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DOI:10.1051/ps/2009012 www.esaim-ps.org

EXPANSIONS FOR THE DISTRIBUTION OF M -ESTIMATES WITH APPLICATIONS TO THE MULTI-TONE PROBLEM

Christopher S. Withers

1

and Saralees Nadarajah

2

Abstract. We give a stochastic expansion for estimates θthat minimise the arithmetic mean of (typically independent) random functions of a known parameter θ. Examples include least squares estimates, maximum likelihood estimates and more generally M-estimates. This is used to obtain leading cumulant coefficients ofθneeded for the Edgeworth expansions for the distribution and density ofn1/2(θ−θ0) to magnituden−3/2 (or ton−2for the symmetric case), whereθ0 is the true parameter value and n is typically the sample size. Applications are given to least squares estimates for both real and complex models. An alternative approach is given when the linear parameters of the model are nuisance parameters. The methods are illustrated with the problem of estimating the frequencies when the signal consists of the sum of sinusoids of unknown amplitudes.

Mathematics Subject Classification. 62E17, 62E20.

Received November 13, 2008. Revised March 9, 2009 and June 4, 2009.

1. Introduction and summary

Letθdenote an estimate of θ inRp based on a random sample of size n. There is a large amount of work on expansions for θ−θ0, where θ0 is the true value of θ. However, most of the work to date are for the sample mean and functions of it. For example, Monti [5] obtains an expansion for the sample mean up to the second order by expanding the saddlepoint approximation. Boothet al. [1] give tilted expansions of a sample mean from a distribution on kpoints. Kakizawa and Taniguchi [4] obtain expansions for P(θ < x) under the assumption thatθhas a cumulant expansion in powers ofn−1. Gatto and Ronchetti [3] provide approximations forP(m( ¯X)< x) up to 1 +O(n−1) form(·) a smooth function. For a comprehensive review of the known work, we refer the readers to [8].

The aim of this paper is to provide expansions for those θthat minimise the arithmetic mean of random functions of θ. Maximum likelihood estimates (MLEs), least squares estimates (LSEs), and more generally M-estimates are examples of θwhich minimise a random mean function

Λ = Λ(θ) =n−1 n N=1

ΛN(θ)

Keywords and phrases. Bias, edgeworth, maximum likelihood,M-estimates, Skewness.

1 Applied Mathematics Group, Industrial Research Limited, Lower Hutt, New Zealand.

2 School of Mathematics, University of Manchester, Manchester M13 9PL, UK;mbbsssn2@manchester.ac.uk

Article published by EDP Sciences c EDP Sciences, SMAI 2011

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for θin Rp. If E ∂Λ/∂θ = 0 andE ∂2Λ/∂θ∂θ >0, thenθ→ p θ0 as n→ ∞. (We use Rand C to denote the real and complex numbers).

The contents of this paper are organized as follows. In Section 2 we give a stochastic expansion forθ−θ0 of the form

θ−θ0

a=1

δa,

whereδa=Op(n−a/2). (Theainδa is a superscript not a power).

In Section 3 we use this to obtain the leading coefficients in the expansions for the cumulants ofθ:

κ(θi1, . . . ,θir)

j=r−1

kij1...irn−j (1.1)

for r 1 with k0i = θ0i. This implies that Yn = n1/2(θ−θ0) → Np(0, V) with V = (k1i1i2) = A−1 for A=E ∂2Λ/∂θ∂θ. The leading bias and skewness coefficientsk0i1 andki21i2i3 give the Edgeworth expansion of the distribution (and its derivatives) ofYn toO(n−1), while the coefficientsk2i1i2 andki31...i4 give the Edgeworth expansion of the distribution ofYn toO(n−3/2) andP(Yn∈S) toO(n−2) forS =−S⊂ Rp.

Section 4 applies these results to the LSEs for the general signal plus noise model

YN =SN(θ) +eN inRorC, 1≤N ≤n (1.2)

with ΛN(θ) =|YN−SN(θ)|2/2, where the residualse1, . . . , en are assumed independent with mean zero. While the complex formulation can also be dealt with by the real formulation, there are some significant simplifications in staying with the complex model. The M-estimate with respect to a given convex functionρonRor C for the model (1.2) isθfor ΛN(θ) =ρ(YN−ρ(θ)). For smoothρthe leading cumulant coefficients were essentially found by this method in the real case in [7].

