DeepAI

# Moments of Maximum: Segment of AR(1)

Let X_t denote a stationary first-order autoregressive process. Consider five contiguous observations (in time t) of the series (e.g., X_1, ..., X_5). Let M denote the maximum of these. Let ρ be the lag-one serial correlation, which satisfies |ρ| < 1. For what value of ρ is E(M) maximized? How does V(M) behave for increasing ρ? Answers to these questions lie in Afonja (1972), suitably decoded.

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## 1 Partial Correlations

Set for simplicity.  Fix

.  There exists a unique ordered pair

such that .  Define a matrix

 P=⎛⎜ ⎜⎝1ri,mnri,jmri,mn1ri,jnri,jmri,jn1⎞⎟ ⎟⎠

which captures all correlations among , , .  Let be the cofactor of the element in the expansion of the determinant of .  The partial correlation between and , given , is prescribed by

 ri,mn.j=−P12√P11P22=ri,mn−ri,jmri,jn√(1−r2i,jm)(1−r2i,jn).

The notation here  differs from that used elsewhere [2, 3, 4].  In words, measures the linear dependence of and in which the influence of is removed.  Define finally a matrix

 Ri,j=(1ri,mn.jri,mn.j1)

as was to be done.

Set now .  Fix .  There exists a unique ordered triple such that .  The preceding discussion extends naturally, supplementing the case by additional possible cases and .  Define finally a matrix

 Ri,j=⎛⎜ ⎜⎝1ri,mn.jri,mo.jri,mn.j1ri,no.jri,mo.jri,no.j1⎞⎟ ⎟⎠.

We could go on for larger , but this is all that is needed for our purposes.

Again, set .  Fix .  There exists a unique ordered pair such that .  Define a matrix

 Q=⎛⎜ ⎜ ⎜ ⎜ ⎜⎝1ri,mnri,jmri,kmri,mn1ri,jnri,knri,jmri,jn1ri,jkri,kmri,knri,jk1⎞⎟ ⎟ ⎟ ⎟ ⎟⎠

which captures all correlations among , , , .  Let be the cofactor of the element in the expansion of the determinant of .  The partial correlation between and , given and , is prescribed by

 ri,mn.jk =−Q12√Q11Q22 =ri,mn−ri,jmri,jn−ri,kmri,kn+ri,kmri,jnri,jk+ri,jmri,knri,jk−ri,mnr2i,jk√(1−r2i,jm−r2i,km+2ri,jmri,kmri,jk−r2i,jk)(1−r2i,jn−r2i,kn+2ri,jnri,knri,jk−r2i,jk).

In words, measures the linear dependence of and in which the influence of and is removed.  Define finally a matrix

 Ri,jk=(1ri,mn.jkri,mn.jk1)

as was to be done.

Set now .  Fix .  There exists a unique ordered triple such that .  The preceding discussion extends naturally, supplementing the case by additional possible cases and .  Define finally a matrix

 Ri,jk=⎛⎜ ⎜⎝1ri,mn.jkri,mo.jkri,mn.jk1ri,no.jkri,mo.jkri,no.jk1⎞⎟ ⎟⎠.

We could go on for larger , but this is all that is needed for our purposes.

## 2 Small Segments

For convenience, define

 h(x,y,z) =1−x4π⋅1+x−y−z√4(1−x)(1−y)−(1−x−y+z)2 =1−x4π⋅1+x−y−z√4(1−x)(1−z)−(1−x−z+y)2 =1−x4π⋅1+x−y−z√(1−x)(3+x)−y(2+y)−z(2+z)+2(xy+xz+yz).

The latter expression, while more cumbersome, exhibits symmetry in , .

If , then and .  We have

 E(M)=2√1−ρ124π=√1−ρ12π,E(M2)=1.

If , then and .  We have

 E(M)=1√4π(√1−ρ12+√1−ρ13+√1−ρ23),
 E(M2)=1+2h(ρ12,ρ13,ρ23)+2h(ρ13,ρ12,ρ23)+2h(ρ23,ρ12,ρ13).

