Analyzing Nonlinear Time Series
with
Hilbert-Huang Transform
Sai-Ping Li
Lunch Seminar
December 7, 2011
 Introduction
 Hilbert-Huang Transform
 Empirical Mode Decomposition
 Some Examples and Applications
 Summary and Future Works
Some Examples of Wave Forms
Stationary and Non-Stationary Time Series
Time series: random data plus trend, with best-fit line and different smoothings
Examples of Non-Stationary Time Series
Tidal data of Kahului Harbor, Maui, October 4-9, 1994.
Blood pressure of a rat.
Earthquake data of El Centro in 1985.
Difference of daily Non-stationary annual cycle and 30year mean annual cycle of surface temperature at Victoria
Station, Canada.
Consider the non-dissipative Duffing equation:
𝑑2 𝑥
𝑑𝑡 2
+ 𝑥 + 𝜖𝑥 3 = 𝛾 cos 𝜔𝑡
γ is the amplitude of a periodic forcing function with a frequency ω.
If 𝜖 = 0, the system is linear and the solution can be found easily.
If 𝜖 ≠ 0, the system becomes nonlinear.
If 𝜖 is not small, perturbation method cannot be used. The system is
highly nonlinear and new phenomena such as bifurcation and chaos
can occur.
Rewrite the above equation as:
𝑑2 𝑥
𝑑𝑡 2
+ 𝑥(1 + 𝜖𝑥 2 ) = 𝛾 cos 𝜔𝑡
The quantity in the parenthesis above can be regarded as a varying
spring constant, or varying pendulum length.
The frequency of the system can change from location to
location, and also from time to time, even within one
oscillation cycle.
Numerical Solution for the Duffing Equation
Left: The numerical solution of x and dx/dt as a function of time.
Right: The phase diagram with continuous winding indicates no
fixed period of oscillations.
Financial Time Series
Stylized Facts :
Stylized Facts :
An Empirical data analysis method:
Hilbert-Huang Transform
Empirical Mode Decomposition (EMD)
+
Hilbert Spectral Analysis (HSA)
The Hilbert-Huang Transform
Empirical Mode Decomposition:
Based on the assumption that any dataset consists of different
simple intrinsic modes of oscillations. Each of these intrinsic
oscillatory modes is represented by an intrinsic mode function
(IMF) with the following definition:
(1) in the whole dataset, the number of extrema and the number
of zero-crossings must either equal or differ at most by one,
and
(2) at any point, the mean value of the envelope defined by the
local maxima and the envelope defined by the local minima
is zero.
Example: A test Dataset
Identify all local extrema, then connect all the local maxima by a cubic spline
line in the upper envelope. Repeat the procedure for the local minima to
produce the lower envelope. The upper and lower envelopes should cover all
the data between them. Their mean is designated as 𝑚1 .
The difference between the data 𝒙 𝒕 and the mean 𝒎𝟏 is the first component 𝒉𝟏 ,
i.e.,
𝒉𝟏 = 𝒙 𝒕 − 𝒎𝟏
The data (blue) upper and lower
envelopes (green) defined by the local
maxima and minima, respectively, and
the mean value of the upper and lower
envelopes given in red.
The data (red) and ℎ1 (blue).
Repeat the Sifting Process:
Left: Repeated sifting steps with 𝒉𝟐 and 𝒎𝟐 . Right: Repeated sifting steps with
𝒉𝟑 and 𝒎𝟑 .
The Sifting Process for the first IMF:
𝒉𝟏 = 𝒙 𝒕 − 𝒎𝟏
In the next step,
𝒉𝟏𝟏 = 𝒉𝟏 − 𝒎𝟏𝟏
Repeat Sifting,
●
●
●
𝒉𝟏𝒌 = 𝒉𝟏(𝒌−𝟏) − 𝒎𝟏𝒌
And is designated as,
𝒄𝟏 = 𝒉𝟏𝒌
The first IMF component 𝒄𝟏 after 12 steps.
Separate 𝒄𝟏 from the original data, and call the residue 𝒓𝟏 ,
𝒓𝟏 = 𝒙 𝒕 − 𝒄𝟏
The original data (blue) and the residue 𝒓𝟏 .
The procedure is repeated to get all the subsequent 𝒓𝒋 ’s ,
𝒓𝟐 = 𝒓𝟏 − 𝒄𝟐
●
●
●
𝒓𝒌 = 𝒓(𝒌−𝟏) − 𝒄𝒌
The original dataset is decomposed into a sum of IMF’s and a
residue,
𝒙 𝒕 =
𝒏
𝒋=𝟏 𝒄𝒋
+ 𝒓𝒏
A decomposition of the data into n-empirical modes is achieved,
and a residue 𝒓𝒏 obtained which can either be the mean trend or a
constant.
A test dataset shown in the above.
On the right is the original dataset
decomposed into 8 IMF’s and a
trend (𝒄𝟗 ).
Empirical Mode Decomposition
Example: Gold Price Data Analysis
Statistics: LME gold prices
res.
IMF8
IMF7
IMF6
IMF5
IMF4
IMF3
IMF2
IMF1
data
1400
0
50
-50
50
-50
50
-50
100
-100
100
-100
100
-100
100
-100
200
-200
900
0
1968M1
1976M5
1984M9
1992M1
Date
2000M5
2008M9
Composition: LME monthly gold prices
• Dividing the components into high, low and trend by reasonable boundary
(12 months), and analyze the factors or economic meanings in different
time scales.