Section 5 considers two examples on the signal frequency problem

yk=s(θ) +nk inCM, k= 1, . . . , K, (1.3) wherenk is complex normal with covariance not depending onθ, and themth component ofs(θ) is

sm(θ) = R r=1

arexp(jφr+jwr(m+m0)/M) form= 0,1, . . . , M 1,

where j =

1. The p= 3R parameters areθ = (a, φ, w). So, (1.3) can be written in the form (1.2) with n= 2kM. Changing the covariance constantm0is equivalent to reparameterisingφr: we shall see that ifR= 1 then takingm0=(M1)/2 makes the asymptotic covariance ofθdiagonal.

In Section 6 we give a variation of the method for the case when the linear parameters of the model are nuisance parameters.

Appendix A provides a list of summation notations used throughout the paper. Some technical details required for the two examples in Section 5 are given in Appendices B and C. The proofs of all theorems are given in Appendix D.

Forxa complex matrix we shall usexto denote its transpose, ¯xits complex conjugate, andxthe transpose of its complex conjugate.

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2. The stochastic expansion

Suppose {ΛN(θ)} are real random functions of θ in Rp. Here we show that θ minimising Λ = Λ(θ) = n−1n

N=1ΛN(θ) in Rp has the stochastic expansion δ=θ−θ0=

a=1

δa (2.1)

withδa =Op(n−a/2). To avoid excessive subscripts we shall fix 1≤i0, i1, . . .≤pand set

δaj = (δa)ij, ∂j =∂/∂θij, δj = (δ)ij,Λ·12...=12. . .Λ(θ), (2.2) A12... = EΛ·12..., and Δ12...= Λ·12...−A12.... (2.3) Theorem 2.1 gives the first threeδa explicitly in terms of these Δ’s andA’s. By (2.1) this givesθexplicitly in terms of Δ’s andA’s toOp(n−2).

Forθto be a consistent estimate we need to assume that

A1 = ·1= 0, (2.4)

Δ1...r = Op(n−1/2) asn→ ∞. (2.5)

Typically the model contains a location parameter, and the constraint (2.4) effectively specifies how it is defined, as well as identifying the other parameters of the model. The constraint (2.5) generally follows by the Central Limit Theorem, if the{ΛN(θ)}are independent or weakly dependent.

Theorem 2.1. Suppose θ is the estimate as defined above satisfying (2.4) and (2.5). Suppose also that the eigenvalues of the p×pmatrix A= (A12 : 1≤i1, i2 ≤p) are bounded away from zero as n→ ∞, so A has bounded inverseA−1= (A12)asn→ ∞. Then (2.1) holds. Furthermore,

δ10 = −A01Δ1, (2.6)

δ20 = A01Δ12A23Δ3−B045Δ4Δ5, (2.7)

δ30 = −A0112δ22−A123 2

23

A24Δ4δ32/2 + Δ123A24Δ4A35Δ5/2−C1567Δ5Δ6Δ7),

= −A01

Δ12(A23Δ34A45Δ5−B234Δ3Δ4)

−2−1A123 2

23

A24Δ4(A35Δ56A67Δ7−B356Δ5Δ6) +2−1A24A35Δ123Δ4Δ5−C1234Δ2Δ3Δ4

, (2.8)

and so on, whereB123=A14A25A36A456/2 andC1567=A1234A25A36A47/6.

Note that B045,B234,B356 andC1234 in (2.6)–(2.8) follow from the definitions given forB123andC1567. For example,B045=A04A45A56A456/2,B234=A24A35A46A456/2 andC1234=A1234A22A33A44/6. The first three δa given will be sufficient to obtain the cumulant coefficients needed.

3. The leading cumulant coefficients

In this section we give the cumulant coefficients of (1.1) needed for the distribution ofn1/2(θ−θ0) toO(n−3/2), namely k.112, k.10, k123.2 , k.212, k.31234, where we extend the notation of (D.1), (2.3) by settingk.j1...r =kji1...ir. We

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now assume Λ1(θ), . . . ,Λn(θ) independent. Forπ a sequence of integers in {1, . . . , p}, using the dot notation of (2.2) for partial derivatives,

Λ·π =n−1 n N=1

ΛN·π.

So, forπ1, π2, . . .such sequences, the joint cumulants of the Λ·π are given by

κ(Λ·π1, . . . ,Λ·πr) =n1−r1, . . . , πr], (3.1) where

1, . . . , πr] =n−1 n N=1

κ(ΛN·π1, . . . ,ΛN·πr). (3.2)

For example, [1. . . r] = ·1...r =A1...r. We shall give the leading cumulant coefficients we need in terms of these [.] functions.