In formula (3.6) for in , should be replaced by .

If , then

 Φℓ−2(Ri,j)=14+12πarcsin(ri,mn.j),Φℓ−3(Ri,,jk)=12.

In general, and thus symmetry fails for .  We have

 E(M) =1√4π[√1−ρ12⋅(14+12πarcsin(r1,34.2))+√1−ρ12⋅(14+12πarcsin(r2,34.1)) +√1−ρ13⋅(14+12πarcsin(r1,24.3))+√1−ρ13⋅(14+12πarcsin(r3,24.1)) +√1−ρ14⋅(14+12πarcsin(r1,23.4))+√1−ρ14⋅(14+12πarcsin(r4,23.1)) +√1−ρ23⋅(14+12πarcsin(r2,14.3))+√1−ρ23⋅(14+12πarcsin(r3,14.2)) +√1−ρ24⋅(14+12πarcsin(r2,13.4))+√1−ρ24⋅(14+12πarcsin(r4,13.2)) +√1−ρ34⋅(14+12πarcsin(r3,12.4))+√1−ρ34⋅(14+12πarcsin(r4,12.3))],
 E(M2) =1+h(ρ12,ρ13,ρ23)+h(ρ12,ρ14,ρ24)+h(ρ13,ρ12,ρ23)+h(ρ13,ρ14,ρ34) +h(ρ14,ρ12,ρ24)+h(ρ14,ρ13,ρ34)+h(ρ23,ρ12,ρ13)+h(ρ23,ρ24,ρ34) +h(ρ24,ρ12,ρ14)+h(ρ24,ρ23,ρ34)+h(ρ34,ρ13,ρ14)+h(ρ34,ρ23,ρ24).

In formula (3.7) for in , a factor should be inserted in front of the summation.

If , then

 Φℓ−2(Ri,j)=12−14πarccos(ri,mn.j)−14πarccos(ri,mo.j)−14πarccos(ri,no.j),
 Φℓ−3(Ri,,jk)=14+12πarcsin(ri,mn.jk).

Symmetry now fails for both and .  We have

 E(M) =1√4π[√1−ρ12⋅(12−14πarccos(r1,34.2)−14πarccos(r1,35.2)−14πarccos(r1,45.2)) +√1−ρ12⋅(12−14πarccos(r2,34.1)−14πarccos(r2,35.1)−14πarccos(r2,45.1)) +√1−ρ13⋅(12−14πarccos(r1,24.3)−14πarccos(r1,25.3)−14πarccos(r1,45.3)) +√1−ρ13⋅(12−14πarccos(r3,24.1)−14πarccos(r3,25.1)−14πarccos(r3,45.1))+⋯],
 E(M2) =1+h(ρ12,ρ13,ρ23)⋅(14+12πarcsin(r1,45.23))+h(ρ12,ρ23,ρ13)⋅(14+12πarcsin(r2,45.13)) +h(ρ12,ρ14,ρ24)⋅(14+12πarcsin(r1,35.24))+h(ρ12,ρ24,ρ14)⋅(14+12πarcsin(r2,35.14)) +h(ρ12,ρ15,ρ25)⋅(14+12πarcsin(r1,34.25))+h(ρ12,ρ25,ρ15)⋅(14+12πarcsin(r2,34.15))+⋯,

a total of terms and terms, respectively.

If , then contains non-elementary functions which require numerical integration (beyond our present scope).  In contrast,

