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Asymptotic expansions for bivariate normal extremes
Statistics and Probability Letters 119 (2016) 124–133
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Statistics and Probability Letters journal homepage: www.elsevier.com/locate/stapro
Asymptotic expansions for bivariate normal extremes Saralees Nadarajah School of Mathematics, University of Manchester, Manchester M13 9PL, UK
article
abstract
info
Article history: Received 16 July 2016 Received in revised form 25 July 2016 Accepted 26 July 2016 Available online 3 August 2016
Nadarajah (2015) derived a complete asymptotic expansion for normal extremes. Here, we extend the expansion for bivariate normal extremes. © 2016 Elsevier B.V. All rights reserved.
Keywords: Bell polynomials Bivariate normal distribution Expansions
1. Introduction Let Φ (·) denote the cumulative distribution function (cdf) of a standard normal random variable. It is well known that Φ (·) belongs to the max domain of attraction of the Gumbel extreme value distribution, i.e.,
Φ n (un (x)) → exp (− exp(−x))
(1)
as n → ∞ and for any −∞ < x < +∞, where un (x) = an x + bn
(2)
and an cn 1 bn = a− n − 2 for n ≥ 1, where cn = log log n + log(4π ). Nadarajah (2015) provided a complete asymptotic expansion for (1). In particular, an expansion was provided for an = (2 log n)−1/2 ,
Φ n (un (x)) −
(3)
n (−1)m exp(−mx)
m!
m=0
as n → ∞, where un , an and bn are given by (2)–(3). The aim of this note is to extend Nadarajah (2015)’s work for the bivariate normal distribution given by the joint cdf F (x, y) =
1 2π
1 − ρ2
y
x
exp − −∞
−∞
u2 + v 2 − 2ρ uv 2 1 − ρ2
dudv
for −∞ < x < +∞, −∞ < y < +∞ and −1 < ρ < 1. It is well known that F n (un (x), un (y)) → exp (− exp(−x) − exp(−y)) as n → ∞, where un , an and bn are given by (2)–(3). E-mail address: [email protected]. http://dx.doi.org/10.1016/j.spl.2016.07.023 0167-7152/© 2016 Elsevier B.V. All rights reserved.
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S. Nadarajah / Statistics and Probability Letters 119 (2016) 124–133
125
To the best of our knowledge, we are aware of no studies on convergence aspects of (4). However, there have been studies on convergence aspects of the extremes of other multivariate distributions and multivariate processes, see Hashorva and Ji (2014a,b, 2015, 2016), Hashorva and Kortschak (2014), Hashorva (2015), Hashorva et al. (2015, 2016), Hashorva and Li (2015) and Hashorva and Ling (2016). Asymptotic expansions for (4) could have both practical and theoretical appeal given the wide applicability of the bivariate normal distribution. In a practical sense, they could lead to better approximations for the limit exp(− exp(−x) − exp(−y)). Theoretically, such expansions can be used to derive expansions for the corresponding joint probability density function (pdf), moments, cumulants, quantiles, etc. Complete asymptotic expansions for (4) are derived in Section 2. Technical lemmas for their proof are given and proved in Section 3. The expansions involve Bell polynomials. In-built routines for Bell polynomials are available in most computer algebra packages. For example, see BellY in Mathematica. The following notation is used throughout this note: 0k denotes a k × 1 vector of zeros; 1k denotes a k × 1 vector of ones; ∞ ∞k denotes a k × 1 vector of infinities; |a| = a1 + a2 + · · · + ak for a k × 1 vector a = (a1 , a2 , . . . , ak )′ ; m=k 0k denotes the k fold summation m1 =0 m2 =0 · · · mk =0 ; (a)k = a(a + 1) · · · (a + k − 1) denotes the ascending factorial; Br ,k (x) denotes the partial exponential Bell polynomial defined by
∞
∞
xr t / r ! r
∞
k k! =
∞
∞
Br ,k (x)t r /r !
r =k
r =1
for x = (x1 , x2 , . . .). This polynomial is tabled on page 307 of Comtet (1974) for r ≤ 12. Our results in Sections 2 and 3 can in principle be extended to any other joint cdf. We have illustrated our results for the bivariate normal distribution because of its universality. 2. Main results Theorem 1 derives an asymptotic expansion for F n (un (x), un (y)). Theorem 2 uses Theorem 1 to prove (4). Theorems 3 and 4 derive complete asymptotic expansions for
Φ n (un (x)) Φ n (un (y)) −
n n (−1)m1 +m2 exp (−m1 x − m2 y) m1 !m2 ! m =0 m =0 2
1
and F n (un (x), un (y)) −
n n (−1)m1 +m2 exp (−m1 x − m2 y) , m1 !m2 ! m =0 m =0 2
1
respectively, where un , an and bn are given by (2)–(3). Proofs are given for Theorems 1 and 2. These proofs involve the use of Lemmas 1–9 in Section 3. Theorems 3 and 4 are straightforward consequences of Theorems 1 and 2, so their proofs are not given. Theorem 1. For un , an and bn given by (2)–(3), F n (un (x), un (y)) =
(1)
(1)
(2)
n n π k
m1 ,p1 ,q1 ,i,i m2 ,p2 ,q2 ,j,j k=0
k
2
(2)
(3)
m1 ,q3 ,p3 ,ℓ,ℓ m2 ,q4 ,p4 ,r ,r p,q,s,u,v
· λ (m1 , p1 , q1 , i, i, m2 , p2 , q2 , j, j, m1 , p3 , q3 , ℓ, ℓ, m2 , p4 , q4 , r , r, p, q, s, u, v, x, y) · cn2i2 +i3 +2j2 +j3 +2ℓ2 +ℓ3 +2r2 +r3 · n−m1 −m2 −|m1 |−|m2 |−2k · a2n|i|−m1 +i+2|j|−m2 +j+ℓ+2|ℓ|−|m1 |+r +2|r|−|m2 |−2k−u−v m1 −2q1 −i−m2 −2q2 −j−|m1 |−2p3 −ℓ−|m2 |−2p4 −r −2|p|−2|q|−s1 −s2 +u+v · b− n
(5)
as n → ∞, where λ = γ (m1 , p1 , q1 , i, i, x) γ (m2 , p2 , q2 , j, j, y) δ (m1 , p3 , q3 , ℓ, ℓ, x) δ (m2 , p4 , q4 , r , r, y) ϵ(p, q, s, u, v, x, y) and γ (m1 , p1 , q1 , i, i, x), γ (m2 , p2 , q2 , j, j, y), δ (m1 , p3 , q3 , ℓ, ℓ, x), δ (m2 , p4 , q4 , r , r, y), ϵ (p, q, s, u, v, x, y), (1) (1) (2) (2) (3) m2 ,q4 ,p4 ,r ,r , p,q,s,u,v are as defined in Lemmas 1–9. m1 ,p1 ,q1 ,i,i , m2 ,p2 ,q2 ,j,j , m ,q ,p ,ℓ,ℓ , 1
3
3
Proof. By Balakrishnan and Lai (2009, page 493, Equation (11.30)), F (x, y) = Φ (x)Φ (y) + φ(x)φ(y)