Setδrs= 1 forr=sand 0 otherwise, andab1...r1...br =κ(δb11, . . . , δrbr). This has an expansion of the form ab1...r1...br =

j≥(b1+...+br)/2

ab1...r·j1...br n−j.

Also, the left hand side of (1.1) is equal to θ0i1δr1+κ(δ1, . . . , δr). Substituting (2.1) into this gives k·j1...r=θ0i1δr1δj0+

b1+...+br≤2j

ab1...r.j1...br, (3.3)

whereb1, . . . , br are positive integers.

Theorem 3.1. The coefficients k12.1,k0.1,k123.2 ,k12.2 andk1234.3 of (3.3) can be expressed as

k.112=a1112.1, k.10 =a21.1, k.2123=a111123.2+ 3 112

a112123.2, (3.4)

k12.2 = 2

12

a1212.2+ 2

13

a1312.2+a2212.2 (3.5)

and

k.31234=a11111234·3+ 4 1112

a11121234·3+ 4 1113

a11131234·3+ 6 1122

a11221234·3, (3.6)

where

k.112=a1112.1=A13A24[3,4], (3.7)

k.10 =a21.1=A01A23[12,3]−B045[4,5], (3.8)

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a111123·2=−A14A25A36[4,5,6], (3.9)

a112123·2=A14A25

A36A78 2

45

[4,67] [5,8]−B367 2

45

[4,6] [5,7]

, (3.10)

a1212·2=−A13A24A56[3,45,6] +A13B245[3,4,5], (3.11)

a1312·2 = A13A24

A56A78 3

[3,45] [67,8]−B567 3

[3,45] [6,7]

−2−1A456 2

56

A57 A68A9,10 3

[3,7] [89,10]−B689 3

[3,7] [8,9]

+2−1A57A68 3

[3,456] [7,8]−C4567 3

[3,5] [6,7]

, (3.12)

a2212·2 = A13A45A26A78 2 34,5

[5,8] [34,67]2

12

A13A45B267 2

6,7

[5,6] [7,34]

+B134B267 2

34

[3,6] [4,7], (3.13)

a11111234·3=A15A26A37A48[5,6,7,8], (3.14)

a11121234·3=A15A26A37

−A48A9,10 6

[89,10] [5,6,7] +B489 6

[8,9] [5,6,7]

, (3.15)

a11131234·3 = A15A26A37A48

A9,10A11,12 15

[5,6] [7,89] [10.11,12]

−B9,10,11 15

[5,6] [7,89] [10,11]

−2−1A89,10 2 9,10

A9,11 A10,12A13,14 15

[5,6] [7,11] [12.13,14]

−B10,12,13 15

[5,6] [7,11] [12,13]

+2−1A9,11A10,12 15

[5,6] [7,89.10] [11,12]

−C89,10,1115

[5,6] [7,9] [10,11]

(3.16)

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and

a11221234·3 = A15A26

A37A89

A4,10A11,12 15

[5,6] [78,9] [10.11,12]

−B4,10,11 15

[5,6] [78,9] [10,11]

−B378

A4,10A11,12 15

[5,6] [7,8] [10.11,12]

−B4,10,11 15

[5,6] [7,8] [10,11]

. (3.17)

4. Least squares estimates

Here we apply the previous section to LSEs for both real and complex models. We begin with therealmodel.

Suppose we observe

Y =S(θ) +ein Rn, that isYN =SN(θ) +eN in Rfor 1≤N ≤n (4.1) with e1, . . . , en independent and identically distributed (i.i.d.) with mean zero. Denote theirrth cumulant by λr=κr(e1). The LSE isθminimising

Λ =n−1|Y −S(θ)|2/2 =n−1 n N=1

ΛN(θ), (4.2)

where ΛN(θ) =|YN −SN(θ)|2/2. So,

ΛN·1...r =1. . . ∂rSN2/2−YNSN·1...r, A1...r = [1. . . r] =n−1 n

N=1

{∂1. . . ∂rSN2/2−SNSN·1...r}.

Forπ1, π2, . . .sequences of integers in 1. . . p, set π1, π2, . . .=n−1

n N=1

SN·π1SN·π2. . . (4.3)

Theorem 4.1 uses the previous section to express the cumulant coefficients we need in terms of these functions.

