# Wavelet Analysis of Red Noise and Its Application in Climate Diagnosis

Authors: Zhihua Zhang
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

Signals are often destroyed by various kinds of noises. A common way to statistically assess the significance of a broad spectral peak in signals and the synchronization between signals is to compare with simple noise processes. At present, wavelet analysis of red noise is studied limitedly and there is no general formula on the distribution of the wavelet power spectrum of red noise. Moreover, the distribution of the wavelet phase of red noise is also unknown. In this paper, for any given real/analytic wavelet, we will use a rigorous statistical framework to obtain the distribution of the wavelet power spectrum and wavelet phase of red noise and apply these formulas in climate diagnosis.


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## Body

## 1. Introduction

Signals are often destroyed by various kinds of noises during the process of generation, transportation, and processing [1, 2]. A common way to statistically assess the significance of a broad Fourier/wavelet spectral peak in signals [3, 4] and the synchronization between signals is to compare with a white/red noise process [4, 5]. White noise has zero mean, has constant variance, and is uncorrelated in time. As its name suggests, white noise has a power spectrum which is uniformly spread across all allowable frequencies. Different from white noise, red noise has a power spectrum weighted toward low frequencies and is serially correlated in time. Red noise can describe climatic background noise with relatively enhanced low-frequency fluctuations arising from the interaction of white noise forcing with the slow-response components in the earth system (e.g., the thermal inertia of the oceans provides memory, effectively integrating atmospheric “weather” forcing [6]). In practice, the red noise model has always provided a reasonable description of the noise spectra for a variety of climatic and hydrological time series [5–9]. The red noise is also an important noise model in outputs of the feedback system [4], the neural network coupled with genetic algorithm [10], the optimization system in random scenarios [11, 12], and the big data processing system [5, 13].Wavelet analysis is a very popular analysis tool for a wide range of applications, including time-frequency analysis, feature extraction, statistical estimation, and denoising. A wavelet is a waveform-like function which has zero mean and is localized in both time and frequency space. The wavelet transform is the set of inner products of all dilated and translated wavelets with a signal; in detail, the wavelet transform of a discrete signalX=xkk=0,…,N−1 with time step δt is defined as follows [14, 15]:(1)Wns=∑m=0N−1xmψ¯m−nδts,where ψ¯ is the conjugate of ψ. Wns2 is called the wavelet power spectrum, and ArgWns is called the wavelet phase. Due to time and frequency localization of the wavelet, the wavelet transform can extract localized intermittent periodicity of any signal very well.Wavelet analysis of white noise has been widely studied (e.g., [16–19]), while wavelet analysis of red noise is studied limitedly, only using some specific wavelets (Morlet/Paul/DOG wavelets [15] and modulated Haar wavelet [20]). Since 1980s, a large family of wavelets with nice properties has been constructed [14]. However, there is no general formula on the distribution of the wavelet power spectrum of red noise. On the other hand, the wavelet phase can be used to test the synchronization between signals, but the distribution of the wavelet phase of red noise is also unknown. In this paper, for any given real/analytic wavelet, we will use a rigorous statistical framework to derive a general formula for the distribution of the wavelet power spectrum and wavelet phase of red noise.


## 2. Background

The simplest red noise model is the lag-1 autoregressive (AR (1)) red noise process [1, 4–6, 15]: Let X=xkk=0,1,…,N−1 with time step δt be an AR (1) red noise with parameters λ0≤λ<1 and σ2; then,(2)x0=0,xk=λxk−1+Zk,k∈ℤ+,where Zkk∈ℤ+ is a Gaussian white noise with mean 0 and variance σ2 and EZkZl=0k≠l. The discrete Fourier transform of an AR (1) red noise is(3)x^k=1N∑n=0N−1xne−inθk,k=0,1,…,N−1,where θk=2πk/N. Again, by equation (1), the wavelet transform of an AR (1) red noise can be expressed as [15](4)Wns=∑k=0N−1x^kψ^¯sωkeinθk,where ωk=θk/δt,k≤N/2, and ωk=−θN−k/δt,k>N/2 (here, we have fixed some typos according to Torrence and Compo’s Matlab codes [15]).Torrence and Compo [15] gave an empirical distribution of the Morlet wavelet power spectrum of an AR (1) red noise:(5)Wns2is distributed as1−λ2σ˜221−2λcos2πk/N+λ2χ22,where χ22 is the chi-square distribution with two degrees of freedom and σ˜2 is the variance of the AR (1) red noise σ˜2≈σ2/1−λ2. For Paul/DOG wavelet power spectrum of an AR (1) red noise, Torrence and Compo [15] empirically gave the following:(6)Wns2is distributed as1−λ2σ˜21−2λcos2πk/N+λ2χ12.Formulas (5) and (6) are empirically obtained by assuming that “the wavelet spectrum of a red noise is distributed as its Fourier spectrum” [15].Zhang and Jorgensen [20] showed that under the rigorous statistical framework, the modulated Haar wavelet power spectra of AR (1) red noise is distributed as(7)2σ2s1−α2σR2χ12+σI2χ22,where χ1 and χ2 are two independent standard Gaussian distributions. Also, Pkλ=1−λ2/1−2λcos4kπ/s+λ2 and(8)σR2=s2Pkλ+4λ1−λs/21−λ2λ−cos4kπs1−λcos2kπs∗Pk2λ−2λ21−λs/221−λ2λ−cos4kπs2Pk2λ,σI2=s2Pkλ+Pk2λ4λ21−λs/21−λ22sin24kπs−Pk2λ2λ21−λs/221−λ22sin24kπs.Until now, wavelet analysis of red noise is studied limitedly, only using some specific wavelets. There is no general formula on the distribution of the wavelet power spectrum and wavelet phase of an AR (1) red noise.