∞ ρj j =1
j
Hj−1 (x)Hj−1 (y),
126
S. Nadarajah / Statistics and Probability Letters 119 (2016) 124–133
where φ(·) denotes the pdf of a standard normal random variable and Hν (·) denotes the Hermite polynomial of order ν . So, we can write
F (un (x), un (y)) = n
Φ (un (x)) Φ (un (y)) + φ (un (x)) φ (un (y))
∞ ρj j =1
j
n Hj−1 (un (x)) Hj−1 (un (y))
n ∞ j φ u ( x )) φ u ( y )) ρ ( ( n n = Φ n (un (x)) Φ n (un (y)) · 1 + Hj−1 (un (x)) Hj−1 (un (y)) Φ (un (x)) Φ (un (y)) j=1 j n n φ (un (x)) k φ (un (y)) k n n = Φ (un (x)) Φ (un (y)) k Φ (un (x)) Φ (un (y)) k=0 k ∞ ρj Hj−1 (un (x)) Hj−1 (un (y)) . ·
j=1
j
k k Φ n (un (x)) and Φ n (un (y)) can be expanded using Lemma 7. Φφ((uunn((xx)))) and Φφ((uunn((yy)))) can be expanded using Lemma 8. k ∞ ρj can be expanded using Lemma 9. The result follows. j=1 j Hj−1 (un (x)) Hj−1 (un (y)) The term in (5) corresponding to k = 0 is
Φ n (un (x)) Φ n (un (y)) =
µ (m1 , p1 , q1 , i, i, m2 , p2 , q2 , j, j, x, y)
m1 ,p1 ,q1 ,i,i m2 ,p2 ,q2 ,j,j
· cn2i2 +i3 +2j2 +j3 n−m1 −m2 an2|i|−m1 +i+2|j|−m2 +j bn−m1 −2q1 −i−m2 −2q2 −j ,
(6)
where
µ (m1 , p1 , q1 , i, i, m2 , p2 , q2 , j, j, x, y) = γ (m1 , p1 , q1 , i, i, x) γ (m2 , p2 , q2 , j, j, y) . The terms in (5) corresponding to k > 0 all approach zero because of the presence of n−m1 −m2 −|m1 |−|m2 |−2k . Theorem 2 shows that (6) approaches exp (− exp(−x) − exp(−y)) as n → ∞. Theorem 2. For un , an and bn given by (2)–(3),
Φ n (un (x)) Φ n (un (y)) → exp (− exp(−x) − exp(−y)) as n → ∞. Proof. The terms in (6) corresponding to q1 > 0, i > 0, i ̸= 03 , q2 > 0, j > 0 or j ̸= 03 each approaches zero as n → ∞. So, we consider the terms corresponding to q1 = 0, i = 0, i = 03 , q2 = 0, j = 0 and j = 03 : n n
(an bn )
−m1 −m2
n
−m1 −m2
n n
m1
m1 =0 m2 =0
=
n
m1 −1
(an bn )−m1 −m2
m1 =0 m2 =0
i=0
1−
n m2
i
(−1)m1 +m2 exp (−m1 x − m2 y)
m 2 −1
n
j =0
1−
j n
(−1)m1 +m2 exp (−m1 x − m2 y) . m1 !m2 !
Since an bn → 1 as n → ∞, the above limits to ∞ ∞ (−1)m1 +m2 exp (−m1 x − m2 y) = exp (− exp(−x) − exp(−y)) . m1 !m2 ! m =0 m =0 1
2
The proof is complete.
It is clear from the proof of Theorem 2 that the leading term in the expansion for Φ n (un (x)) Φ n (un (y)) is n n (−1)m1 +m2 exp (−m1 x − m2 y) . m1 !m2 ! m =0 m =0 1
2
Theorem 3 gives the complete expansion of Φ n (un (x)) Φ n (un (y)) excluding this leading term.
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127
Theorem 3. Let ωi (p) be defined by p
1−
i =0
i
=1+
n
p
ωi (p)n−i .
i=1
Then, for un , an and bn given by (2)–(3), we can write
Φ n (un (x)) Φ n (un (y)) −
n n (−1)m1 +m2 exp (−m1 x − m2 y) m1 !m2 ! m =0 m =0 1
(1)∗
(1)∗
=
2
µ (m1 , p1 , q1 , i, i, m2 , p2 , q2 , j, j, x, y)
m1 ,p1 ,q1 ,i,i m2 ,p2 ,q2 ,j,j
· cn2i2 +i3 +2j2 +j3 n−m1 −m2 a2n|i|−m1 +i+2|j|−m2 +j bn−m1 −2q1 −i−m2 −2q2 −j +
n n m 1 −1
ωi1 (m1 − 1) n−i1 +
ωi1 (m2 − 1) n−i1
m1 =0 m2 =0 i1 =1
m1 =0 m2 =0 i1 =1
+
n n m 2 −1
i n n ∞ −m1 − m2 1 1 2i1 i1 − an cn i1
m1 =0 m2 =0 i1 =1
2
m1 −1 m2 −1
+
n n
ωi1 (m1 − 1) ωi2 (m2 − 1) n−i1 −i2
m1 =0 m2 =0 i1 =1 i2 =1
+
n n m ∞ 1 −1
ωi1 (m1 − 1)
−m1 − m2 i2
m1 =0 m2 =0 i1 =1 i2 =1 m2 −1 ∞
i 1 2 2i2 −i1 i2 an n cn − 2
i n n −m1 − m2 1 2 2i2 −i1 i2 an n cn + ωi1 (m2 − 1) −
+
2
i2
m1 =0 m2 =0 i1 =1 i2 =1
∞ n n m 2 −1 1 −1 m
ωi1 (m1 − 1) ωi2 (m2 − 1)
−m1 − m2 i3
m1 =0 m2 =0 i1 =1 i2 =1 i3 =1
where
(1)∗
denotes
(1)
i 1 3 2i3 −i1 −i2 i3 − an n cn , 2
excluding q1 = 0, i = 0, i = 03 , q2 = 0, j = 0, j = 03 .
Since (7) is the leading term in the expansion for Φ n (un (x)) Φ n (un (y)), it is also the leading term in the expansion for F (un (x), un (y)). Theorem 4 gives the complete expansion of F n (un (x), un (y)) excluding this leading term. n
Theorem 4. For un , an and bn given by (2)–(3), F n (un (x), un (y)) −
n n (−1)m1 +m2 exp (−m1 x − m2 y) m1 !m2 ! m =0 m =0 1
=
2
2
(1)
π k
(1)
(2)
n n
m1 ,p1 ,q1 ,i,i m2 ,p2 ,q2 ,j,j k=1
k
(2)
(3)
·λ
m1 ,q3 ,p3 ,ℓ,ℓ m2 ,q4 ,p4 ,r ,r p,q,s,u,v
× (m1 , p1 , q1 , i, i, m2 , p2 , q2 , j, j, m1 , p3 , q3 , ℓ, ℓ, m2 , p4 , q4 , r , r, p, q, s, u, v, x, y) · cn2i2 +i3 +2j2 +j3 +2ℓ2 +ℓ3 +2r2 +r3 · n−m1 −m2 −|m1 |−|m2 |−2k · an2|i|−m1 +i+2|j|−m2 +j+ℓ+2|ℓ|−|m1 |+r +2|r|−|m2 |−2k−u−v · bn−m1 −2q1 −i−m2 −2q2 −j−|m1 |−2p3 −ℓ−|m2 |−2p4 −r −2|p|−2|q|−s1 −s2 +u+v +
(1)∗
(1)∗
µ (m1 , p1 , q1 , i, i, m2 , p2 , q2 , j, j, x, y)
m1 ,p1 ,q1 ,i,i m2 ,p2 ,q2 ,j,j
· cn2i2 +i3 +2j2 +j3 n−m1 −m2 a2n|i|−m1 +i+2|j|−m2 +j bn−m1 −2q1 −i−m2 −2q2 −j +
n n m 1 −1
ωi1 (m1 − 1) n−i1 +
m1 =0 m2 =0 i1 =1
n n m 2 −1 m1 =0 m2 =0 i1 =1
i n n ∞ −m1 − m2 1 1 2i1 i1 an cn + − m1 =0 m2 =0 i1 =1
i1
2
ωi1 (m2 − 1) n−i1
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S. Nadarajah / Statistics and Probability Letters 119 (2016) 124–133
+
n n m 1 −1 m 2 −1
ωi1 (m1 − 1) ωi2 (m2 − 1) n−i1 −i2
m1 =0 m2 =0 i1 =1 i2 =1
+
n n m ∞ 1 −1
ωi1 (m1 − 1)
−m1 − m2
2
i2
m1 =0 m2 =0 i1 =1 i2 =1
i 1 2 2i2 −i1 i2 − an n cn
i n n m ∞ 2 −1 1 2 2i2 −i1 i2 −m1 − m2 − + ωi1 (m2 − 1) an n cn
i2
m1 =0 m2 =0 i1 =1 i2 =1
+
n n m ∞ 1 −1 m 2 −1
2
ωi1 (m1 − 1) ωi2 (m2 − 1)
−m 1 − m 2
m1 =0 m2 =0 i1 =1 i2 =1 i3 =1
(1)∗
i3
i 1 3 2i3 −i1 −i2 i3 − cn an n 2
(1)
as n → ∞, where denotes excluding q1 = 0, i = 0, i = 03 , q2 = 0, j = 0, j = 03 , λ = γ (m1 , p1 , q1 , i, i, x) γ (m2 , p2 , q2 , j, j, y) δ (m1 , p3 , q3 , ℓ, ℓ, x) δ (m2 , p4 , q4 , r , r, y) ϵ (p, q, s, u, v, x, y) and γ (m1 , p1 , q1 , i, i, x), γ (m2 , p2 , q2 , (1) (1) (2) (2) j, j, y), δ (m1 , p3 , q3 , ℓ, ℓ, x), δ (m2 , p4 , q4 , r , r, y), ϵ (p, q, s, u, v, x, y), m1 ,p1 ,q1 ,i,i , m2 ,p2 ,q2 ,j,j , m ,q ,p ,ℓ,ℓ , m2 ,q4 ,p4 ,r ,r , 1
(3)
3
3
p,q,s,u,v are as defined in Lemmas 1–9.