 E(M2) =1+h(ρ12,ρ13,ρ23)⋅(12−14πarccos(r1,45.23)−14πarccos(r1,46.23)−14πarccos(r1,56.23)) +h(ρ12,ρ23,ρ13)⋅(12−14πarccos(r2,45.13)−14πarccos(r2,46.13)−14πarccos(r2,56.13)) +h(ρ12,ρ14,ρ24)⋅(12−14πarccos(r1,35.24)−14πarccos(r1,36.24)−14πarccos(r1,56.24)) +h(ρ12,ρ24,ρ14)⋅(12−14πarccos(r2,35.14)−14πarccos(r2,36.14)−14πarccos(r2,56.14)) +h(ρ12,ρ15,ρ25)⋅(12−14πarccos(r1,34.25)−14πarccos(r1,36.25)−14πarccos(r1,46.25)) +h(ρ12,ρ25,ρ15)⋅(12−14πarccos(r2,34.15)−14πarccos(r2,36.15)−14πarccos(r2,46.15)) +h(ρ12,ρ16,ρ26)⋅(12−14πarccos(r1,34.26)−14πarccos(r1,35.26)−14πarccos(r1,45.26)) +h(ρ12,ρ26,ρ16)⋅(12−14πarccos(r2,34.16)−14πarccos(r2,35.16)−14πarccos(r2,45.16))+⋯,

a total of terms.  In formula (3.9) for in , a constant term should be inserted in front of the first summation; further, the last summation should be taken over both and (not merely ).

## 3 Time Series

Consider a discrete-time stationary first-order autoregressive process

 Xt=ρXt−1+√1−ρ2⋅εt,−∞

where is white noise. The covariance matrix has element

 ρij=ρ|j−i|

which leads to certain simplifications.  Let us make the reliance of on explicit.  We have

 E(M2)=√1−ρπ,
 E(M3)=√1−ρπ+√1−ρ24π,
 E(M4)=14π√1−ρπ[π+2arcsin(1−23−ρ)]+12π√1−ρπ[π+2arcsin(1+2ρ−ρ2√(3−ρ)(3+ρ+ρ2−ρ3))] +12π√1−ρ2π[π+2arcsin((1−ρ)2√(3−ρ)(3+ρ+ρ2−ρ3))]+14π√1−ρ3π[π+2arcsin(1−23+ρ+ρ2−ρ3)]

but is too lengthy to record here.  In the limit as , we obtain

 1√π,32√π,3√π[1−1πarcsec(3)],5√π[1−32πarcsec(3)]

for and these are consistent with well-known values  corresponding to independent .  Figure 1 displays as functions of .  The left-hand endpoint is at and is unsurprisingly the mean of a standard half-normal distribution.  The right-hand endpoint is at .  Associated with are maximum points with equal to

 1−√52=−0.6180339887498948482045868...,
 −0.4973597615161907364022217...,−0.4336476843162656141275672...

respectively.  Closed-form expressions for the latter two quantities remain open.

We also have

 E(M23)=1+(1−ρ)√(3−ρ)(1+ρ)2π,
 E(M24)=1+3+√(3−ρ)(3+ρ+ρ2−ρ3)+ρ[1−2ρ−2ρ2−ρ3+ρ4−ρ√(3−ρ)(3+ρ+ρ2−ρ3)]2π√(1+ρ)(3+ρ+ρ2−ρ3)

but and are too lengthy to record here.  In the limit as , we obtain

 1+√32π,1+√3π,1+5√32π[1−1πarcsec(4)],1+5√3π[1−32πarcsec(4)]

for and these again are consistent with well-known values .  Associated with are maximum points with equal to

 1−√2=−0.4142135623730950488016887,
 −0.3879232988398265768779440...,−0.3599267104829689555367968...
 −0.3406053067160525788737944...

respectively.  Closed-form expressions for the latter three quantities remain open.  Figure 2 displays as functions of .  The left-hand endpoint is at ; the right-hand endpoint is at .  Unlike or , the variance is strictly increasing throughout the interval.  An intuitive reason for such behavior would be good to establish someday.

## 4 Proof of Revision

Our general formula for looks somewhat different from that presented by Afonja .  To demonstrate the equivalence of the two formulas, it suffices to prove that if , and , then

 12πri,ji⋅ri,ki−ri,jkri,ji√1−r2i,jk=h(ρij,ρik,ρjk).