## 3. Main Results

We will use a rigorous statistical framework to establish the distribution of the wavelet power spectrum and wavelet phase of an AR (1) red noise with time stepδt, length N, and parameters λ and σ2. Here, we assume a very weak condition: λN≈0 and 1/N2≈0. In practice, all climatic background noises satisfy this condition, e.g., if λ=0.5 and N=50, then λN≈10−8 and 1/N2=0.0004.A wavelet is real if it is a real-valued function. All of spline wavelets and most of compactly supported wavelets are real wavelets.Theorem 1.
For any real wavelet, the wavelet power spectrumWns2 of an AR (1) red noise is distributed as(9)2σ2N∑k=0N/2ψ^sωk21−2λcosθk+λ2χ12.
A waveletψ is said to be analytic if its Fourier transform satisfies ψ^ω=0,ω<0. Different from real wavelets, analytic wavelets can extract not only spectral information but also phase information from any given signal.Theorem 2.
For any analytic wavelet, the wavelet power spectrum of an AR (1) red noise is distributed as(10)σ22N∑k=0N/2ψ^sωk21−2λcosθk+λ2χ22.Theorem 3.
For any analytic wavelet, the wavelet phase of an AR (1) red noise is distributed uniformly on0,2π.


## 4. Proofs

The AR (1) red noise with parametersλ,λ<1 and σ2 can be expressed as(11)xk=∑ν=1kλk−νZν,k∈ℤ+.ByEZν=0,(12)Exk=∑ν=1kλk−νEZν=0,and so the real part of the wavelet transform is(13)ReWns=∑m=0N−1xmReψm−nδts.SinceReWns is a linear combination of xm, ReWns is a Gaussian random variable and(14)EReWns=0,VarReWns=EReWns2.Similarly,ImWns is a Gaussian random variable and(15)EImWns=0,VarImWns=EImWns2.DenoteArgψ^¯sωk=φk. Then, ψ^¯sωkeinθk=ψ^sωkeiηn,k, where ηn,k=φk+nθk. By equation (4), we have(16)ReWns=∑k=0N−1ψ^sωkRex^keiηn,k=∑k=0N−1ψ^sωkRex^kcosηn,k−Imx^ksinηn,k,ImWns=∑k=0N−1ψ^sωkRex^ksinηn,k+Imx^kcosηn,k.Let(17)αk,j=ERex^kRex^j,βk,j=ERex^kImx^j,γk,j=EImx^kImx^j.Then, the variances ofReWns and ImWns and their correlation are(18)VarReWns=∑k,j=0N−1ψ^sωjψ^sωjαk,jcosηn,kcosηn,j−βk,jsinηn,kcosηn,j−βj,kcosηn,ksinηn,j+γk,jsinηn,ksinηn,j,VarImWns=∑k,j=0N−1ψ^sωjψ^sωjαk,jsinηn,ksinηn,j+βk,jcosηn,ksinηn,j+βj,ksinηn,ksinηn,j+γk,jcosηn,kcosηn,j,EReWnsImWns=∑k,j=0N−1ψ^sωkψ^sωjαk,jcosηn,ksinηn,j+βk,jsinηn,ksinηn,j−βj,kcosηn,kcosηn,j−γk,jsinηn,kcosηn,j.From these, we see that the computations of variances and the correlation ofReWns and ImWns are reduced to the computations of αk,j, βk,j, and γk,j.