3. Technical lemmas The following lemmas are needed to prove the main results. Lemma 1. For un , an and bn given by (2)–(3), exp −pu2n (x) =
∞3
2i +i3
b (i, p, x)
i=03
cn 2
2p−2|i|
n2p an
,
where i = (i1 , i2 , i3 ), p > 0, −∞ < x < +∞, and b (i, p, x) = (2π )p exp(−2px)
(−1)i1 +i2 p|i| x2i1 +i3 . 23i2 +i3 −|i| i1 !i2 !i3 !
Proof. Note that exp −pu2n (x) = exp −pa2n x2 − pb2n − 2pan bn x
px2
= exp −
cn
+ pcn exp −2px + px 2 log n 2 2 pcn cn px exp − exp px = (4π )p exp(−2px)n−2p (log n)p exp − 2 log n
exp −2p log n −
pcn2
8 log n
2 log n
The result now follows by using the series expansion for exponential. Lemma 2. For un , an and bn given by (2)–(3), um n (x) =
m m i =0
i
−i xi ain bm , n
where m > 0 is an integer and −∞ < x < +∞. Proof. Follows by binomial theorem.
Lemma 3. For un , an and bn given by (2)–(3), un (x) = −m
∞ −m i =0
i
m−i xi ain b− , n
where m > 0 and −∞ < x < +∞. Proof. Follows by binomial theorem.
8 log n
2 log n
.
S. Nadarajah / Statistics and Probability Letters 119 (2016) 124–133
129
Lemma 4. As x → ∞, n m ∞
Φ ( x) = n
c (m, p, q)x
−m−2q
exp −
mx2
2
m=0 p=0 q=p
,
where
n m (−1)m+q 2q− m2 p!
c (m, p, q) =
Bq,p (α),
m
p
π 2 q!
m
where
α=
1
1
1!,
2
2
1
2!, . . . ,
1 2
2
(q − p + 1)! .
q−p+1
Proof. Using the fact
Φ (x) =
1
1 + erf
2
x
,
√
2
where erf(·) denotes the error function defined by 2 erf(x) = √
x
π
exp −t 2 dt ,
0
we can write
Φ n ( x) =
1 2n
1 + erf
x
n
√
2
.
As x → ∞, 1
erf(x) = 1 − √
πx
exp −x
2
2 F0
1,
1 2
;;−
1
x2
,
where 2 F0 (a, b; ; x) denotes a hypergeometric function defined by 2 F0
(a, b; ; x) =
∞ (a)k (b)k xk
k!
k=0
.
So, as x → ∞, we can write
Φ (x) = n
1 2n
2− √
= 1− √
=
=
=
=
√
2
πx
1 2π x
exp −
exp −
x2 2
x2 2
∞ 1
2
k=0
∞ 1
k=0
2
k
−
k
−
2
2
k n
x2
k n
x2
k m n ∞ mx2 2 n (−1)m −m 1 exp − − 2 m x m (2π ) 2 2 2 k x m=0 k=0 k m ∞ n mx2 2 n (−1)m −m 1 exp − 1+ − 2 m x 2 2 k x m (2π ) 2 m=0 k=1 k p n m ∞ n (−1)m −m mx2 m 1 2 exp − − 2 m x m (2π ) 2 2 p 2 k x m=0 p=0 k=1 n m ∞ n (−1)m −m mx2 m (−2)q −2q exp − p! Bq,p (α) x , m x m (2π ) 2 2 p q! q =p m=0 p=0
where the last step follows by the definition of the Bell polynomial. The result follows by rearranging.
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S. Nadarajah / Statistics and Probability Letters 119 (2016) 124–133
Lemma 5. As x → ∞,
φ(x) Φ (x)
∞k |m| ∞ d (m, p, q)
k =
exp −
x|m|+2p
m=0k q=0 p=q
(|m| + k) x2
2
,
where m = (m1 , m2 , . . . , mk ) and d (m, p, q) = (2π )−
|m|+k 2
|m| q
(−2)p Bp,q (β), p!
where
1
β=
1
1!,
2
2
1
2!, . . . ,
1
2
2
(p − q + 1)! . p−q+1
Proof. As in the proof of Lemma 4, we can write
φ(x) = Φ (x)
2
π
exp −
x2
= √
2π
exp −
1
∞
2π
m=0
= √
x
1 + erf
2
1
2
√
2
x2
1
1− √ exp − 2 2π x
2
−m −m 2
(2π )
−1
x
x
(m + 1)x2
exp −
∞ 1 2
k=0
−
1
2
p=0
k −1
x2
k
∞
2
2
2
−
p m
x2
p
.
Taking the kth power, we obtain
φ(x) k = (2π )− 2 Φ (x) =
∞k
∞
−m −m 2
(2π )
x
exp −
=
− |m2|+k −|m|
(2π )
x
exp −
− |m2|+k −|m|
(2π )
x
exp −
∞k
(2π )−
|m|+k 2
x−|m| exp −
p m k ∞ 1 2 − 2
2
2
2
2
−
p |m|
x2
p
p |m| ∞ 1 2 1+ − 2 2
p=1
|m| 2
(|m| + k) x 2
x
p
∞ 1 p=0
(|m| + k) x2
m=0k
2
p=0
(|m| + k) x2
m=0k
=
2
m =0
m=0k
∞k
(m + 1)x2
|m| q
q =0
p
∞ p=q
x
Bp,q (β)
(−2)p , x2p p!
where the last step follows by the definition of the Bell polynomial. The result follows by rearranging. Lemma 6. We have
∞ ρj j
j=1
k Hj−1 (x)Hj−1 (y)
=
∞k ∞k k2 π k
2
f (p, q, m) x2|p|+m1 y2|q|+m2 ,
p=0k q=0k m=02
where m = (m1 , m2 ), p = (p1 , p2 , . . . , pk ), q = (q1 , q2 , . . . , qk ), and f (p, q, m) =
∞k (2ρ)|j| e (j, p, q, m) , j · · · jk j =1 1 k
where j = (j1 , j2 , . . . , jk ), and e (j, p, q, m) satisfies
k ∞ ∞ i=1
α1 (ji , p, q) x2p y2q +
p=0 q=0
+
∞ ∞ p=0 q=0
∞ ∞
α2 (ji , p, q) x2p y2q+1
p=0 q=0
α3 (ji , p, q) x
2p+1 2q
y
+
∞ ∞ p=0 q=0
α4 (ji , p, q) x
2p+1 2q+1
y
S. Nadarajah / Statistics and Probability Letters 119 (2016) 124–133
∞k ∞k k2
=
131
e (j, p, q, m) x2|p|+m1 y2|q|+m2 ,
p=0k q=0k m=02
where
1−j 1−j α1 (j, p, q) =
2
Γ2
2
p
q
2 −j 1 1 2
2 p
2p+q p!q! q
2
,
1−j 2−j α2 (j, p, q) = −
2
Γ
2−j 2
Γ
2
p
q
1−j 1 3 2
2 p
2 q
1
,
1
,
2p+q+ 2 p!q!
