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Return Interval Distribution of Extreme Events in Long memory Time Series With Two Scaling Exponents Smrati Kumar Katiyar Department of Physics IISER, Pune May 3, 2011 Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 1 / 27

Master's Thesis defence presentation

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Return Interval Distribution of Extreme Events in Long

memory Time Series With Two Scaling Exponents

Smrati Kumar Katiyar

Department of PhysicsIISER, Pune

May 3, 2011

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 1 / 27

1 Explanation for the title

2 Statistical test for long memory

3 Foundation stone for our work

4 our workAnalytical approachNumerical approach to the problemComparison of analytical and numerical results

5 Long memory probability process with two scaling exponents

6 conclusion

7 future direction

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 2 / 27

What 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 / 27

What are the key terms?

Return interval distribution of extreme events in long memory time serieswith two scaling exponents.

1 Return interval and extreme events

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 3 / 27

What are the key terms?

Return interval distribution of extreme events in long memory time serieswith two scaling exponents.

1 Return interval and extreme events

2 Long memory time series

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 3 / 27

What are the key terms?

Return interval distribution of extreme events in long memory time serieswith two scaling exponents.

1 Return interval and extreme events

2 Long memory time series

3 Scaling exponents

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 3 / 27

Return interval and extreme events

Given a time series X (t)

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 4 / 27

Return interval and extreme events

Given a time series X (t)

0 20 40 60 80 100t

-4

-3

-2

-1

0

1

2

3x(

t)

threshold

r1 r2 r3

Figure: Return intervals and extreme events

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 4 / 27

Aim of our project work

Example : 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, 16

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 5 / 27

Aim of our project work

Example : 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 3

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 5 / 27

Aim of our project work

Example : 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) = 5

10 ,similarly P(2) = 4

10 and P(3) = 110

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 5 / 27

Long memory time series

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 6 / 27

Long memory time series

Plot of sample autocorrelation function (ACF) ρk 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 / 27

Long memory time series

Plot of sample autocorrelation function (ACF) ρk against lag k is one ofthe most useful tool to analyse a given time series.

ρk =∑n

t=k+1(xt−x)(xt−k−x)∑n

t=1(xt−x)2

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 6 / 27

Long memory time series

Plot of sample autocorrelation function (ACF) ρk against lag k is one ofthe most useful tool to analyse a given time series.

ρk =∑n

t=k+1(xt−x)(xt−k−x)∑n

t=1(xt−x)2

For long memory processes

ρk → Cρk−γ as k → ∞

where Cρ > 0 and γ ∈ (0, 1)

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 6 / 27

Long memory time series

Plot of sample autocorrelation function (ACF) ρk against lag k is one ofthe most useful tool to analyse a given time series.

ρk =∑n

t=k+1(xt−x)(xt−k−x)∑n

t=1(xt−x)2

For long memory processes

ρk → Cρk−γ as k → ∞

where Cρ > 0 and γ ∈ (0, 1)

0 10 20 30 40 50 60

0.0

0.2

0.4

0.6

0.8

1.0

Lag

AC

F

Series a

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 6 / 27

Long memory time series

Plot of sample autocorrelation function (ACF) ρk against lag k is one ofthe most useful tool to analyse a given time series.

ρk =∑n

t=k+1(xt−x)(xt−k−x)∑n

t=1(xt−x)2

For long memory processes

ρk → Cρk−γ as k → ∞

where Cρ > 0 and γ ∈ (0, 1)

0 10 20 30 40 50 60

0.0

0.2

0.4

0.6

0.8

1.0

Lag

AC

F

Series a

A 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 / 27

Scaling exponents

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 7 / 27

Scaling exponents

The most common power laws relate two variables and have the form

f (x) ∝ xα

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 7 / 27

Scaling exponents

The most common power laws relate two variables and have the form

f (x) ∝ xα

Here α is called the scaling exponent. where the word ”scaling” denotesthe fact that a power-law function satisfies

f (cx) = cαf (x) ∝ f (x)

Here c is a constant.