The left-hand side is equal to

 12π√1−ρij2⋅√1−ρik2−1−ρij−ρik+ρjk√4(1−ρij)(1−ρik)√1−ρij2√1−(1−ρij−ρik+ρjk)24(1−ρij)(1−ρik) =14π√1−ρij⋅√1−ρik−1−ρij−ρik+ρjk√4(1−ρik)√1−(1−ρij−ρik+ρjk)24(1−ρij)(1−ρik) =14π√1−ρij⋅2(1−ρik)−(1−ρij−ρik+ρjk)√4(1−ρik)−(1−ρij−ρik+ρjk)21−ρij =14π(1−ρij)⋅1+ρij−ρik−ρjk√4(1−ρij)(1−ρik)−(1−ρij−ρik+ρjk)2

which is the right-hand side, as was to be shown.

## 5 Proof from First Principles

An exercise in  suggests that formulas for and should be derived from

 max{X1,X2}=12(X1+X2)+12|X1−X2|.

It is instructive to similarly prove our formula for , using instead

 max{X1,X2,X2} =max{max{X1,X2},max{X2,X3}} =12(max{X1,X2}+max{X2,X3})+12|max{X1,X2}−max{X2,X3}| =14((X1+X2)+(X2+X3)+|X1−X2|+|X2−X3|) +14|(X1+X2)−(X2+X3)+|X1−X2|−|X2−X3||.

Define and .  Clearly is bivariate normally distributed with vector mean zero and covariance matrix

Also

 14E|Y|=σy4√2π=√1−ρ124π,
 14E|Z|=σz4√2π=√1−ρ234π.

The four integrals (depending on signs of and ) underlying

 14E|(Y+|Y|)−(Z+|Z|)|=√1−ρ134π

can all be evaluated (however tediously).  Because

 X1−X3=(X1−X2)−(X3−X2)=Y−Z

we suspect that a more elegant proof ought to be available.  Ideas on bridging this gap would be welcome.

In more detail, letting

 f(y,z)=12π√1−ξ2σyσzexp[−12(1−ξ2)(y2σ2y−2ξyzσyσz+z2σ2z)]

denote the bivariate normal density, we obtain

 ∞∫0∞∫02|y−z|f(y,z)dydz=1√2π[−(1−ξ)(σy+σz)+2√σ2y−2ξσyσz+σ2z]

when and ;

 ∞∫00∫−∞2zf(y,z)dydz=(1−ξ)σz√2π

when and ;

 0∫−∞∞∫02yf(y,z)dydz=(1−ξ)σy√2π

when and ; and when and .  Adding these contributions and dividing by , we verify

 12√2π√σ2y−2ξσyσz+σ2z =12√π√(1−ρ12)−(ρ13−ρ12−ρ23+1)+(1−ρ23) =√1−ρ134π

as was desired.

Calculating the variance of from first principles has not been attempted.  The variance of the median (-tile) is also of interest, appearing explicitly in  for but under the assumption of independence.

An alternative probability density-based derivation of

and can be found in [8, 9].  See also  for the expected range of a normal sample,  for the expected absolute maximum, and  for other aspects of AR(1).

## 6 Large Segments

Assuming depends only on , Berman [13, 14, 15] proved that if either

 limd→∞ρ(d)⋅ln(d)=0or∞∑d=1ρ(d)2<∞,

then

 limℓ→∞P{√2ln(ℓ)(Mℓ−aℓ)≤x}=exp(−e−x)

where

 aℓ=√2ln(ℓ)−12ln(ln(ℓ))+ln(4π)√2ln(ℓ).

Further, the two hypotheses on cannot be significantly weakened.  This theorem clearly applies for a first-order autoregressive process, although we note that does not incorporate lag-one correlation at all.  A more precise asymptotic result might do so.

Other relevant works in the literature include [16, 17, 18, 19, 20, 21, 22, 23, 24].  In particular, Figure 2 of  depicts the density of AR(1) maximum for and , , …, , .

## 7 Acknowledgements

Raymond Kan  symbolically evaluated the integrals in Section 5, at my request, for the special case