1−j 2−j α3 (j, p, q) = −
2
Γ
2−j 2
Γ
2
q
p
1−j 1 3 2
2 q
2 p
2p+q+ 2 p!q!
2−j 2−j α4 (j, p, q) =
2
Γ
2
p
2
1 −j 3 3 2
2 p
2 q
q
2p+q+1 p
!q!
.
Proof. We can write
∞ ρj j =1
j
k Hj−1 (x)Hj−1 (y)
=
∞k j =1 k
k ρ | j| Hji −1 (x)Hji −1 (y) . j1 · · · jk i=1
(8)
Using the fact H n ( x) = 2
n 2
√
π
1
Γ
n 1 x2
1−n 1 F1 − ; ; 2 2
2
2
−√
x 2Γ − 2n
1 F1
1 − n 3 x2 2
; ; 2
2
,
where 1 F1 (a; b; x) denotes the confluent hypergeometric function defined by 1 F1
(a; b; x) =
∞ (a)k xk k=0
(b)k k!
,
we can rewrite (8) as
∞ ρj j =1
j
k Hj−1 (x)Hj−1 (y)
∞k k π k x (2ρ)|j| 1 ji − 1 1 x2 2 − ji 3 x2 1 F1 − 1 F1 = · ; ; −√ ; ; j i −1 2 j · · · jk i=1 Γ 2−ji 2 2 2 2 2 2 2 Γ − j=1k 1 2 2 k y 1 ji − 1 1 y2 2 − ji 3 y2 1 F1 − 1 F1 · ; ; −√ ; ; j −1 2 2 2 2 2 2 Γ 2−2 ji 2Γ − i 2 i=1 ∞k k π k (2ρ)|j| 1 j i − 1 1 x2 ji − 1 1 y2 1 F1 − = ; ; ; ; 1 F1 − 2 j · · · jk i=1 Γ 2 2−ji 2 2 2 2 2 2 j =1 1 k
2
1 j i − 1 1 x2 2 − ji 3 y2 1 F1 − ; ; F ; ; 1 1 j −1 2−ji 2 2 2 2 2 2 2Γ − i 2 Γ 2 x 2 − j i 3 x2 ji − 1 1 y2 1 F1 −√ ; ; F − ; ; 1 1 j −1 2−ji 2 2 2 2 2 2 2Γ − i 2 Γ 2 xy 2 − j i 3 x2 2 − ji 3 y2 1 F1 + ; ; F ; ; 1 1 2 2 2 2 2 2 Γ 2 − j i −1
−√
2
132
S. Nadarajah / Statistics and Probability Letters 119 (2016) 124–133
∞k π k (2ρ)|j| 2 j · · · jk j =1 1
=
k
+
∞ ∞
k ∞ ∞ i=1
α1 (ji , p, q) x2p y2q +
p=0 q=0
α2 (ji , p, q) x2p y2q+1
p=0 q=0
α3 (ji , p, q) x2p+1 y2q +
∞ ∞
p=0 q=0
=
∞ ∞
α4 (ji , p, q) x2p+1 y2q+1
p=0 q=0
∞k ∞k ∞k k2 π k (2ρ)|j| e (j, p, q, m) x2|p|+m1 y2|q|+m2 2 j · · · j 1 k j =1 p=0 q=0 m=0 k
k
k
2
∞k ∞k ∞k k2 π k (2ρ)|j| = e (j, p, q, m) x2|p|+m1 y2|q|+m2 2 j · · · j k p=0 q=0 m=0 j=1 1 k
k
=
2
k
2
∞k ∞k k2 π k
f (p, q, m) x2|p|+m1 y2|q|+m2 .
p=0k q=0k m=02
Hence, the result.
Lemma 7. For un , an and bn given by (2)–(3),
Φ n (un (x)) =
(1)
2i +i3
cn 2
γ (m, p, q, i, i, x)
−2|i|+m−i m+2q+i
nm an
m,p,q,i,i
bn
as n → ∞, where (1)
:=
m,p,q,i,i
∞3 n m ∞ ∞ m=0 p=0 q=p i=0 i=03
and
m −m − 2q , γ (m, p, q, i, i, x) = x b i, , x c (m, p, q) i
2
where b i,
m 2
i
, x and c (m, p, q) are as defined in Lemmas 1 and 4, respectively.
Proof. Follows by Lemmas 1, 3 and 4.
Lemma 8. For un , an and bn given by (2)–(3),
φ (un (x)) Φ (un (x))
k =
(2)
2i +i3
δ (m, p, q, i, i, x)
m,q,p,i,i
cn 2
−i−2|i|+|m|+k |m|+2p+i
n|m|+k an
bn
as n → ∞, where (2)
:=
m,q,p,i,i
∞3 ∞k |m| ∞ ∞ m=0k q=0 p=q i=0 i=03
and
δ (m, p, q, i, i, x) = x b i, i
where b i,
|m|+k 2
| m| + k 2
, x d (m, p, q)
−|m| − 2p i
,
, x and d (m, p, q) are as defined in Lemmas 1 and 5, respectively.
Proof. Follows by Lemmas 1, 3 and 5.
Lemma 9. For un , an and bn given by (2)–(3),
∞ ρj j=1
j
k Hj−1 (un (x)) Hj−1 (un (y))
=
(3) π k
2
p,q,m,i,j
ϵ (p, q, m, i, j, x, y) ani+j bn2|p|+2|q|+m1 +m2 −i−j ,
S. Nadarajah / Statistics and Probability Letters 119 (2016) 124–133
133
where (3) p,q,m,i,j
:=
∞k ∞k |+m1 2|q |+m2 k2 2|p p=0k q=0k m=02
i=0
j =0
and
ϵ (p, q, m, i, j, x, y) = xi yj f (p, q, m)
2|p| + m1 i
2|q| + m2 j
,
where f (p, q, m) is as defined in Lemma 6. Proof. Follows by Lemmas 2 and 6.