### 4.1. Computation ofαk,j

By equation (3), we have(19)Rex^k=1N∑n=0N−1xncosnθk,where θk=2πk/N. Furthermore,(20)ERex^jRex^k=1N2∑l,p=1N−1Exlxpcoslθjcospθk.By equation (11), it follows that(21)xlxp=∑ν=1lλl−νZν∑m=1pλp−mZm=∑ν=1l∑m=1pλl+p−ν−mZνZm,where(22)EZνZm=0,nu≠m,σ2,nu=m.Hence, we have forl>p,(23)Exlxp=∑ν=1p∑m=1pλl+p−ν−mEZνZm=σ2∑ν=1pλl+p−2ν=σ2λl+p−λl−pλ2−1.Similarly, forl≤p,(24)Exlxp=σ2λl+p−λp−lλ2−1.From these, we obtain(25)N2λ2−1σ2ERex^kRex^j=Σ1+Σ2,where(26)Σ1=∑l=1N−1λlcoslθj∑p=1l−1λp−λ−pcospθk,Σ2=∑l=1N−1λl−λ−lcoslθj∑p=lN−1λpcospθk.We first computeΣ1. By using Euler formula cospθk=Reeipθk, the inner summation becomes(27)∑p=1l−1λp−λ−pcospθk=Pkλ+μl1+μl2,where Pkλ=1−λ2/1−2λcosθk+λ2 and(28)μl1=−λl1−λ2Pkλcoslθk−λcosl−1θk,μl2=λ−l+11−λ2Pkλλcoslθk−cosl−1θk.From this and equation (25), we obtain(29)Σ1=Pkλ∑l=1N−1λlcoslθj+∑l=1N−1λlcoslθjμl1+∑l=1N−1λlcoslθjμl2≕Σ11+Σ12+Σ13.ForΣ11, when λN≈0,(30)Σ11=Pkλλcosθj−λ21−2λcosθj+λ2.ForΣ13, we have(31)Σ13=λ21−2λcosθk+λ2∑l=1N−1coslθk−λ−1cosl−1θkcoslθj=Aλ21−2λcosθk+λ2,where(32)Aλ=λ2∑l=1N−1coslθk+j+coslθk−j−λ−1coslθk+j−θk+coslθk−j−θk.Noticing thatθk=2πk/N, we consider four cases to compute Aλ and then estimate Σ13:(i)
Fork≠j and k+j≠N, by θk=2πk/N and ∑l=0N−1coslθk=Re∑l=0N−1eilθk=0, we have(33)Aλ=−2λ2+2λcosθk,and thenΣ13=−λ2+λcosθk/1−2λcosθk+λ2.(ii)
Fork=j≠N/2, we obtain(34)Aλ=λ2∑l=1N−1cos2lθk+1−λ−1cos2lθk−θk+cosθk,and then(35)Aλ=N−2λ2−λcosθk.Therefore,(36)Σ13=N−2λ2−λcosθk21−2λcosθk+λ2.(iii)
Fork+j=N and k≠j, we obtain(37)Aλ=λ2∑l=1N−11+coslθk−j−λ−1cosθk+coslθk−j−θk=N−2λ2−λcosθk.Therefore,(38)Σ13=N−2λ2−λcosθk21−2λcosθk+λ2.(iv)
Fork=j=N/2, we obtain(39)Aλ=λ2∑l=1N−12−2cosθkλ=2N−1λ2−λcosθk,and so(40)Σ13=N−1λ2−λcosθk1−2λcosθk+λ2.Now, we computeΣ2 in equation (25). When λN≈0, it follows that(41)∑p=lN−1λpcospθk=λlcoslθk−λl+1cosl−1θk1−2λcosθk+λ2.By equation (25), we obtain(42)Σ2=∑l=1N−1λl−λ−lcoslθjλlcoslθk−λcosl−1θk1−2λcosθk+λ2=Σ21+Σ22,where(43)Σ21=11−2λcosθk+λ2∑l=1N−1λ2lcoslθjcoslθk−λcosl−1θk,Σ22=−11−2λcosθk+λ2∑l=1N−1coslθjcoslθk−λcosl−1θk.Furthermore,(44)Σ22=−121−2λcosθk+λ2∑l=1N−1coslθk+j+coslθk−j−λcoslθk+j−θk+coslθk−j−θk.Similar to the estimation process ofΣ13, we have(45)Σ22=2−2λcosθk21−2λcosθk+λ2,k≠j&,k+j≠N,Σ22=−N−21−λcosθk21−2λcosθk+λ2,k=j≠N2,Σ22=−N−21−λcosθk21−2λcosθk+λ2,k+j=N&,k≠j,Σ22=−N−11−λcosθk1−2λcosθk+λ2,k=j=N2.Combining this and equations (29)–(42), noticing that Σ12+Σ21=0, it follows that(46)Σ1+Σ2=PkλPjλ,k≠j&,k+j≠N,Σ1+Σ2=−N1−λ221−2λcosθk+λ2+Pkλλcosθj−λ21−2λcosθj+λ2−1,k=j≠N2,Σ1+Σ2=−N1−λ221−2λcosθk+λ2+Pkλλcosθj−λ21−2λcosθj+λ2−1,k+j=N&,k≠j,Σ1+Σ2=−N1−λ21−2λcosθk+λ2+Pkλλcosθj−λ21−2λcosθj+λ2−1,k=j=N2.Noticingλ<1 and 1−2λcosθk+λ2>1−λ2, we have(47)PkλPjλ<41−λ4,Pkλλcosθj−λ21−2λcosθj+λ2−1<121−λ4.Again, by equation (25), we obtain(48)ERex^kRex^j≤CN2,k≠j&k+j≠N,ERex^kRex^j−σ22N1−2λcosθk+λ2≤CN2,k=j≠N2ork+j=N&k≠j,ERex^kRex^j−σ2N1−2λcosθk+λ2≤CN2,k=j=N2,where C=12σ2/1−λ21−λ4 is a constant.Finally, we obtain the following.Lemma 1.
WhenλN≈0 and 1/N2≈0, the series Rex^kk=0,…,N−1 satisfy(49)αk,j=ERex^kRex^j=0,k≠jandk+j≠N,σ22N1−2λcosθk+λ2,k=j≠N2ork+j=N&k≠j,σ2N1−2λcosθk+λ2,k=j=N2.