• Boundary:
12 months
Mean period ≧ 12 months
Trend
Low frequency term
Mean period < 12 months
High frequency term
Composition of LME Gold Prices
Trend: inflation
Pearson
Kendall
correlation
correlation
Trend of CPI
0.939
0.917
Trend of PPI
0.906
0.917
• We assume the gold price trend relates to inflation at first.
• US monthly CPI and PPI are used to quantify the ordering of inflation.
• Since the gold price trend hold high correlation with trend of CPI and PPI, the
economic meanings of trend can be described by “inflation”.
Low frequency term: significant events
2008/09:
Lehman Brothers
bankruptcy
1979/11～1980/01:
Iranian hostage crisis
1980/09:
Iran/Iraq War
2007/02:
USA Subprime
Mortgage Crisis
1987/10:
New York Stock
Market Crash
1996 to 2006:
Booming economic in USA
1982/08:
Mexico External
Debt Crisis
1973/10:
4th Middle East War
• The six obvious variations correspond to six significant events.
• The six significant events include wars, panic international situation, and financial
crisis.
data
C1
20
0
-20
C2
1500
1000
500
20
0
-20
Residue
C7
C6
C5
C4
C3
50
0
-50
50
0
-50
50
0
-50
50
0
-50
200
0
-200
1500
1000
500
2-Jan-07
2-Jan-08
2-Jan-09
Day
4-Jan-10
30-Dec-10
Example:
Electroencephalography (EEG)
and
Heart Rate Variability (HRV)
Example: Electroencephalography (EEG)
Active Wake Stage
Example: Electroencephalography (EEG)
Stage I:
Light Sleep
Stage II
Stage III
Slow Wave
Sleep
Stage IV
Quiet Sleep
Example: Electroencephalography (EEG)
Left: A dataset of active wake stage.
Right: Its corresponding IMF’s after EMD.
Electrocardiography (ECG)
Example: Heart Rate Variability (HRV)
Example: Heart Rate Variability (HRV)
Left: Original dataset
Right: HRV after EMD
Relation of EEG and
HRV of an Obstructive
Sleep Apnea (OSA)
patient using EMD
analysis
Partial List of Application of HHT
•
•
•
•
•
•
•
•
•
•
•
Biomedical applications: Huang et al. [1999]
Chemistry and chemical engineering: Phillips et al. [2003]
Financial applications: Huang et al. [2003b]
Image processing: Hariharan et al. [2006]
Meteorological and atmospheric applications: Salisbury and
Wimbush [2002]
Ocean engineering:Schlurmann [2002]
Seismic studies: Huang et al. [2001]
Solar Physics: Barnhart and Eichinger [2010]
Structural applications: Quek et al. [2003]
Health monitoring: Pines and Salvino [2002]
System identification: Chen and Xu [2002]
Comparison of Fourier, Wavelet and HHT
Problems related to Hilbert-Huang Transform
(1) Adaptive data analysis methodology in general
(2) Nonlinear system identification methods
(3) Prediction problem for non-stationary processes (end effects)
(4) Spline problems (best spline implementation for the HHT,
convergence and 2-D)
(5) Optimization problems (the best IMF selection and uniqueness)
(6) Approximation problems (Hilbert transform and quadrature)
(7) Other miscellaneous questions concerning the HHT…….
Application: Price Fluctuations in Financial Time Series
𝑑𝑆𝑡 = 𝜇𝑡 𝑆𝑡 𝑑𝑡 + 𝜎𝑡 𝑆𝑡 𝑑𝑊𝑡
⇒ 𝑑𝑆𝑡 /𝑆𝑡 = 𝜇𝑡 𝑑𝑡 + 𝜎𝑡 𝑑𝑊𝑡
⇒ 𝑑ln(𝑆𝑡 ) = 𝜇𝑡 𝑑𝑡 + 𝜎𝑡 𝑑𝑊𝑡
All parameters are time dependent. 𝜇𝑡 is the drift, 𝜎𝑡 is the
volatility and 𝑑𝑊𝑡 is a standard Wiener process with zero
mean and unit rate of variance.
Stoppage Criterion: The criterion to stop the sifting of IMF’s.
Historically, there are two methods:
(1) The normalized squared difference between two successive sifting operations
defined as
𝑆𝐷𝑘 =
𝑇 |ℎ
2
𝑡=0 𝑘−1 𝑡 −ℎ𝑘 (𝑡)|
2
𝑇 ℎ
𝑡=0 𝑘−1
If this squared difference 𝑆𝐷𝑘 is less than a preset value, the sifting process
will stop.
(2) The sifting process will stop only after S consecutive times, when the numbers
of zero-crossings and extrema stay the same and are equal or differ at most by
one. The number S is preset.
Orthogonality:
The orthogonality of the EMD components should also be checked a posteriori
numerically as follows: let us first write equation
𝒙 𝒕 =
𝒏
𝒋=𝟏 𝒄𝒋
as
+ 𝒓𝒏
𝒏+𝟏
𝒙 𝒕 =
𝒄𝒋
𝒋=𝟏
Form the square of the signal as
𝒏+𝟏
𝒙𝟐 𝒕 =
𝒏+𝟏 𝒏+𝟏
𝒄𝒋 𝟐 + 𝟐
𝒋=𝟏
𝒄𝒋 𝒕 𝒄𝒌 𝒕
𝒋=𝟏 𝒌=𝟏
And define IO as
𝑻
𝒏+𝟏 𝒏+𝟏
𝒄𝒋 𝒕 𝒄𝒌 𝒕 /𝒙𝟐 𝒕
𝑰𝑶 =
𝒕=𝟎
𝒋=𝟏 𝒌=𝟏
If the IMF’s are orthogonal, IO should be zero.