Acknowledgments The author would like to thank the Editor and the referee for careful reading and comments which greatly improved the paper. References Balakrishnan, N., Lai, C.-D., 2009. Continuous Bivariate Distributions. Springer Verlag, New York. Comtet, L., 1974. Advanced Combinatorics. Reidel, Dordrecht. Hashorva, E., 2015. Extremes of aggregated Dirichlet risks. J. Multivariate Anal. 133, 334–345. Hashorva, E., Ji, L., 2014a. Extremes and first passage times of correlated fractional Brownian motions. Stoch. Models 30, 272–299. Hashorva, E., Ji, L., 2014b. Asymptotics of the finite-time ruin probability for the Sparre Andersen risk model perturbed by an inflated stationary chi-process. Comm. Statist.-Theory Methods 43, 2540–2548. Hashorva, E., Ji, L., 2015. Piterbarg theorems for chi-processes with trend. Extremes 18, 37–64. Hashorva, E., Ji, L., 2016. Extremes of α(t )-locally stationary Gaussian random fields. Trans. Amer. Math. Soc. 368, 1–26. Hashorva, E., Korshunov, D., Piterbarg, V.I., 2015. Asymptotic expansion of Gaussian chaos via probabilistic approach. Extremes 18, 315–347. Hashorva, E., Kortschak, D., 2014. Tail asymptotics of random sum and maximum of log-normal risks. Statist. Probab. Lett. 87, 167–174. Hashorva, E., Li, J., 2015. Tail behavior of weighted sums of order statistics of dependent risks. Stoch. Models 31, 1–19. Hashorva, E., Ling, C., 2016. Maxima of skew elliptical triangular arrays. Comm. Statist.-Theory Methods 45, 3692–3705. Hashorva, E., Peng, Z., Weng, Z., 2016. Higher-order expansions of distributions of maxima in a Husler-Reiss model. Methodol. Comput. Appl. Probab. 18, 181–196. Nadarajah, S., 2015. Complete asymptotic expansions for normal extremes. Statist. Probab. Lett. 103, 127–133.
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Statistics and Probability Letters journal homepage: www.elsevier.com/locate/stapro
Asymptotic expansions for bivariate normal extremes Saralees Nadarajah School of Mathematics, University of Manchester, Manchester M13 9PL, UK
article
abstract
info
Article history: Received 16 July 2016 Received in revised form 25 July 2016 Accepted 26 July 2016 Available online 3 August 2016
Nadarajah (2015) derived a complete asymptotic expansion for normal extremes. Here, we extend the expansion for bivariate normal extremes. © 2016 Elsevier B.V. All rights reserved.
Keywords: Bell polynomials Bivariate normal distribution Expansions
1. Introduction Let Φ (·) denote the cumulative distribution function (cdf) of a standard normal random variable. It is well known that Φ (·) belongs to the max domain of attraction of the Gumbel extreme value distribution, i.e.,
Φ n (un (x)) → exp (− exp(−x))
(1)
as n → ∞ and for any −∞ < x < +∞, where un (x) = an x + bn
(2)
and an cn 1 bn = a− n − 2 for n ≥ 1, where cn = log log n + log(4π ). Nadarajah (2015) provided a complete asymptotic expansion for (1). In particular, an expansion was provided for an = (2 log n)−1/2 ,
Φ n (un (x)) −
(3)
n (−1)m exp(−mx)
m!
m=0
as n → ∞, where un , an and bn are given by (2)–(3). The aim of this note is to extend Nadarajah (2015)’s work for the bivariate normal distribution given by the joint cdf F (x, y) =
1 2π
1 − ρ2
y
x
exp − −∞
−∞
u2 + v 2 − 2ρ uv 2 1 − ρ2
dudv
for −∞ < x < +∞, −∞ < y < +∞ and −1 < ρ < 1. It is well known that F n (un (x), un (y)) → exp (− exp(−x) − exp(−y)) as n → ∞, where un , an and bn are given by (2)–(3). E-mail address: [email protected]. http://dx.doi.org/10.1016/j.spl.2016.07.023 0167-7152/© 2016 Elsevier B.V. All rights reserved.
(4)
S. Nadarajah / Statistics and Probability Letters 119 (2016) 124–133
125
To the best of our knowledge, we are aware of no studies on convergence aspects of (4). However, there have been studies on convergence aspects of the extremes of other multivariate distributions and multivariate processes, see Hashorva and Ji (2014a,b, 2015, 2016), Hashorva and Kortschak (2014), Hashorva (2015), Hashorva et al. (2015, 2016), Hashorva and Li (2015) and Hashorva and Ling (2016). Asymptotic expansions for (4) could have both practical and theoretical appeal given the wide applicability of the bivariate normal distribution. In a practical sense, they could lead to better approximations for the limit exp(− exp(−x) − exp(−y)). Theoretically, such expansions can be used to derive expansions for the corresponding joint probability density function (pdf), moments, cumulants, quantiles, etc. Complete asymptotic expansions for (4) are derived in Section 2. Technical lemmas for their proof are given and proved in Section 3. The expansions involve Bell polynomials. In-built routines for Bell polynomials are available in most computer algebra packages. For example, see BellY in Mathematica. The following notation is used throughout this note: 0k denotes a k × 1 vector of zeros; 1k denotes a k × 1 vector of ones; ∞ ∞k denotes a k × 1 vector of infinities; |a| = a1 + a2 + · · · + ak for a k × 1 vector a = (a1 , a2 , . . . , ak )′ ; m=k 0k denotes the k fold summation m1 =0 m2 =0 · · · mk =0 ; (a)k = a(a + 1) · · · (a + k − 1) denotes the ascending factorial; Br ,k (x) denotes the partial exponential Bell polynomial defined by
∞
∞
xr t / r ! r
∞
k k! =
∞
∞
Br ,k (x)t r /r !
r =k
r =1