Theorem 4.2 notes what form these take for the special case whenA is diagonal, having in mind the example of the next section for the caseR= 1.

Theorem 4.1. For the model given by (4.1), the cumulant coefficients of Theorem 3.1 are

ki11i2 =k12·1 =λ2A12, (4.4)

ki10=k0·1= 2−1λ2A01A23{−1,23 − 2,13+3,12}, (4.5)

ki21i2i3=k123·2 =A14A25A36

λ34,5,6+λ22 3

456

4,56

22B123, (4.6)

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a1212·2=λ3A13

A24A563,45,6 −B2453,4,5

, (4.7)

a1312·222 = A13A24A56 A78 2

38

3,45 8,67+34,56

−2−1A243,45A56A678(A13A78+ 2A18A37) +A13A24

2−13,45B56,45B356

+2A346B6+ 2−1A456A378A57A68+ 2−13,456A56 +6,345A562−1A3456A56

, (4.8)

a2212·222=A13A26(A4734,67+A45A7834,8 5,67)2 2

12

A13B2466,34+ 2B613B263 , (4.9) a11111234·3=λ4A15A26A37A485,6,7,8, (4.10)

a11121234·3 = [λ2λ3]

−A15A26A37A48A9,10 3

567

5,89 10,6,7

−A48[A26A37A1989,6,7+A15A37A2989,5,7+A15A26A3989,5,6]

+A15A26A37A48A9,10 3

567

A58,106,7,9

, (4.11)

a11131234·3 = λ32A15A26A37A48

A9,10A11,12 15

5,6 7,89 12,10.11

−B9,10,1115

5,6 7,89 10,11

−2−1A89,10 2 9,10

A9,11 A10,12A13,14 15

5,6 7,11 12.13,14

−B10,12,1315

5,6 7,11 12,13

+ 2−1A9,11A10,12 15

5,6 7,89.10 11,12

−C89,10,11 15

5,6 7,9 10,11

(4.12) and

a11221234·3 = λ32A15A26

A37A89

A4,10A11,12 15

5,678,910.11,12

−B4,10,1115

5,6 78,910,11

−B378

A4,10A11,12 15

5,67,810.11,12

−B4,10,1115

5,6 7,8 10,11

, (4.13)

whereB1=B123A23= 2−1A12A234A34 andB123=B234A14= 2−1A145A24A35.

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Theorem 4.2. The corresponding expressions of Theorem 4.1 for A diagonal are

k1i0 =k·10 =2−1λ2A00A220,22, (4.14)

ki21i2i3=k123·2 =A11A22A33

λ31,2,3+λ22 3

123

1,23 −3A123

, (4.15)

a1212·2=λ3A11

A22A551,25,5 −B2451,4,5

, (4.16)

a1312·222 = A11A22A33[A77(1,25 7,57+1,57 7,25) +12,55]

2−1A11A22A55(1,25A577A77

+23,25A531A33) +A11A22[−2−11,25B5

−6,25B156+ 2A126B6+ 2−1A156A256A55A66

+2−11,255A55+5,125A552−1A1255A55], (4.17)

a2212·222=A11A22A44(14,24+A777,14 4,27) 2

12

A11B2466,14+ 2B613B263 , (4.18)

a11111234·3=λ4A11A22A33A441,2,3,4, (4.19)

a11121234·3/(λ2λ3) = A11. . . A44

−A55 3

123

1,452,3,5 3

123

1,2,34

+A55 3

123

A1452,3,5

, (4.20)

a11131234·332 = A11A22A33A44

A99A11,11 15

1,23,4911,9.11

−B9,10,11 15

1,23,4910,11

−2−1A49,10 2 9,10

A9,9 A10,10A13,13 15

1,23,910·13,13

−B10,12,13 15

1,23,912,13

+ 2−1A99A10,10 15

1,23,49.109,10

−C49,10,1115

1,23,910,11

(4.21)

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and

a11221234·332 = A11A22

A33A88

A44A11,11 15

1,238,84.11,11

−B4,10,11 15

1,238,810,11

−B378

A44A11,11 15

1,27,84.11,11

−B4,10,1115

1,27,810,11

, (4.22)

whereB123=A11A22A33A123/2,C1234=A1234A22A33A44/6, and

B1= 2−1A11A22A122, B123= 2−1A123A22A33. (4.23) Note that the implicit summations are over the i’s not on the left hand side. For example, in (4.14) the summation is overi2, not i0.