### 4.2. Computation ofγk,j

By equation (3), we have(50)EImx^jImx^k=1N2∑l,p=1N−1Exlxpsinlθjsinpθk.Similar to the deducing process of equation (25), we have(51)N2λ2−1σ2EImx^jImx^k=∑l=1N−1λlsinlθj∑p=1l−1λp−λ−psinpθk+∑l=1N−1λl−λ−lsinlθj∑p=lN−1λpsinpθk≕Σ˜1+Σ˜2.Noticing that(52)∑p=1l−1λpsinpθk=−λlsinlθk+λsinθk+λl+1sinl−1θk1−2λcosθk+λ2,we obtain(53)∑p=1l−1λp−λ−psinpθk=μ˜l1+μ˜l2,where(54)μ˜l1=−λlsinlθk−λsinl−1θk1−2λcosθk+λ2,μ˜l2=λ−l+2sinlθk−λ−1sinl−1θk1−2λcosθk+λ2.We estimateΣ˜1 in equation (51):(55)Σ˜1=∑l=1N−1λlμ˜l1sinlθj+∑l=1N−1λlμ˜l2sinlθj≕Σ˜11+Σ˜12.Notice that(56)Σ˜12=λ221−2λcosθk+λ2∑l=1N−1−coslθk+j+coslθk−j+1λcoslθk+j−θk−coslθk−j−θk.Fork≠j and k+j≠N, we have(57)∑l=0N−1coslθk+j=∑l=0N−1coslθk−j=∑l=0N−1coslθk+j−θk=∑l=0N−1coslθk−j−θk=0,and so Σ˜12=0.Fork=j≠N/2, we have(58)Σ˜12=Nλ2−λcosθk21−2λcosθk+λ2.Fork+j=N and k≠N/2, we have(59)Σ˜12=−Nλ2−λcosθk21−2λcosθk+λ2.Fork=j=N/2, Σ˜12=0.We estimateΣ˜2 in equation (51). Since λN≈0, it follows that(60)∑p=lN−1λpsinpθk=λlsinlθk−λsinl−1θk1−2λcosθk+λ2.Furthermore,(61)Σ˜2=11−2λcosθk+λ2∑l=1N−1λ2lsinlθjsinlθk−λsinl−1θk−11−2λcosθk+λ2∑l=1N−1sinlθjsinlθk−λsinl−1θk≕Σ˜21+Σ˜22.By equation (55), it follows that Σ˜11+Σ˜12=0.ForΣ˜22,(62)Σ˜22=121−2λcosθk+λ2∑l=1N−1coslθk+j−coslθk−j−λcoslθk+j−θk−coslθk−j−θk0,k≠j&k+j≠Nork=j=N2,−N1−λcosθk21−2λcos2πk/N+λ2,k=j≠N2,N1−λcosθk21−2λcosθk+λ2,k+j=N&k≠N2.From all of the above, it follows that(63)Σ˜1+Σ˜2=0,k≠j&,k+j≠Nork=j=N2,Nλ2−121−2λcos2πk/N+λ2,k=j≠N2,N1−λ221−2λcosθk+λ2,k+j=N&,k≠N2.Therefore, we have the following.Lemma 2.
WhenλN≈0, the series Imx^kk=0,…,N−1 satisfy(64)γk,j=EImx^kImx^j=0,k≠j&k+j≠Nork=j=N2,σ22N1−2λcosθk+λ2,k=j≠N2,−σ22N1−2λcosθk+λ2k+j=N,&k≠N2.


### 4.3. Computation ofβk,j

By equation (3), we have(65)ERex^jImx^k=−1N2∑l,p=1N−1Exlxpcoslθjsinpθk.Similar to the deducing process of equation (25), we obtain(66)N21−λ2σ2ERex^jImx^k=∑l=1N−1λlcoslθj∑p=1l−1λp−λ−psinsθk+∑l=1N−1λl−λ−lcoslθj∑p=lN−1λpsinpθk≕Σ¯1+Σ¯2.Since(67)∑p=1l−1λp−λ−psinpθk=μ¯l1+μ¯l2,where(68)μ¯l1=−λlsinlθk−λsinl−1θk1−2λcosθk+λ2,μ¯l2=λ−l+2sinlθk−λ−1sinl−1θk1−2λcosθk+λ2,we have(69)Σ¯1=−11−2λcosθk+λ2∑l=1N−1λ2lcoslθjsinlθk−λsinl−1θk+λ21−2λcosθk+λ2∑l=1N−1coslθjsinlθk−λ−1sinl−1θk≕Σ¯11+Σ¯12.Furthermore,(70)Σ¯12=λ221−2λcosθk+λ2∑l=1N−1sinlθk+j+sinlθk−j−λ−1sinlθk+j−θk+sinlθk−j−θk.Noticing that∑l=1N−1sinlθk+j=∑l=1N−1sinlθk−j=0, we obtain(71)Σ¯12=−λ21−2λcosθk+λ2∑l=1N−1sinlθk+j−θk+sinlθk−j−θk.ForΣ¯2, we obtain(72)Σ¯2=11−2λcosθk+λ2∑l=1N−1λ2lcoslθjsinlθk−λsinl−1θk−11−2λcosθk+λ2∑l=1N−1coslθjsinlθk−λsinl−1θk≕Σ¯21+Σ¯22,where(73)Σ¯22=−121−2λcosθk+λ2∑l=1N−1sinlθk+j+sinlθk−j−λsinlθk+j−θk+sinlθk−j−θk=λ21−2λcosθk+λ2∑l=1N−1sinlθk+j−θk+sinlθk−j−θk.Since(74)Σ¯11+Σ¯21=0,Σ¯12+Σ¯22=0,we have the following.Lemma 3.
WhenλN≈0, the series ERex^k,Imx^jk,j=0,…,N−1 satisfy(75)βk,j=ERex^kImx^j=0,k,j=0,…,N−1.


### 4.4. Proof of Theorem1

Without loss of generalization, we assume thatN is an odd number. By equation (4), we obtain(76)ReWns=∑k=0N/2+∑k=N/2+1N−1rkRex^kcosηn,k−Imx^ksinηn,k=I1+I2,where(77)ηn,k=φk+nθk,φk=Argψ^¯sωk,rk=ψ^sωk.Sincex^k is real, we have x^¯k=x^N−k. This implies that(78)Rex^N−k=Rex^k,Imx^N−k=−Imx^k.Sinceψ is real-valued, ψ^ξ=ψ^¯−ξ; by ωk=θk/δt,k≤N/2, and ωk=−θN−k/δt,k>N/2, we obtain(79)ψ^sωN−k=ψ^−sωk=ψ^¯sωk.Therefore, by equation (77),(80)rN−k=ψ^sωN−k=ψ^sωk=rk,φN−k=Argψ^¯sωN−k=Argψ^¯sωk=−Argψ^sωk=−φk.Hence,ηn,N−k=φN−k+nθN−k=−φk−nθk=−ηn,k. From this, ψ^0=0, and equation (76), it follows that(81)I2=∑k=N/2+1N−1rkRex^kcosηn,k−Imx^ksinηn,k=∑k=1N/2rN−kRex^N−kcosηn,N−k−Imx^N−ksinηn,N−k=I1,Wns2=4∑k,j=0N/2rkrjRex^kcosηn,k−Imx^ksinηn,kRex^jcosηn,j−Imx^jsinηn,j.By Lemmas1–3, we obtain(82)EWns2=4∑k=0N/2rk2ERex^k2cos2ηn,k+EImx^k2sin2ηn,k=4∑k=0N/2rk2Rex^k2cos2ηn,k+sin2ηn,k=4∑k=0N/2rk2Rex^k2=2σ2N∑k=0N/2ψ^sωk21−2λcosθk+λ2.SinceWns=ReWns is a Gaussian random variable with mean 0, we obtain Theorem 1.