for x = (x1 , x2 , . . .). This polynomial is tabled on page 307 of Comtet (1974) for r ≤ 12. Our results in Sections 2 and 3 can in principle be extended to any other joint cdf. We have illustrated our results for the bivariate normal distribution because of its universality. 2. Main results Theorem 1 derives an asymptotic expansion for F n (un (x), un (y)). Theorem 2 uses Theorem 1 to prove (4). Theorems 3 and 4 derive complete asymptotic expansions for
Φ n (un (x)) Φ n (un (y)) −
n n (−1)m1 +m2 exp (−m1 x − m2 y) m1 !m2 ! m =0 m =0 2
1
and F n (un (x), un (y)) −
n n (−1)m1 +m2 exp (−m1 x − m2 y) , m1 !m2 ! m =0 m =0 2
1
respectively, where un , an and bn are given by (2)–(3). Proofs are given for Theorems 1 and 2. These proofs involve the use of Lemmas 1–9 in Section 3. Theorems 3 and 4 are straightforward consequences of Theorems 1 and 2, so their proofs are not given. Theorem 1. For un , an and bn given by (2)–(3), F n (un (x), un (y)) =
(1)
(1)
(2)
n n π k
m1 ,p1 ,q1 ,i,i m2 ,p2 ,q2 ,j,j k=0
k
2
(2)
(3)
m1 ,q3 ,p3 ,ℓ,ℓ m2 ,q4 ,p4 ,r ,r p,q,s,u,v
· λ (m1 , p1 , q1 , i, i, m2 , p2 , q2 , j, j, m1 , p3 , q3 , ℓ, ℓ, m2 , p4 , q4 , r , r, p, q, s, u, v, x, y) · cn2i2 +i3 +2j2 +j3 +2ℓ2 +ℓ3 +2r2 +r3 · n−m1 −m2 −|m1 |−|m2 |−2k · a2n|i|−m1 +i+2|j|−m2 +j+ℓ+2|ℓ|−|m1 |+r +2|r|−|m2 |−2k−u−v m1 −2q1 −i−m2 −2q2 −j−|m1 |−2p3 −ℓ−|m2 |−2p4 −r −2|p|−2|q|−s1 −s2 +u+v · b− n
(5)
as n → ∞, where λ = γ (m1 , p1 , q1 , i, i, x) γ (m2 , p2 , q2 , j, j, y) δ (m1 , p3 , q3 , ℓ, ℓ, x) δ (m2 , p4 , q4 , r , r, y) ϵ(p, q, s, u, v, x, y) and γ (m1 , p1 , q1 , i, i, x), γ (m2 , p2 , q2 , j, j, y), δ (m1 , p3 , q3 , ℓ, ℓ, x), δ (m2 , p4 , q4 , r , r, y), ϵ (p, q, s, u, v, x, y), (1) (1) (2) (2) (3) m2 ,q4 ,p4 ,r ,r , p,q,s,u,v are as defined in Lemmas 1–9. m1 ,p1 ,q1 ,i,i , m2 ,p2 ,q2 ,j,j , m ,q ,p ,ℓ,ℓ , 1
3
3
Proof. By Balakrishnan and Lai (2009, page 493, Equation (11.30)), F (x, y) = Φ (x)Φ (y) + φ(x)φ(y)
∞ ρj j =1
j
Hj−1 (x)Hj−1 (y),
126
S. Nadarajah / Statistics and Probability Letters 119 (2016) 124–133
where φ(·) denotes the pdf of a standard normal random variable and Hν (·) denotes the Hermite polynomial of order ν . So, we can write
F (un (x), un (y)) = n
Φ (un (x)) Φ (un (y)) + φ (un (x)) φ (un (y))
∞ ρj j =1
j
n Hj−1 (un (x)) Hj−1 (un (y))
n ∞ j φ u ( x )) φ u ( y )) ρ ( ( n n = Φ n (un (x)) Φ n (un (y)) · 1 + Hj−1 (un (x)) Hj−1 (un (y)) Φ (un (x)) Φ (un (y)) j=1 j n n φ (un (x)) k φ (un (y)) k n n = Φ (un (x)) Φ (un (y)) k Φ (un (x)) Φ (un (y)) k=0 k ∞ ρj Hj−1 (un (x)) Hj−1 (un (y)) . ·
j=1
j
k k Φ n (un (x)) and Φ n (un (y)) can be expanded using Lemma 7. Φφ((uunn((xx)))) and Φφ((uunn((yy)))) can be expanded using Lemma 8. k ∞ ρj can be expanded using Lemma 9. The result follows. j=1 j Hj−1 (un (x)) Hj−1 (un (y)) The term in (5) corresponding to k = 0 is
Φ n (un (x)) Φ n (un (y)) =
µ (m1 , p1 , q1 , i, i, m2 , p2 , q2 , j, j, x, y)
m1 ,p1 ,q1 ,i,i m2 ,p2 ,q2 ,j,j
· cn2i2 +i3 +2j2 +j3 n−m1 −m2 an2|i|−m1 +i+2|j|−m2 +j bn−m1 −2q1 −i−m2 −2q2 −j ,
(6)
where
µ (m1 , p1 , q1 , i, i, m2 , p2 , q2 , j, j, x, y) = γ (m1 , p1 , q1 , i, i, x) γ (m2 , p2 , q2 , j, j, y) . The terms in (5) corresponding to k > 0 all approach zero because of the presence of n−m1 −m2 −|m1 |−|m2 |−2k . Theorem 2 shows that (6) approaches exp (− exp(−x) − exp(−y)) as n → ∞. Theorem 2. For un , an and bn given by (2)–(3),
Φ n (un (x)) Φ n (un (y)) → exp (− exp(−x) − exp(−y)) as n → ∞. Proof. The terms in (6) corresponding to q1 > 0, i > 0, i ̸= 03 , q2 > 0, j > 0 or j ̸= 03 each approaches zero as n → ∞. So, we consider the terms corresponding to q1 = 0, i = 0, i = 03 , q2 = 0, j = 0 and j = 03 : n n
(an bn )
−m1 −m2
n
−m1 −m2
n n
m1
m1 =0 m2 =0
=
n
m1 −1
(an bn )−m1 −m2
m1 =0 m2 =0
i=0
1−
n m2
i
(−1)m1 +m2 exp (−m1 x − m2 y)
m 2 −1
n
j =0
1−
j n
(−1)m1 +m2 exp (−m1 x − m2 y) . m1 !m2 !
Since an bn → 1 as n → ∞, the above limits to ∞ ∞ (−1)m1 +m2 exp (−m1 x − m2 y) = exp (− exp(−x) − exp(−y)) . m1 !m2 ! m =0 m =0 1
2
The proof is complete.
It is clear from the proof of Theorem 2 that the leading term in the expansion for Φ n (un (x)) Φ n (un (y)) is n n (−1)m1 +m2 exp (−m1 x − m2 y) . m1 !m2 ! m =0 m =0 1
2
Theorem 3 gives the complete expansion of Φ n (un (x)) Φ n (un (y)) excluding this leading term.
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S. Nadarajah / Statistics and Probability Letters 119 (2016) 124–133
127
Theorem 3. Let ωi (p) be defined by p
1−
i =0
i
=1+
n
p
ωi (p)n−i .
i=1
Then, for un , an and bn given by (2)–(3), we can write
Φ n (un (x)) Φ n (un (y)) −
n n (−1)m1 +m2 exp (−m1 x − m2 y) m1 !m2 ! m =0 m =0 1
(1)∗
(1)∗
=
2
µ (m1 , p1 , q1 , i, i, m2 , p2 , q2 , j, j, x, y)
m1 ,p1 ,q1 ,i,i m2 ,p2 ,q2 ,j,j
· cn2i2 +i3 +2j2 +j3 n−m1 −m2 a2n|i|−m1 +i+2|j|−m2 +j bn−m1 −2q1 −i−m2 −2q2 −j +
n n m 1 −1
ωi1 (m1 − 1) n−i1 +
ωi1 (m2 − 1) n−i1
m1 =0 m2 =0 i1 =1
m1 =0 m2 =0 i1 =1
+
n n m 2 −1
i n n ∞ −m1 − m2 1 1 2i1 i1 − an cn i1
m1 =0 m2 =0 i1 =1
2
m1 −1 m2 −1
+
n n
ωi1 (m1 − 1) ωi2 (m2 − 1) n−i1 −i2
m1 =0 m2 =0 i1 =1 i2 =1
+
n n m ∞ 1 −1
ωi1 (m1 − 1)
−m1 − m2 i2
m1 =0 m2 =0 i1 =1 i2 =1 m2 −1 ∞
i 1 2 2i2 −i1 i2 an n cn − 2
i n n −m1 − m2 1 2 2i2 −i1 i2 an n cn + ωi1 (m2 − 1) −
+
2
i2
m1 =0 m2 =0 i1 =1 i2 =1
∞ n n m 2 −1 1 −1 m
ωi1 (m1 − 1) ωi2 (m2 − 1)
−m1 − m2 i3
m1 =0 m2 =0 i1 =1 i2 =1 i3 =1
where
(1)∗
denotes
(1)
i 1 3 2i3 −i1 −i2 i3 − an n cn , 2
excluding q1 = 0, i = 0, i = 03 , q2 = 0, j = 0, j = 03 .
