# Master's Thesis defence presentation

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Return Interval Distribution of Extreme Events in Longmemory Time Series With Two Scaling ExponentsSmrati Kumar KatiyarDepartment of PhysicsIISER, PuneMay 3, 2011Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 1 / 271 Explanation for the title2 Statistical test for long memory3 Foundation stone for our work4 our workAnalytical approachNumerical approach to the problemComparison of analytical and numerical results5 Long memory probability process with two scaling exponents6 conclusion7 future directionKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 2 / 27What are the key terms?Return interval distribution of extreme events in long memory time serieswith two scaling exponents.Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 3 / 27What are the key terms?Return interval distribution of extreme events in long memory time serieswith two scaling exponents.1 Return interval and extreme eventsKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 3 / 27What are the key terms?Return interval distribution of extreme events in long memory time serieswith two scaling exponents.1 Return interval and extreme events2 Long memory time seriesKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 3 / 27What are the key terms?Return interval distribution of extreme events in long memory time serieswith two scaling exponents.1 Return interval and extreme events2 Long memory time series3 Scaling exponentsKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 3 / 27Return interval and extreme eventsGiven a time series X (t)Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 4 / 27Return interval and extreme eventsGiven a time series X (t)0 20 40 60 80 100t-4-3-2-10123x(t)thresholdr1 r2 r3Figure: Return intervals and extreme eventsKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 4 / 27Aim of our project workExample : let say we are given a time series X(t) and there are total 11time instants at which the value of X is more than the threshold(q).Those time instants are,t = 0, 1, 3, 5, 6, 7, 10, 11, 12, 14, 16Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 5 / 27Aim of our project workExample : let say we are given a time series X(t) and there are total 11time instants at which the value of X is more than the threshold(q).Those time instants are,t = 0, 1, 3, 5, 6, 7, 10, 11, 12, 14, 16So the return intervals will be :return intervals = 1, 2, 2, 1, 1, 3, 1, 1, 2, 2out of these 10 return intervals we have5 return intervals of length 14 return intervals of length 2and 1 return interval of length 3Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 5 / 27Aim of our project workExample : let say we are given a time series X(t) and there are total 11time instants at which the value of X is more than the threshold(q).Those time instants are,t = 0, 1, 3, 5, 6, 7, 10, 11, 12, 14, 16So the return intervals will be :return intervals = 1, 2, 2, 1, 1, 3, 1, 1, 2, 2out of these 10 return intervals we have5 return intervals of length 14 return intervals of length 2and 1 return interval of length 3so the probability of occurance of return interval of length 1 will beP(1) = 510 ,similarly P(2) = 410 and P(3) =110Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 5 / 27Long memory time seriesKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 6 / 27Long memory time seriesPlot of sample autocorrelation function (ACF) k against lag k is one ofthe most useful tool to analyse a given time series.Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 6 / 27Long memory time seriesPlot of sample autocorrelation function (ACF) k against lag k is one ofthe most useful tool to analyse a given time series.k =nt=k+1(xtx)(xtkx)nt=1(xtx)2Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 6 / 27Long memory time seriesPlot of sample autocorrelation function (ACF) k against lag k is one ofthe most useful tool to analyse a given time series.k =nt=k+1(xtx)(xtkx)nt=1(xtx)2For long memory processesk Ck as k where C > 0 and (0, 1)Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 6 / 27Long memory time seriesPlot of sample autocorrelation function (ACF) k against lag k is one ofthe most useful tool to analyse a given time series.k =nt=k+1(xtx)(xtkx)nt=1(xtx)2For long memory processesk Ck as k where C > 0 and (0, 1)0 10 20 30 40 50 600.00.20.40.60.81.0LagACFSeries aKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 6 / 27Long memory time seriesPlot of sample autocorrelation function (ACF) k against lag k is one ofthe most useful tool to analyse a given time series.k =nt=k+1(xtx)(xtkx)nt=1(xtx)2For long memory processesk Ck as k where C > 0 and (0, 1)0 10 20 30 40 50 600.00.20.40.60.81.0LagACFSeries aA Long memory process is trend reinforcing, which means the direction(up or down compared to the last value) of the next value is more likelythe same as current value.Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 6 / 27Scaling exponentsKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 7 / 27Scaling exponentsThe most common power laws relate two variables and have the formf (x) xKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 7 / 27Scaling exponentsThe most common power laws relate two variables and have the formf (x) xHere is called the scaling exponent. where the word scaling denotesthe fact that a power-law function satisfiesf (cx) = cf (x) f (x)Here c is a constant.Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 7 / 27Statistical test for long memoryHow to find whether a given time series x(t) has long memory or not?Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 8 / 27Statistical test for long memoryHow to find whether a given time series x(t) has long memory or not?Detrended fluctuation analysisKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 8 / 27Statistical test for long memoryHow to find whether a given time series x(t) has long memory or not?Detrended fluctuation analysisx(t) is the time series. (t = 1, 2, 3, .......Nmax )Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 8 / 27Statistical test for long memoryHow to find whether a given time series x(t) has long memory or not?Detrended fluctuation analysisx(t) is the time series. (t = 1, 2, 3, .......Nmax )y(k) =ki=1(xi x) cumulative sum or profileKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 8 / 27Statistical test for long memoryHow to find whether a given time series x(t) has long memory or not?Detrended fluctuation analysisx(t) is the time series. (t = 1, 2, 3, .......Nmax )y(k) =ki=1(xi x) cumulative sum or profileDivide y(k) into time window of length n samples. In each box of lengthn, we fit y(k), using a polynomial function of order l , which represents thetrend in that box. The y coordinate of the fit line in each box is denotedby yn(k). Since we use a polynomial fit of order l , we denote the algorithmas DFA-l .0 100 200 300 400 500 600 700 800 900 1000050100150200250300350KYkKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 8 / 27The integrated signal y(k) is detrended by subtracting the local trendyn(k) in each box of length n.For a given box size n, the root-mean-square (rms) fluctuation for thisintegrated and detrended signal is calculated:F (n) =1NmaxNmaxk=1 [y(k) yn(k)]2Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 9 / 27The integrated signal y(k) is detrended by subtracting the local trendyn(k) in each box of length n.For a given box size n, the root-mean-square (rms) fluctuation for thisintegrated and detrended signal is calculated:F (n) =1NmaxNmaxk=1 [y(k) yn(k)]2The above computation is repeated for a broad range of scales (box sizen) to provide a relationship between F (n) and the box size n.Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 9 / 27The integrated signal y(k) is detrended by subtracting the local trendyn(k) in each box of length n.For a given box size n, the root-mean-square (rms) fluctuation for thisintegrated and detrended signal is calculated:F (n) =1NmaxNmaxk=1 [y(k) yn(k)]2The above computation is repeated for a broad range of scales (box sizen) to provide a relationship between F (n) and the box size n.A power-law relation between the average root-meansquare fluctuationfunction F (n) and the box size n indicates the presence of scalingF (n) nKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 9 / 27We can fit the log-log plot with a straight line and the slope of that linewill be the scaling exponent.Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 10 / 27We can fit the log-log plot with a straight line and the slope of that linewill be the scaling exponent.0 0.2 0.4 0.6 0.8 1ln n00.20.40.60.81ln F(n)Figure: log-log plot of F (n) Vs nKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 10 / 27We can fit the log-log plot with a straight line and the slope of that linewill be the scaling exponent.0 0.2 0.4 0.6 0.8 1ln n00.20.40.60.81ln F(n)Figure: log-log plot of F (n) Vs nWhen slope of line is in range (1/2,1) ,The time series displays longmemory.Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 10 / 27Foundation stone for our workfor a long memory time series with one scaling exponent, the probabilitydistribution of return intervals are known (Santhanam et. al, 2008)Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 11 / 27Foundation stone for our workfor a long memory time series with one scaling exponent, the probabilitydistribution of return intervals are known (Santhanam et. al, 2008)P(R) = a R(1) e( a)RHere R is the scaled return intervalR = rr ,where r are the actual return intervals r = 1, 2, 3, 4, ......Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 11 / 27our workWhat about time series with two scaling exponent?How to calculate their return interval distributions?Examples of these kind of time series are:Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 12 / 27our workWhat about time series with two scaling exponent?How to calculate their return interval distributions?Examples of these kind of time series are:high frequency financial data,Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 12 / 27our workWhat about time series with two scaling exponent?How to calculate their return interval distributions?Examples of these kind of time series are:high frequency financial data,network traffic of a web server etc.Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 12 / 27our workWhat about time series with two scaling exponent?How to calculate their return interval distributions?Examples of these kind of time series are:high frequency financial data,network traffic of a web server etc.0 0.2 0.4 0.6 0.8 1ln n00.20.40.60.81ln F(n)Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 12 / 27our workWhat about time series with two scaling exponent?How to calculate their return interval distributions?Examples of these kind of time series are:high frequency financial data,network traffic of a web server etc.0 0.2 0.4 0.6 0.8 1ln n00.20.40.60.81ln F(n)Figure: Podobnik et al. PHYSICA A,2002Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 12 / 27our approach to solve the problemanalytical approachKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 13 / 27our approach to solve the problemanalytical approach We will consider a probability model for a stationaryprocess with long memory, given an extreme event at time t = 0, theprobability to find an extreme event at time t = r is given byKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 13 / 27our approach to solve the problemanalytical approach We will consider a probability model for a stationaryprocess with long memory, given an extreme event at time t = 0, theprobability to find an extreme event at time t = r is given byPex (r) ={a1r(211) = a1r(11) for 0 < r < nxa2r(221) = a2r(12) for nx < r < where 0.5 < 1, 2 < 1 are DFA exponents and 0 < 1, 2 < 1 areautocorrelation exponentsnx is the crossover scaleKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 13 / 27After a very long algebra we find the return interval distributionP(r) ={a1r(11)e(a1/1)r1 for 0 < r < nxCa2r(12)e(a2/2)r2 for nx < r < Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 14 / 27After a very long algebra we find the return interval distributionP(r) ={a1r(11)e(a1/1)r1 for 0 < r < nxCa2r(12)e(a2/2)r2 for nx < r < How to find three unknowns a1, a2 and C ?Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 14 / 27After a very long algebra we find the return interval distributionP(r) ={a1r(11)e(a1/1)r1 for 0 < r < nxCa2r(12)e(a2/2)r2 for nx < r < How to find three unknowns a1, a2 and C ? we need three equations........Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 14 / 27After a very long algebra we find the return interval distributionP(r) ={a1r(11)e(a1/1)r1 for 0 < r < nxCa2r(12)e(a2/2)r2 for nx < r < How to find three unknowns a1, a2 and C ? we need three equations........normalization equation0P(r)dr = 1.Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 14 / 27After a very long algebra we find the return interval distributionP(r) ={a1r(11)e(a1/1)r1 for 0 < r < nxCa2r(12)e(a2/2)r2 for nx < r < How to find three unknowns a1, a2 and C ? we need three equations........normalization equation0P(r)dr = 1.normalizing r to unity 0rP(r)dr = 1Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 14 / 27After a very long algebra we find the return interval distributionP(r) ={a1r(11)e(a1/1)r1 for 0 < r < nxCa2r(12)e(a2/2)r2 for nx < r < How to find three unknowns a1, a2 and C ? we need three equations........normalization equation0P(r)dr = 1.normalizing r to unity 0rP(r)dr = 1using continuity condition{a1r(11) = a2r(12) at r = nxa1n(11)x = a2n(12)xKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 14 / 27final equations for a1, a2 and Ca1n(11)x = a2n(12)xKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 15 / 27final equations for a1, a2 and Ca1n(11)x = a2n(12)xCe(a2/2)n2x = e(a1/1)n1xKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 15 / 27final equations for a1, a2 and Ca1n(11)x = a2n(12)xCe(a2/2)n2x = e(a1/1)n1xC (2/a2)1/2nxE212(n2x )2 (1/a1)1/1nxE111(n1x )1= 1Here En(x) =1exttndt = 10 ex/(n2)dEn(x) is known as exponential integral function.Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 15 / 27numerical approach to the problemfirst and only challenge with this approach : to get a long memory timeseries which contains two different scaling exponents.Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 16 / 27numerical approach to the problemfirst and only challenge with this approach : to get a long memory timeseries which contains two different scaling exponents.The modelKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 16 / 27numerical approach to the problemfirst and only challenge with this approach : to get a long memory timeseries which contains two different scaling exponents.The modelStep 1:set the length of time series, say, l = 105.Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 16 / 27numerical approach to the problemfirst and only challenge with this approach : to get a long memory timeseries which contains two different scaling exponents.The modelStep 1:set the length of time series, say, l = 105.Step 2:generate a series of random numbers yi i = 0......(l 1) which followgaussian distribution with mean 0 and variance 1Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 16 / 27numerical approach to the problemfirst and only challenge with this approach : to get a long memory timeseries which contains two different scaling exponents.The modelStep 1:set the length of time series, say, l = 105.Step 2:generate a series of random numbers yi i = 0......