### 4.5. Proof of Theorems2 and 3

Without loss of generalization, we assume thatN is an old number. By equation (4), for n=0,…,N−1, we have(83)Wns=∑k=0N/2x^kψ^¯sωkeinθk+∑k=N/2+1N−1x^kψ^¯sωkeinθk.Sinceψ is an analytic wavelet, ψ^ω=0,ω<0. Notice that ωk<0,k>N/2. The second term on the right-hand side of equation (83) is vanishing. Therefore,(84)Wns=∑k=0N/2x^kψ^¯sωkeinθk.Furthermore,(85)EReWns2=∑k,j=0N/2rkrjERex^keiηn,kRex^jeiηn,j.However, by Lemma3 and equation (77), it follows that(86)EReWns2=∑k,j=0N/2rkrjERex^kRex^jcosηn,kcosηn,j+EImx^kImx^jsinηn,ksinηn,j.By Lemmas1 and 2, ERex^k2=EImx^k2,0≤k≤N/2, and so(87)EReWns2=∑k=0N/2rk2ERex^k2cos2ηn,k+EImx^k2sin2ηn,k=∑k=0N/2rk2ERex^k2cos2ηn,k+sin2ηn,k=∑k=0N/2rk2ERex^k2.Similarly, we have(88)EImWns2=∑k,j=0N/2rkrjEImx^keiηn,kImx^jeiηn,j.By Lemma3,(89)EImx^keiηn,kImx^jeiηn,j=ERex^kRex^jsinηn,ksinηn,j+EImx^kImx^jcosηn,kcosηn,j.This implies that(90)EImWns2=∑k=0N/2−1rk2ERex^k2sin2ηn,k+EImx^k2cos2ηn,k=∑k=0N/2−1rk2ERex^k2.Therefore, we have(91)VarReWns=VarImWns=∑k=0N/2rk2ERex^k2=σ22N1−λ2∑k=0N/2Pkλψ^sωk2.Now, we prove thatReWns and ImWns are independent.By Lemmas1–3,(92)EReWnsImWns=∑k,j=0N/2rkrjERex^kRex^jcosηn,ksinηn,j−EImx^kImx^jsinηn,kcosηn,j=12∑k=0N/2rk2ERex^k2sin2ηn,k−EImx^k2sin2ηn,k=0.Since bothReWns and ImWns are Gaussian random variables with mean 0 and(93)VarReWns=VarImWns≕σ2s,where(94)σ2s=σ22N1−λ2∑k=0N/2Pkλψ^sωk2,we obtain Theorem 2.Consider the phaseΦ=ArgWns. Notice that the p.d.f.s of ReWns and ImWns are, respectively,(95)pRx=12πσse−x2/2σ2s,pIx=12πσse−y2/2σ2s,and ReWns and ImWns are independent, the joint p.d.f. of ReWns and ImWns is(96)px,y=12πσ2se−x2+y2/2σ2s.The probability(97)Pβ1<Φ<β2=12πσ2s∬Ωe−x2+y2/2σ2sdxdy,where the domain Ω:=0<r<∞,0≤β1<φ<β2<2π. Hence,(98)Pβ1<Φ<β2=12πσ2s∫β1β2dφ∫0∞re−r2/2σ2sdr=12πβ2−β1,i.e., Theorem 3 holds.