Since (7) is the leading term in the expansion for Φ n (un (x)) Φ n (un (y)), it is also the leading term in the expansion for F (un (x), un (y)). Theorem 4 gives the complete expansion of F n (un (x), un (y)) excluding this leading term. n
Theorem 4. For un , an and bn given by (2)–(3), F n (un (x), un (y)) −
n n (−1)m1 +m2 exp (−m1 x − m2 y) m1 !m2 ! m =0 m =0 1
=
2
2
(1)
π k
(1)
(2)
n n
m1 ,p1 ,q1 ,i,i m2 ,p2 ,q2 ,j,j k=1
k
(2)
(3)
·λ
m1 ,q3 ,p3 ,ℓ,ℓ m2 ,q4 ,p4 ,r ,r p,q,s,u,v
× (m1 , p1 , q1 , i, i, m2 , p2 , q2 , j, j, m1 , p3 , q3 , ℓ, ℓ, m2 , p4 , q4 , r , r, p, q, s, u, v, x, y) · cn2i2 +i3 +2j2 +j3 +2ℓ2 +ℓ3 +2r2 +r3 · n−m1 −m2 −|m1 |−|m2 |−2k · an2|i|−m1 +i+2|j|−m2 +j+ℓ+2|ℓ|−|m1 |+r +2|r|−|m2 |−2k−u−v · bn−m1 −2q1 −i−m2 −2q2 −j−|m1 |−2p3 −ℓ−|m2 |−2p4 −r −2|p|−2|q|−s1 −s2 +u+v +
(1)∗
(1)∗
µ (m1 , p1 , q1 , i, i, m2 , p2 , q2 , j, j, x, y)
m1 ,p1 ,q1 ,i,i m2 ,p2 ,q2 ,j,j
· cn2i2 +i3 +2j2 +j3 n−m1 −m2 a2n|i|−m1 +i+2|j|−m2 +j bn−m1 −2q1 −i−m2 −2q2 −j +
n n m 1 −1
ωi1 (m1 − 1) n−i1 +
m1 =0 m2 =0 i1 =1
n n m 2 −1 m1 =0 m2 =0 i1 =1
i n n ∞ −m1 − m2 1 1 2i1 i1 an cn + − m1 =0 m2 =0 i1 =1
i1
2
ωi1 (m2 − 1) n−i1
128
S. Nadarajah / Statistics and Probability Letters 119 (2016) 124–133
+
n n m 1 −1 m 2 −1
ωi1 (m1 − 1) ωi2 (m2 − 1) n−i1 −i2
m1 =0 m2 =0 i1 =1 i2 =1
+
n n m ∞ 1 −1
ωi1 (m1 − 1)
−m1 − m2
2
i2
m1 =0 m2 =0 i1 =1 i2 =1
i 1 2 2i2 −i1 i2 − an n cn
i n n m ∞ 2 −1 1 2 2i2 −i1 i2 −m1 − m2 − + ωi1 (m2 − 1) an n cn
i2
m1 =0 m2 =0 i1 =1 i2 =1
+
n n m ∞ 1 −1 m 2 −1
2
ωi1 (m1 − 1) ωi2 (m2 − 1)
−m 1 − m 2
m1 =0 m2 =0 i1 =1 i2 =1 i3 =1
(1)∗
i3
i 1 3 2i3 −i1 −i2 i3 − cn an n 2
(1)
as n → ∞, where denotes excluding q1 = 0, i = 0, i = 03 , q2 = 0, j = 0, j = 03 , λ = γ (m1 , p1 , q1 , i, i, x) γ (m2 , p2 , q2 , j, j, y) δ (m1 , p3 , q3 , ℓ, ℓ, x) δ (m2 , p4 , q4 , r , r, y) ϵ (p, q, s, u, v, x, y) and γ (m1 , p1 , q1 , i, i, x), γ (m2 , p2 , q2 , (1) (1) (2) (2) j, j, y), δ (m1 , p3 , q3 , ℓ, ℓ, x), δ (m2 , p4 , q4 , r , r, y), ϵ (p, q, s, u, v, x, y), m1 ,p1 ,q1 ,i,i , m2 ,p2 ,q2 ,j,j , m ,q ,p ,ℓ,ℓ , m2 ,q4 ,p4 ,r ,r , 1
(3)
3
3
p,q,s,u,v are as defined in Lemmas 1–9.
3. Technical lemmas The following lemmas are needed to prove the main results. Lemma 1. For un , an and bn given by (2)–(3), exp −pu2n (x) =
∞3
2i +i3
b (i, p, x)
i=03
cn 2
2p−2|i|
n2p an
,
where i = (i1 , i2 , i3 ), p > 0, −∞ < x < +∞, and b (i, p, x) = (2π )p exp(−2px)
(−1)i1 +i2 p|i| x2i1 +i3 . 23i2 +i3 −|i| i1 !i2 !i3 !
Proof. Note that exp −pu2n (x) = exp −pa2n x2 − pb2n − 2pan bn x
px2
= exp −
cn
+ pcn exp −2px + px 2 log n 2 2 pcn cn px exp − exp px = (4π )p exp(−2px)n−2p (log n)p exp − 2 log n
exp −2p log n −
pcn2
8 log n
2 log n
The result now follows by using the series expansion for exponential. Lemma 2. For un , an and bn given by (2)–(3), um n (x) =
m m i =0
i
−i xi ain bm , n
where m > 0 is an integer and −∞ < x < +∞. Proof. Follows by binomial theorem.
Lemma 3. For un , an and bn given by (2)–(3), un (x) = −m
∞ −m i =0
i
m−i xi ain b− , n
where m > 0 and −∞ < x < +∞. Proof. Follows by binomial theorem.
8 log n
2 log n
.
S. Nadarajah / Statistics and Probability Letters 119 (2016) 124–133
129
Lemma 4. As x → ∞, n m ∞
Φ ( x) = n
c (m, p, q)x
−m−2q
exp −
mx2
2
m=0 p=0 q=p
,
where
n m (−1)m+q 2q− m2 p!
c (m, p, q) =
Bq,p (α),
m
p
π 2 q!
m
where
α=
1
1
1!,
2
2
1
2!, . . . ,
1 2
2
(q − p + 1)! .
q−p+1
Proof. Using the fact
Φ (x) =
1
1 + erf
2
x
,
√
2
where erf(·) denotes the error function defined by 2 erf(x) = √
x
π
exp −t 2 dt ,
0
we can write
Φ n ( x) =
1 2n
1 + erf
x
n
√
2
.
As x → ∞, 1
erf(x) = 1 − √
πx
exp −x
2
2 F0
1,
1 2
;;−
1
x2
,
where 2 F0 (a, b; ; x) denotes a hypergeometric function defined by 2 F0
(a, b; ; x) =
∞ (a)k (b)k xk
k!
k=0
.
So, as x → ∞, we can write
Φ (x) = n
1 2n
2− √
= 1− √
=
=
=
=
√
2
πx
1 2π x
exp −
exp −
x2 2
x2 2
∞ 1
2
k=0
∞ 1
k=0
2
k
−
k
−
2
2
k n
x2
k n
x2
k m n ∞ mx2 2 n (−1)m −m 1 exp − − 2 m x m (2π ) 2 2 2 k x m=0 k=0 k m ∞ n mx2 2 n (−1)m −m 1 exp − 1+ − 2 m x 2 2 k x m (2π ) 2 m=0 k=1 k p n m ∞ n (−1)m −m mx2 m 1 2 exp − − 2 m x m (2π ) 2 2 p 2 k x m=0 p=0 k=1 n m ∞ n (−1)m −m mx2 m (−2)q −2q exp − p! Bq,p (α) x , m x m (2π ) 2 2 p q! q =p m=0 p=0
where the last step follows by the definition of the Bell polynomial. The result follows by rearranging.
130
S. Nadarajah / Statistics and Probability Letters 119 (2016) 124–133
Lemma 5. As x → ∞,
φ(x) Φ (x)
∞k |m| ∞ d (m, p, q)
k =
exp −
x|m|+2p
m=0k q=0 p=q
(|m| + k) x2
2
,
where m = (m1 , m2 , . . . , mk ) and d (m, p, q) = (2π )−
|m|+k 2
|m| q
(−2)p Bp,q (β), p!
where
1
β=
1
1!,
2
2
1
2!, . . . ,
1
2
2
(p − q + 1)! . p−q+1
Proof. As in the proof of Lemma 4, we can write
φ(x) = Φ (x)
2
π
exp −
x2
= √
2π
exp −
1
∞
2π
m=0
= √
x
1 + erf
2
1
2
√
2
x2
1
1− √ exp − 2 2π x
2
−m −m 2
(2π )
−1
x
x
(m + 1)x2
exp −
∞ 1 2
k=0
−
1
2
p=0
k −1
x2
k
∞
2
2
2
−
p m
x2
p
.