Note thatB5,B6,B356,B613 andB326 in Theorem 4.1 follow from the definitions given forB1 andB123. For example, B5 = B523A23, B6 = B623A23, B356 = B564A34 and B613 = B134A64. Similar comments apply to Theorem 4.2.

Now suppose we replace the real model (4.1) by the complexmodel

Y =S(θ) +einCn, that isYN =SN(θ) +eN in C for 1≤N ≤n (4.24) witheN =eN1+jeN2forj=

−1 and{eN1, eN2}independent and identically distributed with mean zero and cumulantsr}. The LSE is again given by (4.2). This can be put in the framework of (4.1) with nreplaced by 2n, (so that Λ andA1···r are half what they are for the complex version with eN1 =eN of (4.1)), but it is simpler to adapt the preceding as follows:

ΛN·1...r = 1. . . ∂rS¯NSN/2−Y¯NSN·1...r/2−YNS¯N·1...r/2, A1...r = [1. . . r] =n−1

n N=1

{∂1. . . ∂r|SN|2/2−ReS¯NSN·1...r}

= n−1(∂1. . . ∂r|S|2/2−Re SS·1...r).

(Recall that ¯YN is the complex conjugate ofYN, and S= ¯S.) Let us extend the notation of (4.3) by writing

¯π1, π2, π3=n−1 n N=1

S¯N·π1SN·π2SN·π3

and so on forπ1, π2, . . .sequences of integers in 1. . . p. One obtains

·1 = −Re eS·1, nΛ·12=Re(S·1S·2−eS·12),

A12 = Re B12= (B12+ ¯B12)/2 forB12=S·1S·2/n=¯1,2, (4.25) Λ·1...r = Re(B1...r−eS·1...r/n), A1...r=Re B1...r,

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where

B123 = 3

123

S·1S·23 n= 3

123

¯1,23,

B1234 = 4 1234

S·1S·234+ 3

123

S·12 S·34 n= 4 1234

¯1,234+ 3

123

12,¯ 34.

Note that B1...r can be written down using the form for the partial Bell polynomialBr2on page 307 of [2]; his B22=x21, his B32= 3x1x2, hisB42= 4x1x3+ 3x22, hisB52= 5x1x4+ 10x2x3 so that

B12345= 5

S·1S·2345+ 10

S·12 S·345 n= 5

¯1,2345+ 10

12,345,

and so on. We now give the complex form of (D.5). Set γi=S·πi andTi=eγi. Ifr >1 then (−2)r1, . . . , πr] = 2rn−1κ(Re T1, . . . , Re Tr) =n−1κ(T1+ ¯T1, . . . , Tr+ ¯Tr)

= r i=0

k(1i1r−i) (ri)

π1, . . . , πi, πi+1, . . . , πr, (4.26)

wherek(1i1j) =κ(e1,· · · , e1, e1,· · ·, e1), countinge1 itimes ande1 jtimes. Also the inner summation is over all such (π1, . . . , πi) giving different terms. These joint cumulants can be written in terms of the real cumulants r}: k(1i1r−i) = [1 + (−1)r−ijrr so thatk(12) = 0, k(1¯1) = 2λ2. For example,

4[π1, π2] = E T1T2+ 2

T1T¯2+ ¯T1T¯2 n= 2λ2 2

π1, π2= 4λ2Real (π1, π2),

8[π1, π2, π3] = 2λ3Real (1 +j)

π1, π2, π3+ 3

π1¯2¯3

,

16[π1, π2, π3, π4] = 4λ4Real π1, π2, π3, π4+ 3

234

π1, π2, π3, π4

.

So, by (D.5) the cumulant coefficients for the complex case are obtained from those for the real case by replacing π1, π2 by Real(¯π1, π2),

π1, π2, π3 by Real (1 +j)

π1, π2, π3+ 3

π1¯2¯3 4, (4.27)

π1, π2, π3, π4 by Real π1, π2, π3, π4+ 3

234

π1, π2, π3, π4 4,

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and so on. A simpler way of seeing this – without having to involve the joint cumulantsk(1i1r−i), is as follows.

