School of Mathematics and Statistics
Solutions to Extreme Value Theory Sample Quiz Week 13
STAT5611: Statistical Methodology Semester 2, 2021
Lecturers: Neville Weber and Michael Stewart
Please write out your responses, scan and upload before 2pm.
You do not need to derive again results stated and/or derived
in lectures or tutorials, just refer to them as needed.
1. The Kumaraswamy distribution with parameters c > 0 and d > 0 has CDF
F (x; c, d) =

0 for x ≤ 0
1− (1− xc)d for 0 < x < 1
1 for x ≥ 1.
(1)
For the rest of this question, we assume c > 0 and d > 0 are fixed, and therefore just write F (·).
(a) Derive the high quantile or upper- 1n quantile function U(n) = F
−1 (1− 1n).
Solution: Write Un = U(n), set F (Un) = 1− 1n and solve for Un:
F (Un) = 1− (1− U cn)d = 1−
1
n
(1− U cn)d =
1
n
1− U cn =
1
n1/d
= n−1/d
U cn = 1− n−1/d
U(n) = Un =
(
1− n−1/d
)1/c
(b) Show that U(·) is regularly varying at infinity and find its exponent.
Solution: Fix x > 0. Then
U(tx)
U(t)
=
(
1− (xt)−1/d)1/c(
1− t−1/d)1/c → 1
as t→∞ since both numerator and denominator tend to 1. Therefore U(·) is slowly varying
at infinity, i.e. regularly varying at infinity with exponent 0.
(c) Show that U(·) is of extended regular variation at infinity, find its exponent and an appro-
priate scaling function.
Solution: Fix x > 0. Then since (1 + x)1/c = 1 + (x/c) + o(x) as x→ 0, as n→∞,
U(n) =
(
1− n−1/d
)1/c
= 1− (n−1/d/c) + o
(
n−1/d
)
Therefore, for fixed x > 0,
U(nx) = 1− [(nx)−1/d]/c+ o
(
n−1/d
)
and so
U(nx)− U(n) = −[(nx)−1/d]/c+ (n−1/d/c) + o
(
n−1/d
)
= −n
−1/d
c
[
x−1/d − 1 + o(1)
]
as t→∞. Therefore, with scaling function a(n) = n−1/d/(cd),
U(nx)− U(n)
a(n)
→ −d
(
x−1/d − 1
)
= Hγ(x) ,
with γ = −1/d, therefore U(·) is of extended variation at infinity with exponent −1/d.
(d) Suppose X1, . . . , Xn are iid with common CDF F (·) as at (1) above, and write Mn =
maxi=1,...,nXi. Find constants an > 0, bn so that
Mn − bn
an
converges in distribution, and derive the limiting CDF.
Solution: If we take bn = U(n) and an = a(n) (the scaling function in the previous part),
we know we will get the GEVD(γ) as the limiting distribution, for γ = −1/d, but this may
be laborious to derive properly. Alternatively, we can look at the form. of 1−F (·) and based
on that propose a simpler normalisation.
For example, we may take bn ≡ 1 and an = n−1/d. Then we get
P
{
Mn − bn
an
≤ x
}
= P {Mn ≤ bn + anx} = P
{
Mn ≤ 1 + n−1/dx
}
.
Note next that
1− F
(
1 + n−1/dx
)
=
0 for x ≥ 0[1− (1 + n−1/dx)c]d for x < 0.
For x < 0, (
1 + n−1/dx
)c
= 1 + cn−1/dx+ o
(
n−1/d
)
,
so
n
[
1− F
(
1 + n−1/dx
)]
= n
[
−cn−1/dx
]d
= [−cx]d
Therefore, by Proposition 1.3,
P
(
Mn ≤ 1 + n−1/dx
)
→
{
e−[−cx]
d
for x < 0
1 for x ≥ 0
which is a scaled version of a reversed Weibull(d) distribution (which is then also a linear
transformation of GEVD(−1/d)).
2. The (unit scale) Rayleigh distribution has CDF
F (x) =
{
0 for x < 0
1− e−x2/2 for x ≥ 0. (2)
(a) Determine, for each u > 0 and each x > 0, the excess CDF
Fu(x) =
P (u < X ≤ u+ x)
P (X > u)
.
Solution:
Fu(x) =
F (u+ x)− F (u)
1− F (u) =
1− e−(x+u)2/2 −
[
1− e−u2/2
]
e−u2/2
= 1− e−ux−x2/2
2
(b) Find a measurable function a(·) such that for each x > 0
lim
u→∞ 1− Fu(a(u)x) = 1−G(x)
for a CDF G(.) satisfying G(0) = 0. Derive the form. of this CDF G(·).