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 7 / 27

Statistical test for long memory

How 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 / 27

Statistical test for long memory

How to find whether a given time series x(t) has long memory or not?Detrended fluctuation analysis

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 8 / 27

Statistical test for long memory

How to find whether a given time series x(t) has long memory or not?Detrended fluctuation analysis

x(t) is the time series. (t = 1, 2, 3, .......Nmax )

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 8 / 27

Statistical test for long memory

How to find whether a given time series x(t) has long memory or not?Detrended fluctuation analysis

x(t) is the time series. (t = 1, 2, 3, .......Nmax )y(k) =

∑ki=1(xi − 〈x〉) cumulative sum or profile

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 8 / 27

Statistical test for long memory

How to find whether a given time series x(t) has long memory or not?Detrended fluctuation analysis

x(t) is the time series. (t = 1, 2, 3, .......Nmax )y(k) =

∑ki=1(xi − 〈x〉) cumulative sum or profile

Divide 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 10000

50

100

150

200

250

300

350

K

Yk

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 8 / 27

The 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) =√

1Nmax

∑Nmax

k=1 [y(k)− yn(k)]2

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 9 / 27

The 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) =√

1Nmax

∑Nmax

k=1 [y(k)− yn(k)]2

The 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 / 27

The 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) =√

1Nmax

∑Nmax

k=1 [y(k)− yn(k)]2

The 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 scaling

F (n) ∝ nα

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 9 / 27

We 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 / 27

We 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 n

0

0.2

0.4

0.6

0.8

1

ln F

(n)

Figure: log-log plot of F (n) Vs n

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 10 / 27

We 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 n

0

0.2

0.4

0.6

0.8

1

ln F

(n)

Figure: log-log plot of F (n) Vs n

When 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 / 27

Foundation stone for our work

for 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 / 27

Foundation stone for our work

for 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−γ) e−( a

γ)Rγ

Here R is the scaled return intervalR = r

〈r〉 ,where r are the actual return intervals r = 1, 2, 3, 4, ......

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 11 / 27

our work

What 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 / 27

our work

What 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 / 27

our work

What 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 / 27

our work

What 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 n

0

0.2

0.4

0.6

0.8

1

ln F

(n)

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 12 / 27

our work

What 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 n

0

0.2

0.4

0.6

0.8

1

ln F

(n)

Figure: Podobnik et al. PHYSICA A,2002

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 12 / 27

our approach to solve the problem

analytical approach

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 13 / 27

our approach to solve the problem

analytical 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 by

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 13 / 27

our approach to solve the problem

analytical 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 by

Pex (r) =

{

a1r−(2α1−1) = a1r

−(1−γ1) for 0 < r < nx

a2r−(2α2−1) = a2r

−(1−γ2) for nx < r < ∞

where 0.5 < α1, α2 < 1 are DFA exponents and 0 < γ1, γ2 < 1 areautocorrelation exponentsnx is the crossover scale

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 13 / 27

After a very long algebra we find the return interval distribution

P(r) =

{

a1r−(1−γ1)e−(a1/γ1)r

γ1 for 0 < r < nx

Ca2r−(1−γ2)e−(a2/γ2)rγ2 for nx < r < ∞

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 14 / 27

After a very long algebra we find the return interval distribution

P(r) =

{

a1r−(1−γ1)e−(a1/γ1)r

γ1 for 0 < r < nx

Ca2r−(1−γ2)e−(a2/γ2)rγ2 for nx < r < ∞

How to find three unknowns a1, a2 and C ?

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 14 / 27

After a very long algebra we find the return interval distribution

P(r) =

{

a1r−(1−γ1)e−(a1/γ1)r

γ1 for 0 < r < nx

Ca2r−(1−γ2)e−(a2/γ2)rγ2 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 / 27

After a very long algebra we find the return interval distribution

P(r) =

{

a1r−(1−γ1)e−(a1/γ1)r

γ1 for 0 < r < nx

Ca2r−(1−γ2)e−(a2/γ2)rγ2 for nx < r < ∞

How to find three unknowns a1, a2 and C ? we need three equations........normalization equation

∫ ∞

0P(r)dr = 1.

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 14 / 27

After a very long algebra we find the return interval distribution

P(r) =

{

a1r−(1−γ1)e−(a1/γ1)r

γ1 for 0 < r < nx

Ca2r−(1−γ2)e−(a2/γ2)rγ2 for nx < r < ∞

How to find three unknowns a1, a2 and C ? we need three equations........normalization equation

∫ ∞

0P(r)dr = 1.

normalizing 〈r〉 to unity∫ ∞

0rP(r)dr = 1

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 14 / 27

After a very long algebra we find the return interval distribution

P(r) =

{

a1r−(1−γ1)e−(a1/γ1)r

γ1 for 0 < r < nx

Ca2r−(1−γ2)e−(a2/γ2)rγ2 for nx < r < ∞

How to find three unknowns a1, a2 and C ? we need three equations........normalization equation

∫ ∞

0P(r)dr = 1.