## 4.1. Computation ofαk,j

By equation (3), we have(19)Rex^k=1N∑n=0N−1xncosnθk,where θk=2πk/N. Furthermore,(20)ERex^jRex^k=1N2∑l,p=1N−1Exlxpcoslθjcospθk.By equation (11), it follows that(21)xlxp=∑ν=1lλl−νZν∑m=1pλp−mZm=∑ν=1l∑m=1pλl+p−ν−mZνZm,where(22)EZνZm=0,nu≠m,σ2,nu=m.Hence, we have forl>p,(23)Exlxp=∑ν=1p∑m=1pλl+p−ν−mEZνZm=σ2∑ν=1pλl+p−2ν=σ2λl+p−λl−pλ2−1.Similarly, forl≤p,(24)Exlxp=σ2λl+p−λp−lλ2−1.From these, we obtain(25)N2λ2−1σ2ERex^kRex^j=Σ1+Σ2,where(26)Σ1=∑l=1N−1λlcoslθj∑p=1l−1λp−λ−pcospθk,Σ2=∑l=1N−1λl−λ−lcoslθj∑p=lN−1λpcospθk.We first computeΣ1. By using Euler formula cospθk=Reeipθk, the inner summation becomes(27)∑p=1l−1λp−λ−pcospθk=Pkλ+μl1+μl2,where Pkλ=1−λ2/1−2λcosθk+λ2 and(28)μl1=−λl1−λ2Pkλcoslθk−λcosl−1θk,μl2=λ−l+11−λ2Pkλλcoslθk−cosl−1θk.From this and equation (25), we obtain(29)Σ1=Pkλ∑l=1N−1λlcoslθj+∑l=1N−1λlcoslθjμl1+∑l=1N−1λlcoslθjμl2≕Σ11+Σ12+Σ13.ForΣ11, when λN≈0,(30)Σ11=Pkλλcosθj−λ21−2λcosθj+λ2.ForΣ13, we have(31)Σ13=λ21−2λcosθk+λ2∑l=1N−1coslθk−λ−1cosl−1θkcoslθj=Aλ21−2λcosθk+λ2,where(32)Aλ=λ2∑l=1N−1coslθk+j+coslθk−j−λ−1coslθk+j−θk+coslθk−j−θk.Noticing thatθk=2πk/N, we consider four cases to compute Aλ and then estimate Σ13:(i)
Fork≠j and k+j≠N, by θk=2πk/N and ∑l=0N−1coslθk=Re∑l=0N−1eilθk=0, we have(33)Aλ=−2λ2+2λcosθk,and thenΣ13=−λ2+λcosθk/1−2λcosθk+λ2.(ii)
Fork=j≠N/2, we obtain(34)Aλ=λ2∑l=1N−1cos2lθk+1−λ−1cos2lθk−θk+cosθk,and then(35)Aλ=N−2λ2−λcosθk.Therefore,(36)Σ13=N−2λ2−λcosθk21−2λcosθk+λ2.(iii)
Fork+j=N and k≠j, we obtain(37)Aλ=λ2∑l=1N−11+coslθk−j−λ−1cosθk+coslθk−j−θk=N−2λ2−λcosθk.Therefore,(38)Σ13=N−2λ2−λcosθk21−2λcosθk+λ2.(iv)
Fork=j=N/2, we obtain(39)Aλ=λ2∑l=1N−12−2cosθkλ=2N−1λ2−λcosθk,and so(40)Σ13=N−1λ2−λcosθk1−2λcosθk+λ2.Now, we computeΣ2 in equation (25). When λN≈0, it follows that(41)∑p=lN−1λpcospθk=λlcoslθk−λl+1cosl−1θk1−2λcosθk+λ2.By equation (25), we obtain(42)Σ2=∑l=1N−1λl−λ−lcoslθjλlcoslθk−λcosl−1θk1−2λcosθk+λ2=Σ21+Σ22,where(43)Σ21=11−2λcosθk+λ2∑l=1N−1λ2lcoslθjcoslθk−λcosl−1θk,Σ22=−11−2λcosθk+λ2∑l=1N−1coslθjcoslθk−λcosl−1θk.Furthermore,(44)Σ22=−121−2λcosθk+λ2∑l=1N−1coslθk+j+coslθk−j−λcoslθk+j−θk+coslθk−j−θk.Similar to the estimation process ofΣ13, we have(45)Σ22=2−2λcosθk21−2λcosθk+λ2,k≠j&,k+j≠N,Σ22=−N−21−λcosθk21−2λcosθk+λ2,k=j≠N2,Σ22=−N−21−λcosθk21−2λcosθk+λ2,k+j=N&,k≠j,Σ22=−N−11−λcosθk1−2λcosθk+λ2,k=j=N2.Combining this and equations (29)–(42), noticing that Σ12+Σ21=0, it follows that(46)Σ1+Σ2=PkλPjλ,k≠j&,k+j≠N,Σ1+Σ2=−N1−λ221−2λcosθk+λ2+Pkλλcosθj−λ21−2λcosθj+λ2−1,k=j≠N2,Σ1+Σ2=−N1−λ221−2λcosθk+λ2+Pkλλcosθj−λ21−2λcosθj+λ2−1,k+j=N&,k≠j,Σ1+Σ2=−N1−λ21−2λcosθk+λ2+Pkλλcosθj−λ21−2λcosθj+λ2−1,k=j=N2.Noticingλ<1 and 1−2λcosθk+λ2>1−λ2, we have(47)PkλPjλ<41−λ4,Pkλλcosθj−λ21−2λcosθj+λ2−1<121−λ4.Again, by equation (25), we obtain(48)ERex^kRex^j≤CN2,k≠j&k+j≠N,ERex^kRex^j−σ22N1−2λcosθk+λ2≤CN2,k=j≠N2ork+j=N&k≠j,ERex^kRex^j−σ2N1−2λcosθk+λ2≤CN2,k=j=N2,where C=12σ2/1−λ21−λ4 is a constant.Finally, we obtain the following.Lemma 1.
WhenλN≈0 and 1/N2≈0, the series Rex^kk=0,…,N−1 satisfy(49)αk,j=ERex^kRex^j=0,k≠jandk+j≠N,σ22N1−2λcosθk+λ2,k=j≠N2ork+j=N&k≠j,σ2N1−2λcosθk+λ2,k=j=N2.


