ,  ( old )
 ( new)    ( old )   H 

 

  ( new)   ( old )  
Iterate to convergenc e :   ( new)    ( old )  
  
 
 

T
  ( new)   ( old )  
  ( new)    ( old )    k
  
 
 


Maximum Likelihood Estimation (III)
• Step 1, choose starting values, say: (0) and (0) are
assigned the method of moment estimators
• Step 2: Choose a tolerance level for the change is
estimates, say k=10-6
• Step 3: Obtain H-1((0), (0)) and f((0), (0))
• Step 4: Obtain (1) and (1) from Newton-Raphson
algorithm
• Step 5: Check to see if ((1)- (0))2+ ((1)- (0))2 <k.
– If Yes, Stop.
– If No, Return to Step 3, obtaining H-1((1), (1)), f((1), (1))
Rhone River Discharge Data
n  111 Y  2493.351 S  700.2752
~
   0.7977 S  0.7977(700.2752)  545.99
~
   Y  0.4501S  2493.351  0.4501(700.2572)  2178.17
Setting:  (0)  2178.17 and  (0)  545.99 :

n
l
111
1
 ( y  2178.17) 


exp  i
  0.028122357

 545.99 545.99 i 1
545.99



l
111
1



545.99 (545.99)2

n
 ( yi  2178.17) 
  0.084758042
545.99

 ( y  2178.17) 1  exp 
i 1
i

 ( yi  2178.17) 
  0.000423838
545.99

i 1
n
n
 2l
111
1
1
 ( yi  2178.17) 
 ( y  2178.17) 


exp


( yi  2178.17) exp  i


  0.000364134
2
2 
3 

(545.99) (545.99) i 1
545.99
545.99

 (545.99) i 1


 2l
1

2

(545.99) 2
n
 exp 
n

 2l
111
2
1
 ( y  2178.17) 


( yi  2178.17) 1  exp  i
 
2
2
3 
4

(545.99) (545.99) i 1
545.99

 (545.99)

 0.001352124
1
 (1)   2178.17   .000424 .000364   .028122  2161.90
  (1)   

 


    545.99   .000364 .001352  .084758   604.31 
  (0)   (1)     (0)   (1)   3664  106  Continue Iterations
2
2
n
 ( y  2178.17)
i 1
i
2
 ( y  2178.17) 
exp  i

545.99


Rhone River Discharge Data (2)
• Iteration History:
Round
alpha
beta
Delta
<k?
0
2178.157
546.0046
N/A
1
2161.899
604.3112
3663.998
No
2
2156.052
631.3299
764.1949
No
3
2155.18
635.4366
17.62557
No
4
2155.163
635.5164
0.006657
No
5
2155.163
635.5164
8.99E-10
Yes

1
  ( y  2155.2) 
  ( y  2155.2) 
f ( y) 
exp 
 exp  exp 

635.5
635.5
635.5





Rhone River Discharge (3)
Gumbel pdf - Rhone River Data
0.0007
0.0006
0.0005
f(y)
0.0004
0.0003
0.0002
0.0001
0
0
1000
2000
3000
y
4000
5000
6000
Rhone River Discharge Data (4)
^
^
Estimated standard errors of  ,  :
 ^ 
1
V  ^     H 
 
   
^
^

^
 ^ ^ 
COV   ,  
 V  
 

  4076.6

^ ^
^
 ^    936.3
COV   ,  
V 



  
^
      4076.6  63.8
 
^
^
      2001.3  44.7
 
^
Approximat e 95%Confidence Intervals for  and  :
 : 2155.2  2(63.8)  2155.2  127.6  (2027.6 , 2282.8)
 : 635.5  2(44.7)  635.5  89.4  (546.1, 724.9)
936.3 
2001.3