Consider the complex numbersa=a1+ja2, b=b1+jb2, . . .Thena1= (a+ ¯a)/2 anda2=−(a−¯a)j/2. So, a1b1+a2b2 = Real(ab),

a1b1c1+a2b2c2 = Real((1 +j) abc+ 3

abc 4, (4.28)

a1b1c1d1+a2b2c2d2 = Real abcd+ 3

bcd

abcd 4

= Real

abcd+abcd+acbd+adbc 4, and so on. Now takea=SN.π1,b=SN.π2,. . .Thenn−1n

N=1(a1b1c1+a2b2c2) is twice the real version ofπ1, π2, π3for the real version of the complex version with nreplaced by 2n. By (4.28) it equals the right hand side of (4.27). Similarly, one can write downπ1,. . .,πrfor the real version of the complex model in terms of π1,. . .,πrfor the complex model. So, the real versions (4.4), (4.5), (4.6), imply the complex versions

k1i1i2 = k12·1 =λ2A12, ki10 =k0·1= 2−1λ2A01A23Real ({−¯1,23¯2,13+¯3,12}), k2i1i2i3 = k123·2 =A14A25A36Real λ3(1 +j)

4,5,6+ 3

4,¯5,¯6 4 +λ22 3 456

¯4,56

22B123.

Similarly,ki21i2 =k12·2 can be written down from its real form given by (3.5), (3.11), (3.12), (3.13), andki31i2i3i4= k1234·3 can be written down from its real form given by (3.6), (3.14)–(3.17). Note that if e1 is complex normal with components having the same variance, then (4.26) implies that [π1, . . . , πr] = 0 forr >2 just as this holds by (D.5) for the real case (4.1) ife1∼ N(0, λ2).

5. Examples

We now drop the convention of Sections 2–4 of suppressing thei’s to the usual convention that SN·i1...ir =i1. . . ∂irSN

fori =∂/∂θi. That is, we writei1. . . ir, where we had1. . . r, Ai1i2 =i1, i2, where we hadA12=1,2 in (D.3),i1, i2i3, where we had 1,23in (D.4), and so on. So, now

Ars=r, s=n−1 n N=1

SN·rSN·s.

Example 5.1. Consider the RsignalM frequency problem: observe yk =s(θ) +nk inCM fork= 1, . . . , K,

where n1, . . . , nK are independent CNM(0, vIM), that is, with real and imaginary parts independent NM(0, vIM/2). (So, λ2 of Sect. 4 is v/2 and λr = 0 for r = 2.) Counting m = 0,1, . . . , M 1, suppose that themth component of s(θ) has the form

sm(θ) = R r=1

arexp(jαmr) =sm1+jsm2

(12)

say, for j =

−1, where αmr =ϕr+νmwr and νm = (m+m0)/M, and {ar, ϕr, wr} are real so that p= 3R and we can takeθ = (a, ϕ, w). The main parameter isw; (a, ϕ) are nuisance parameters. We shall obtain the leading cumulant coefficients firstly by using the real model (4.1), and then for comparison using the complex model (4.24). For the real model the MLE θminimises

K k=1

|yk−s(θ)|2/2 = n

N=1

(YN−SN(θ))2/2, wheren= 2kM, Y1

S1

, . . . , Yn

Sn

=

ykm1 sm1

,

ykm2 sm2

: 0≤m < M, 1≤k≤K,

and ykm = ykm1 +jykm2. This puts the problem into the real formulation of (4.1) with eN ∼ N(0, λ2).

The constant m0 is arbitrary, since it reparameterises ϕ. Choose m0 = −(M 1)/2, so that m} have arithmetic mean zero. For 1≤r, s≤R, sm1·r = cosαmr, sm1·r+R =−arsinαmr,sm1·r+2R=−νmarsinαmr, sm2·r= sinαmr,sm2·r+R=arcosαmr,sm2·r+2R=νmarcosαmr and

Ars= (2M)−1

M−1

m=0

2 j=1

smj·rsmj·s.

Fix 1≤r, s≤R and set

ϕrs=ϕr−ϕs, wrs=wr−ws, δm=αmr−αms =ϕrs+νmwrs. (5.1) Then the elements ofA={Aab: 1≤a, b≤3R} can be identified as follows.

a, a: r, s= (2M)−1

M−1 m=0

cosδm,

So,r,r= 2−1 and forr=s,

r, s = 2−1 1/2

−1/2

cos(ϕrs+xwrs) dx+O(M−1)

= (cosϕrs)wrs−1sin(wrs/2) +O(M−1) asM → ∞,

a, ϕ:r, s+R = as(2M)−1

M−1

m=0

sinδm=as2−1 1/2

−1/2sin(ϕrs+xwrs) dx+O(M−1)

= as(sinϕrs)wrs−1sin(wrs/2) +O(M−1) ifr=s

= 0 ifr=s,

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