normalizing 〈r〉 to unity∫ ∞

0rP(r)dr = 1

using continuity condition

{

a1r−(1−γ1) = a2r

−(1−γ2) at r = nx

a1n−(1−γ1)x = a2n

−(1−γ2)x

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 14 / 27

final equations for a1, a2 and C

a1n−(1−γ1)x = a2n

−(1−γ2)x

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 15 / 27

final equations for a1, a2 and C

a1n−(1−γ1)x = a2n

−(1−γ2)x

Ce−(a2/γ2)nγ2x = e−(a1/γ1)n

γ1x

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 15 / 27

final equations for a1, a2 and C

a1n−(1−γ1)x = a2n

−(1−γ2)x

Ce−(a2/γ2)nγ2x = e−(a1/γ1)n

γ1x

C (γ2/a2)1/γ2

nxEγ2−1γ2

(nγ2x )

γ2− (γ1/a1)

1/γ1nxEγ1−1

γ1

(nγ1x )

γ1= 1

Here En(x) =∫∞1

e−xt

tndt =

∫ 10 e−x/ηη(n−2)dη

En(x) is known as exponential integral function.

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 15 / 27

numerical approach to the problem

first 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 / 27

numerical approach to the problem

first and only challenge with this approach : to get a long memory timeseries which contains two different scaling exponents.The model

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 16 / 27

numerical approach to the problem

first and only challenge with this approach : to get a long memory timeseries which contains two different scaling exponents.The model

Step 1:

set the length of time series, say, l = 105.

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 16 / 27

numerical approach to the problem

first and only challenge with this approach : to get a long memory timeseries which contains two different scaling exponents.The model

Step 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 1

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 16 / 27

numerical approach to the problem

first and only challenge with this approach : to get a long memory timeseries which contains two different scaling exponents.The model

Step 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:

Cαi =

Γ(i − α)

Γ(−α)Γ(i + 1)= −

α

Γ(1− α)

Γ(i − α)

Γ(i + 1)

α =

{

α1 for 0 < i < nx

α2 for nx < i < ∞

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 16 / 27

Both α1 and α2 belong to the interval (−0.5, 0)The asymptotic behaviour of Cα

i for large i can be written as

Cαi ≃ −

α

Γ(1− α)i−(1+α) for i ≫ 1

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 17 / 27

Both α1 and α2 belong to the interval (−0.5, 0)The asymptotic behaviour of Cα

i for large i can be written as

Cαi ≃ −

α

Γ(1− α)i−(1+α) for i ≫ 1

Step 4:

Now, get a series yαi using yi and Cαi according to the relation

yαi =i

j=0

yi−jCαj i = 0.....(l − 1) (1)

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 17 / 27

DFA of time series generated using previous model

0 1 2 3 4 5 6log (n)

0

1

2

3

4

log

F(n)

DFA analysis of time series

crossover r

egion

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 18 / 27

comparison of analytical and numerical results

try to fit P(r) = ar−(1−γ)e−(c/γ)rγ to each segment, according to theircorresponding γ values

-5 -4 -3 -2 -1 0 1 2ln (R)

-8

-7

-6

-5

-4

-3ln

P(R

)

segment 1

segment 2

break point

discrepancy 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 -1

ln 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 / 27

Long memory probability process with two scaling

exponents

To 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 .

∫ kmax

1Pex(r)dr =

∫ nx

1a1r

−(1−γ1)dr +

∫ kmax

nx

a2r−(1−γ2)dr = 1

Use continuity condition as well and solve for a1 and a2.

a1 =1

[nγ1x

γ1− 1

γ1+ k

γ2maxn

γ1−γ2x

γ2− n

γ1x

γ2]

a2 =1

[nγ2x

γ1− n

γ2−γ1x

γ1+ k

γ2max

γ2− n

γ2x

γ2]

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 21 / 27

Now generate a random number ξr from a uniform distribution at every r

and compare it with the value of Pex(r). A random number is accepted asan extreme event if ξr < Pex(r) at any given value of r . If ξr ≥ 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

-2

ln P

(R)

segment 2

segment 1

break point

Katiyar 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 -1.5

ln 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 2

ln P

(R)

ln(R)

return interval distribution of segment 2

P(r) =

{

a1r−(1−γ1)e−(a1/γ1)rγ1 for 0 < r < nx

Ca2r−(1−γ2)e−(a2/γ2)r

γ2 for nx < r < ∞

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 23 / 27

In 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 / 27

conclusion

for a long memory time series with two different scaling exponent

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 25 / 27

conclusion

for a long memory time series with two different scaling exponent

There is a break point in the return interval distribution graph.

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 25 / 27

conclusion

for a long memory time series with two different scaling exponent

There 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 / 27

conclusion

for a long memory time series with two different scaling exponent

There 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 / 27

future direction

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 26 / 27

future direction

the 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 accurately

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 26 / 27

future direction

the 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 accurately

we should also think of the effects of long memory in return intervals

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 26 / 27

Thank you

Questions????

Katiyar S K (IISER Pune) Thesis Presentation May 3, 2011 27 / 27