Solution:
1− Fu(a(u)x) = e−u[a(u)x]−[a(u)x]2/2
Taking a(u) = 1/u, we get
1− Fu(x/u) = e−x−(x/u)2/2
As u → ∞, this tends to e−x. This is 1 − G(x), for G(·) the unit-mean exponential distri-
bution. Also, note that the function a(u) = u is monotone and thus measurable.
(c) Is F (·) in the domain of attraction of a (generalised) extreme value distribution? If so, which
one?
Solution: The CDF G(·) from the previous part is also the generalised Pareto distribution
with shape γ = 0. Thus by the Proposition in lecture 8, F (·) ∈ DOA of GEVD(0), i.e. the
Gumbel distribution.
(d) Suppose now that X1, . . . , Xn are iid with common CDF F (·) as at (2) above, and write
Mn = maxi=1,...,nXi. Deduce sequences an > 0, bn so that
Mn − bn
an
converges in distribution, and derive the resultant limiting distribution.
Solution: We may take u = un = U(n) = F
−1 (1− 1n), so that (as seen in the previous
part, for x > 0)
1− Fu(x/u) = 1− F (un + x/un)
1− F (un) = n [1− F (un + x/un)]→ e
−x
and indeed this also holds for x ≤ 0 too. Thus by Proposition 1.3, taking an = bn = un, we
get for any real x,
P
{
Mn − bn
an
≤ x
}
= P {Mn ≤ bn + x/an} = P {Mn ≤ un + x/un} → e−e−x .
It remains to determine un. But note that
1
n
= e−u
2
n/2
− log n = −u2n/2
un =
√
2 log n .
3. The log-logistic distribution with unit scale parameter and shape parameter c > 0 has CDF
F (x; c) =
xc
1 + xc
. (3)
For the rest of this question, we assume c > 0 is fixed and just write F (·).
(a) Show that 1− F (·) is regularly varying at infinity and determine its exponent.
Solution:
1− F (x) = 1
1 + xc
=
1
xc(1 + x−c)
= x−c
(
1 + x−c
)−1 ∼ x−c
as x→∞. Thus, for fixed x > 0,
1− F (tx)
1− F (t) ∼
(tx)−c
t−c
= x−c .
So 1− F (·) is regularly varying at infinity with exponent −c.
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(b) Suppose X1, . . . , Xn are iid with common CDF F (·) given at (3) above. Deduce sequences
an > 0, bn such that
Mn − bn
an
converges in distribution, and determine the limiting CDF.
Solution: From the previous part, if un = F
−1 (1− 1n), then as n→∞, un →∞ and so
for x > 0,
1− F (unx)
1− F (un) = n [1− F (unx)]→ x
−c .
For x ≤ 0,
1− F (unx) = 1
n [1− F (unx)] = n→∞
as n→∞. In other words, for all real x,
n [1− F (unx)]→ G¯(x) =
{
+∞ for x ≤ 0
x−c for x > 0.
Therefore, taking bn ≡ 0 and an = un, by Proposition 1.3,
P
{
Mn − bn
an
≤ x
}
= P {Mn/un ≤ x} = P {Mn ≤ unx} →
{
0 for x ≤ 0,
e−(x
−c) for x > 0.
It remains to determine un:
1− F (un) = 1
n
=
1
1 + ucn
1 + ucn = n
ucn = n− 1
un = (n− 1)1/c .
Note also that we may replace this any sequence un such that 1− F (un) ∼ n−1, e.g. un =
n1/c, and get the same limiting distribution. Other normalisations are possible (although
all will give some scaled and/or shifted version of this same limiting Fre´chet distribution).
(c) Show that F (·) is in the Hall-Welsh class, that is find constants γ > 0, β > 0, C > 0 and D
such that, as x→∞,
1− F (x) = Cx−1/γ {1 +Dx−β + o (x−β)} .
Solution: As shown above in (a), as x→∞,
1− F (x) = x−c (1 + x−c)−1 = x−c {1− x−c + o (x−c)}
So F (·) is in the Hall-Welsh class, with
• γ = 1/c;
• β = c;
• C = 1;
• D = −1.
(d) Suppose X(1) ≤ . . . ≤ X(n) are the order statistics of X1, . . . , Xn and denote Hill’s estimator
of γ based on X(n−r), X(n−r+1), . . . , X(n) (for integer 1 ≤ r ≤ n) by
γˆr =
1
r
r∑
i=1
[
logX(n−i+1) − logX(n−r)
]
.
4
Deduce a sequence r = r(n) so that the resultant γˆr has the smallest possible asymptotic
mean-squared error.
Solution: We can use the formula from question 4 in the Week 12 Tutorial:
r =
[
(1 + βγ)2
2D2 (βγ)
3
]1/(2βγ+1)
(Cn)
2βγ/(2βγ+1)
.
Note that in this case, βγ = C = D2 = 1 so this simplifies considerably:
r =
(
4
2
)1/3
n2/3 =
(√
2n
)2/3
.
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