## 4.2. Computation ofγk,j

By equation (3), we have(50)EImx^jImx^k=1N2∑l,p=1N−1Exlxpsinlθjsinpθk.Similar to the deducing process of equation (25), we have(51)N2λ2−1σ2EImx^jImx^k=∑l=1N−1λlsinlθj∑p=1l−1λp−λ−psinpθk+∑l=1N−1λl−λ−lsinlθj∑p=lN−1λpsinpθk≕Σ˜1+Σ˜2.Noticing that(52)∑p=1l−1λpsinpθk=−λlsinlθk+λsinθk+λl+1sinl−1θk1−2λcosθk+λ2,we obtain(53)∑p=1l−1λp−λ−psinpθk=μ˜l1+μ˜l2,where(54)μ˜l1=−λlsinlθk−λsinl−1θk1−2λcosθk+λ2,μ˜l2=λ−l+2sinlθk−λ−1sinl−1θk1−2λcosθk+λ2.We estimateΣ˜1 in equation (51):(55)Σ˜1=∑l=1N−1λlμ˜l1sinlθj+∑l=1N−1λlμ˜l2sinlθj≕Σ˜11+Σ˜12.Notice that(56)Σ˜12=λ221−2λcosθk+λ2∑l=1N−1−coslθk+j+coslθk−j+1λcoslθk+j−θk−coslθk−j−θk.Fork≠j and k+j≠N, we have(57)∑l=0N−1coslθk+j=∑l=0N−1coslθk−j=∑l=0N−1coslθk+j−θk=∑l=0N−1coslθk−j−θk=0,and so Σ˜12=0.Fork=j≠N/2, we have(58)Σ˜12=Nλ2−λcosθk21−2λcosθk+λ2.Fork+j=N and k≠N/2, we have(59)Σ˜12=−Nλ2−λcosθk21−2λcosθk+λ2.Fork=j=N/2, Σ˜12=0.We estimateΣ˜2 in equation (51). Since λN≈0, it follows that(60)∑p=lN−1λpsinpθk=λlsinlθk−λsinl−1θk1−2λcosθk+λ2.Furthermore,(61)Σ˜2=11−2λcosθk+λ2∑l=1N−1λ2lsinlθjsinlθk−λsinl−1θk−11−2λcosθk+λ2∑l=1N−1sinlθjsinlθk−λsinl−1θk≕Σ˜21+Σ˜22.By equation (55), it follows that Σ˜11+Σ˜12=0.ForΣ˜22,(62)Σ˜22=121−2λcosθk+λ2∑l=1N−1coslθk+j−coslθk−j−λcoslθk+j−θk−coslθk−j−θk0,k≠j&k+j≠Nork=j=N2,−N1−λcosθk21−2λcos2πk/N+λ2,k=j≠N2,N1−λcosθk21−2λcosθk+λ2,k+j=N&k≠N2.From all of the above, it follows that(63)Σ˜1+Σ˜2=0,k≠j&,k+j≠Nork=j=N2,Nλ2−121−2λcos2πk/N+λ2,k=j≠N2,N1−λ221−2λcosθk+λ2,k+j=N&,k≠N2.Therefore, we have the following.Lemma 2.
WhenλN≈0, the series Imx^kk=0,…,N−1 satisfy(64)γk,j=EImx^kImx^j=0,k≠j&k+j≠Nork=j=N2,σ22N1−2λcosθk+λ2,k=j≠N2,−σ22N1−2λcosθk+λ2k+j=N,&k≠N2.


## 4.3. Computation ofβk,j

By equation (3), we have(65)ERex^jImx^k=−1N2∑l,p=1N−1Exlxpcoslθjsinpθk.Similar to the deducing process of equation (25), we obtain(66)N21−λ2σ2ERex^jImx^k=∑l=1N−1λlcoslθj∑p=1l−1λp−λ−psinsθk+∑l=1N−1λl−λ−lcoslθj∑p=lN−1λpsinpθk≕Σ¯1+Σ¯2.Since(67)∑p=1l−1λp−λ−psinpθk=μ¯l1+μ¯l2,where(68)μ¯l1=−λlsinlθk−λsinl−1θk1−2λcosθk+λ2,μ¯l2=λ−l+2sinlθk−λ−1sinl−1θk1−2λcosθk+λ2,we have(69)Σ¯1=−11−2λcosθk+λ2∑l=1N−1λ2lcoslθjsinlθk−λsinl−1θk+λ21−2λcosθk+λ2∑l=1N−1coslθjsinlθk−λ−1sinl−1θk≕Σ¯11+Σ¯12.Furthermore,(70)Σ¯12=λ221−2λcosθk+λ2∑l=1N−1sinlθk+j+sinlθk−j−λ−1sinlθk+j−θk+sinlθk−j−θk.Noticing that∑l=1N−1sinlθk+j=∑l=1N−1sinlθk−j=0, we obtain(71)Σ¯12=−λ21−2λcosθk+λ2∑l=1N−1sinlθk+j−θk+sinlθk−j−θk.ForΣ¯2, we obtain(72)Σ¯2=11−2λcosθk+λ2∑l=1N−1λ2lcoslθjsinlθk−λsinl−1θk−11−2λcosθk+λ2∑l=1N−1coslθjsinlθk−λsinl−1θk≕Σ¯21+Σ¯22,where(73)Σ¯22=−121−2λcosθk+λ2∑l=1N−1sinlθk+j+sinlθk−j−λsinlθk+j−θk+sinlθk−j−θk=λ21−2λcosθk+λ2∑l=1N−1sinlθk+j−θk+sinlθk−j−θk.Since(74)Σ¯11+Σ¯21=0,Σ¯12+Σ¯22=0,we have the following.Lemma 3.
WhenλN≈0, the series ERex^k,Imx^jk,j=0,…,N−1 satisfy(75)βk,j=ERex^kImx^j=0,k,j=0,…,N−1.