(l 1) which followgaussian distribution with mean 0 and variance 1Step 3:generate a series of coefficients defined as:Ci =(i )()(i + 1)= (1 )(i )(i + 1) ={1 for 0 < i < nx2 for nx < i < Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 16 / 27Both 1 and 2 belong to the interval (0.5, 0)The asymptotic behaviour of Ci for large i can be written asCi (1 )i(1+) for i 1Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 17 / 27Both 1 and 2 belong to the interval (0.5, 0)The asymptotic behaviour of Ci for large i can be written asCi (1 )i(1+) for i 1Step 4:Now, get a series yi using yi and Ci according to the relationyi =ij=0yijCj i = 0.....(l 1) (1)Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 17 / 27DFA of time series generated using previous model0 1 2 3 4 5 6log (n)01234log F(n)DFA analysis of time seriescrossover regionKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 18 / 27comparison of analytical and numerical resultstry to fit P(r) = ar(1)e(c/)rto each segment, according to theircorresponding values-5 -4 -3 -2 -1 0 1 2ln (R)-8-7-6-5-4-3ln P(R)segment 1segment 2break pointdiscrepancy because of threshold dependence of constants and longmemory in return intervals.Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 19 / 27-4.8-4.6-4.4-4.2-4-3.8-3.6-3.4-4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1ln P(R)ln(R)return interval distribution segment(1)-8-7.5-7-6.5-6-5.5-5-4.5-0.5 0 0.5 1 1.5ln P(R)ln(R)return interval distribution segment(2)Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 20 / 27Long memory probability process with two scalingexponentsTo remove discrepancy because of long memory in return intervals. Wewill generate return intervals such that they have no dependence on eachother.first determine the constants a1 and a2 by normalizing it in the regionkmin = 1 and kmax . kmax1Pex(r)dr = nx1a1r(11)dr + kmaxnxa2r(12)dr = 1Use continuity condition as well and solve for a1 and a2.a1 =1[n1x1 11 +k2maxn12x2 n1x2]a2 =1[n2x1 n21x1+ k2max2 n2x2]Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 21 / 27Now generate a random number r from a uniform distribution at every rand compare it with the value of Pex(r). A random number is accepted asan extreme event if r < Pex(r) at any given value of r . If r Pex(r),then it is not an extreme event. Using this procedure we can generate aseries of extreme events.-6 -4 -2 0 2ln (R)-9-8-7-6-5-4-3-2ln P(R)segment 2segment 1break pointKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 22 / 27-6.5-6-5.5-5-4.5-4-3.5-3-2.5-2-5.5 -5 -4.5 -4 -3.5 -3 -2.5 -2 -1.5ln P(R)ln(R)return interval distribution of segment 1-8.5-8-7.5-7-6.5-6-1 -0.5 0 0.5 1 1.5 2ln P(R)ln(R)return interval distribution of segment 2P(r) ={a1r(11)e(a1/1)r1 for 0 < r < nxCa2r(12)e(a2/2)r2 for nx < r < Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 23 / 27In the previous slide, for segment(1) in place of a1, we have two variables aand b. Why we have two different variables? the possible reason is thatfor normalization integrals we have lower limit as 0 but in reality theminimum size of return interval is 1. So even after scaling, the minimumvalue of lower limit is 1/r .Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 24 / 27conclusionfor a long memory time series with two different scaling exponentKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 25 / 27conclusionfor a long memory time series with two different scaling exponentThere is a break point in the return interval distribution graph.Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 25 / 27conclusionfor a long memory time series with two different scaling exponentThere is a break point in the return interval distribution graph.For each scaling exponent there will be a different segment in returninterval distribution.Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 25 / 27conclusionfor a long memory time series with two different scaling exponentThere is a break point in the return interval distribution graph.For each scaling exponent there will be a different segment in returninterval distribution.Each segment still follow a distribution of the form which is product ofpower law and stretched exponential.Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 25 / 27future directionKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 26 / 27future directionthe model that we have used to generate time series with more than onescaling exponent need a fine tuning so that we can test our analyticalresults more accuratelyKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 26 / 27future directionthe model that we have used to generate time series with more than onescaling exponent need a fine tuning so that we can test our analyticalresults more accuratelywe should also think of the effects of long memory in return intervalsKatiyar S K (IISER Pune) Thesis Presentation May 3, 2011 26 / 27Thank youQuestions????Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 27 / 27Explanation for the titleStatistical test for long memoryFoundation stone for our workour workAnalytical approachNumerical approach to the problemComparison of analytical and numerical resultsLong memory probability process with two scaling exponentsconclusionfuture direction