Taking the kth power, we obtain
φ(x) k = (2π )− 2 Φ (x) =
∞k
∞
−m −m 2
(2π )
x
exp −
=
− |m2|+k −|m|
(2π )
x
exp −
− |m2|+k −|m|
(2π )
x
exp −
∞k
(2π )−
|m|+k 2
x−|m| exp −
p m k ∞ 1 2 − 2
2
2
2
2
−
p |m|
x2
p
p |m| ∞ 1 2 1+ − 2 2
p=1
|m| 2
(|m| + k) x 2
x
p
∞ 1 p=0
(|m| + k) x2
m=0k
2
p=0
(|m| + k) x2
m=0k
=
2
m =0
m=0k
∞k
(m + 1)x2
|m| q
q =0
p
∞ p=q
x
Bp,q (β)
(−2)p , x2p p!
where the last step follows by the definition of the Bell polynomial. The result follows by rearranging. Lemma 6. We have
∞ ρj j
j=1
k Hj−1 (x)Hj−1 (y)
=
∞k ∞k k2 π k
2
f (p, q, m) x2|p|+m1 y2|q|+m2 ,
p=0k q=0k m=02
where m = (m1 , m2 ), p = (p1 , p2 , . . . , pk ), q = (q1 , q2 , . . . , qk ), and f (p, q, m) =
∞k (2ρ)|j| e (j, p, q, m) , j · · · jk j =1 1 k
where j = (j1 , j2 , . . . , jk ), and e (j, p, q, m) satisfies
k ∞ ∞ i=1
α1 (ji , p, q) x2p y2q +
p=0 q=0
+
∞ ∞ p=0 q=0
∞ ∞
α2 (ji , p, q) x2p y2q+1
p=0 q=0
α3 (ji , p, q) x
2p+1 2q
y
+
∞ ∞ p=0 q=0
α4 (ji , p, q) x
2p+1 2q+1
y
S. Nadarajah / Statistics and Probability Letters 119 (2016) 124–133
∞k ∞k k2
=
131
e (j, p, q, m) x2|p|+m1 y2|q|+m2 ,
p=0k q=0k m=02
where
1−j 1−j α1 (j, p, q) =
2
Γ2
2
p
q
2 −j 1 1 2
2 p
2p+q p!q! q
2
,
1−j 2−j α2 (j, p, q) = −
2
Γ
2−j 2
Γ
2
p
q
1−j 1 3 2
2 p
2 q
1
,
1
,
2p+q+ 2 p!q!
1−j 2−j α3 (j, p, q) = −
2
Γ
2−j 2
Γ
2
q
p
1−j 1 3 2
2 q
2 p
2p+q+ 2 p!q!
2−j 2−j α4 (j, p, q) =
2
Γ
2
p
2
1 −j 3 3 2
2 p
2 q
q
2p+q+1 p
!q!
.
Proof. We can write
∞ ρj j =1
j
k Hj−1 (x)Hj−1 (y)
=
∞k j =1 k
k ρ | j| Hji −1 (x)Hji −1 (y) . j1 · · · jk i=1
(8)
Using the fact H n ( x) = 2
n 2
√
π
1
Γ
n 1 x2
1−n 1 F1 − ; ; 2 2
2
2
−√
x 2Γ − 2n
1 F1
1 − n 3 x2 2
; ; 2
2
,
where 1 F1 (a; b; x) denotes the confluent hypergeometric function defined by 1 F1
(a; b; x) =
∞ (a)k xk k=0
(b)k k!
,
we can rewrite (8) as
∞ ρj j =1
j
k Hj−1 (x)Hj−1 (y)
∞k k π k x (2ρ)|j| 1 ji − 1 1 x2 2 − ji 3 x2 1 F1 − 1 F1 = · ; ; −√ ; ; j i −1 2 j · · · jk i=1 Γ 2−ji 2 2 2 2 2 2 2 Γ − j=1k 1 2 2 k y 1 ji − 1 1 y2 2 − ji 3 y2 1 F1 − 1 F1 · ; ; −√ ; ; j −1 2 2 2 2 2 2 Γ 2−2 ji 2Γ − i 2 i=1 ∞k k π k (2ρ)|j| 1 j i − 1 1 x2 ji − 1 1 y2 1 F1 − = ; ; ; ; 1 F1 − 2 j · · · jk i=1 Γ 2 2−ji 2 2 2 2 2 2 j =1 1 k
2
1 j i − 1 1 x2 2 − ji 3 y2 1 F1 − ; ; F ; ; 1 1 j −1 2−ji 2 2 2 2 2 2 2Γ − i 2 Γ 2 x 2 − j i 3 x2 ji − 1 1 y2 1 F1 −√ ; ; F − ; ; 1 1 j −1 2−ji 2 2 2 2 2 2 2Γ − i 2 Γ 2 xy 2 − j i 3 x2 2 − ji 3 y2 1 F1 + ; ; F ; ; 1 1 2 2 2 2 2 2 Γ 2 − j i −1
−√
2
132
S. Nadarajah / Statistics and Probability Letters 119 (2016) 124–133
∞k π k (2ρ)|j| 2 j · · · jk j =1 1
=
k
+
∞ ∞
k ∞ ∞ i=1
α1 (ji , p, q) x2p y2q +
p=0 q=0
α2 (ji , p, q) x2p y2q+1
p=0 q=0
α3 (ji , p, q) x2p+1 y2q +
∞ ∞
p=0 q=0
=
∞ ∞
α4 (ji , p, q) x2p+1 y2q+1
p=0 q=0
∞k ∞k ∞k k2 π k (2ρ)|j| e (j, p, q, m) x2|p|+m1 y2|q|+m2 2 j · · · j 1 k j =1 p=0 q=0 m=0 k
k
k
2
∞k ∞k ∞k k2 π k (2ρ)|j| = e (j, p, q, m) x2|p|+m1 y2|q|+m2 2 j · · · j k p=0 q=0 m=0 j=1 1 k
k
=
2
k
2
∞k ∞k k2 π k
f (p, q, m) x2|p|+m1 y2|q|+m2 .
p=0k q=0k m=02
Hence, the result.
Lemma 7. For un , an and bn given by (2)–(3),
Φ n (un (x)) =
(1)
2i +i3
cn 2
γ (m, p, q, i, i, x)
−2|i|+m−i m+2q+i
nm an
m,p,q,i,i
bn
as n → ∞, where (1)
:=
m,p,q,i,i
∞3 n m ∞ ∞ m=0 p=0 q=p i=0 i=03
and
m −m − 2q , γ (m, p, q, i, i, x) = x b i, , x c (m, p, q) i
2
where b i,
m 2
i
, x and c (m, p, q) are as defined in Lemmas 1 and 4, respectively.
Proof. Follows by Lemmas 1, 3 and 4.
Lemma 8. For un , an and bn given by (2)–(3),
φ (un (x)) Φ (un (x))
k =
(2)
2i +i3
δ (m, p, q, i, i, x)
m,q,p,i,i
cn 2
−i−2|i|+|m|+k |m|+2p+i
n|m|+k an
bn
as n → ∞, where (2)
:=
m,q,p,i,i
∞3 ∞k |m| ∞ ∞ m=0k q=0 p=q i=0 i=03
and
δ (m, p, q, i, i, x) = x b i, i
where b i,
|m|+k 2
| m| + k 2
, x d (m, p, q)
−|m| − 2p i
,
, x and d (m, p, q) are as defined in Lemmas 1 and 5, respectively.
Proof. Follows by Lemmas 1, 3 and 5.
Lemma 9. For un , an and bn given by (2)–(3),
∞ ρj j=1
j
k Hj−1 (un (x)) Hj−1 (un (y))
=
(3) π k
2
p,q,m,i,j
ϵ (p, q, m, i, j, x, y) ani+j bn2|p|+2|q|+m1 +m2 −i−j ,
S. Nadarajah / Statistics and Probability Letters 119 (2016) 124–133
133
where (3) p,q,m,i,j
:=
∞k ∞k |+m1 2|q |+m2 k2 2|p p=0k q=0k m=02
i=0
j =0
and
ϵ (p, q, m, i, j, x, y) = xi yj f (p, q, m)
2|p| + m1 i
2|q| + m2 j
,
where f (p, q, m) is as defined in Lemma 6. Proof. Follows by Lemmas 2 and 6.
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