## 4.4. Proof of Theorem1

Without loss of generalization, we assume thatN is an odd number. By equation (4), we obtain(76)ReWns=∑k=0N/2+∑k=N/2+1N−1rkRex^kcosηn,k−Imx^ksinηn,k=I1+I2,where(77)ηn,k=φk+nθk,φk=Argψ^¯sωk,rk=ψ^sωk.Sincex^k is real, we have x^¯k=x^N−k. This implies that(78)Rex^N−k=Rex^k,Imx^N−k=−Imx^k.Sinceψ is real-valued, ψ^ξ=ψ^¯−ξ; by ωk=θk/δt,k≤N/2, and ωk=−θN−k/δt,k>N/2, we obtain(79)ψ^sωN−k=ψ^−sωk=ψ^¯sωk.Therefore, by equation (77),(80)rN−k=ψ^sωN−k=ψ^sωk=rk,φN−k=Argψ^¯sωN−k=Argψ^¯sωk=−Argψ^sωk=−φk.Hence,ηn,N−k=φN−k+nθN−k=−φk−nθk=−ηn,k. From this, ψ^0=0, and equation (76), it follows that(81)I2=∑k=N/2+1N−1rkRex^kcosηn,k−Imx^ksinηn,k=∑k=1N/2rN−kRex^N−kcosηn,N−k−Imx^N−ksinηn,N−k=I1,Wns2=4∑k,j=0N/2rkrjRex^kcosηn,k−Imx^ksinηn,kRex^jcosηn,j−Imx^jsinηn,j.By Lemmas1–3, we obtain(82)EWns2=4∑k=0N/2rk2ERex^k2cos2ηn,k+EImx^k2sin2ηn,k=4∑k=0N/2rk2Rex^k2cos2ηn,k+sin2ηn,k=4∑k=0N/2rk2Rex^k2=2σ2N∑k=0N/2ψ^sωk21−2λcosθk+λ2.SinceWns=ReWns is a Gaussian random variable with mean 0, we obtain Theorem 1.


## 4.5. Proof of Theorems2 and 3

Without loss of generalization, we assume thatN is an old number. By equation (4), for n=0,…,N−1, we have(83)Wns=∑k=0N/2x^kψ^¯sωkeinθk+∑k=N/2+1N−1x^kψ^¯sωkeinθk.Sinceψ is an analytic wavelet, ψ^ω=0,ω<0. Notice that ωk<0,k>N/2. The second term on the right-hand side of equation (83) is vanishing. Therefore,(84)Wns=∑k=0N/2x^kψ^¯sωkeinθk.Furthermore,(85)EReWns2=∑k,j=0N/2rkrjERex^keiηn,kRex^jeiηn,j.However, by Lemma3 and equation (77), it follows that(86)EReWns2=∑k,j=0N/2rkrjERex^kRex^jcosηn,kcosηn,j+EImx^kImx^jsinηn,ksinηn,j.By Lemmas1 and 2, ERex^k2=EImx^k2,0≤k≤N/2, and so(87)EReWns2=∑k=0N/2rk2ERex^k2cos2ηn,k+EImx^k2sin2ηn,k=∑k=0N/2rk2ERex^k2cos2ηn,k+sin2ηn,k=∑k=0N/2rk2ERex^k2.Similarly, we have(88)EImWns2=∑k,j=0N/2rkrjEImx^keiηn,kImx^jeiηn,j.By Lemma3,(89)EImx^keiηn,kImx^jeiηn,j=ERex^kRex^jsinηn,ksinηn,j+EImx^kImx^jcosηn,kcosηn,j.This implies that(90)EImWns2=∑k=0N/2−1rk2ERex^k2sin2ηn,k+EImx^k2cos2ηn,k=∑k=0N/2−1rk2ERex^k2.Therefore, we have(91)VarReWns=VarImWns=∑k=0N/2rk2ERex^k2=σ22N1−λ2∑k=0N/2Pkλψ^sωk2.Now, we prove thatReWns and ImWns are independent.By Lemmas1–3,(92)EReWnsImWns=∑k,j=0N/2rkrjERex^kRex^jcosηn,ksinηn,j−EImx^kImx^jsinηn,kcosηn,j=12∑k=0N/2rk2ERex^k2sin2ηn,k−EImx^k2sin2ηn,k=0.Since bothReWns and ImWns are Gaussian random variables with mean 0 and(93)VarReWns=VarImWns≕σ2s,where(94)σ2s=σ22N1−λ2∑k=0N/2Pkλψ^sωk2,we obtain Theorem 2.Consider the phaseΦ=ArgWns. Notice that the p.d.f.s of ReWns and ImWns are, respectively,(95)pRx=12πσse−x2/2σ2s,pIx=12πσse−y2/2σ2s,and ReWns and ImWns are independent, the joint p.d.f. of ReWns and ImWns is(96)px,y=12πσ2se−x2+y2/2σ2s.The probability(97)Pβ1<Φ<β2=12πσ2s∬Ωe−x2+y2/2σ2sdxdy,where the domain Ω:=0<r<∞,0≤β1<φ<β2<2π. Hence,(98)Pβ1<Φ<β2=12πσ2s∫β1β2dφ∫0∞re−r2/2σ2sdr=12πβ2−β1,i.e., Theorem 3 holds.


## 5. Climate Diagnosis

Climatic background noises are often modeled as red noise [4–9]. In order to extract instinct features of climatic time series, we can use Theorems 1 and 2 to compare the wavelet power spectrum of climatic time series with that of red noise. If the values of the wavelet power spectrum of climate time series at some regions are all outside the 95% confidence interval for the distribution for the wavelet power spectrum of the red noise, then the wavelet power spectrum at this region contain instinct features of climatic time series. Moreover, we can use Theorem 3 to test the synchronization between the two climatic time series. If the difference of the wavelet phase of the two climatic time series does not change in some significant regions of the wavelet power spectrum, noticing that Theorem 3 indicates that this cannot be caused by climatic background noise effects, these two climatic time series demonstrate synchronization.



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