What is Lévy measure? Why is it needed, and what is $(1wedge|x^2|)$?

3

$begingroup$

A Borel measure $nu$ on $mathbb{R}$ is called a Lévy measure if
$nu({0})=0$ and $int_mathbb{R}(1wedge|x^2|) , nu(dx) < infty .$ (https://en.wikipedia.org/wiki/Financial_models_with_long-tailed_distributions_and_volatility_clustering#Infinitely_divisible_distributions)

So, what exactly is $(1wedge|x^2|)$? (Or rather correctly, what is the definition of levy measure saying?)

Edit: OK, but then why is levy measure needed?

real-analysis measure-theory economics

share|cite|improve this question

edited Jan 11 at 22:00

Did

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asked Mar 7 '13 at 3:24

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6314

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean by "why is it needed"? The Lévy-Khintchine decomposition provides a way of characterizing any Lévy process in terms of three components (the Lévy triplet) - one of which is a measure called the Lévy measure.
  $endgroup$
  – Stefan Hansen
  Mar 7 '13 at 8:20
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Levy measure describes the distribution of the jumps of the process. For example, a Poisson process of parameter $c>0$ has the Levy measure $cdelta(x-1)$, implying that the jump of size 1 occurs with intensity $c$. (In general, the jump part of a Levy process is a compensated sum of the Poisson point process with characteristic measure as the Levy measure.)
  $endgroup$
  – Sangchul Lee
  Mar 7 '13 at 8:30
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SangchulLee You mentioned that the levy measure of poisson process with rate c has levy measure $cdelta(x-1)$, could you please show the calculation explicitly?
  $endgroup$
  – math101
  Feb 25 '16 at 11:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @math101, Following the definition of Lévy measure as in the Lévy-Kintchine formula, you can easily check that $nu(x) = lambda delta(x-1)$ gives $$ phi_{X_1}(xi) = mathrm{e}^{lambda (e^{i xi} - 1)}, $$ which is the characteristic function of the Poisson process.
  $endgroup$
  – Sangchul Lee
  Feb 26 '16 at 0:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes. To match $c(e^{ixi}-1)$ with $int(e^{ixi x}-1)nu(dx)$. The first part, $e^{ixi}=int e^{ixi x}delta_1(dx)=int e^{ixi x} ddelta_1=e^{ixi 1}=e^{ixi}$. The 2nd part, we have $1=intdelta_1(dx).$
  $endgroup$
  – math101
  Feb 26 '16 at 8:26
  
  
  

  
  
  
  

add a comment | 

3

$begingroup$

A Borel measure $nu$ on $mathbb{R}$ is called a Lévy measure if
$nu({0})=0$ and $int_mathbb{R}(1wedge|x^2|) , nu(dx) < infty .$ (https://en.wikipedia.org/wiki/Financial_models_with_long-tailed_distributions_and_volatility_clustering#Infinitely_divisible_distributions)

So, what exactly is $(1wedge|x^2|)$? (Or rather correctly, what is the definition of levy measure saying?)

Edit: OK, but then why is levy measure needed?

real-analysis measure-theory economics

share|cite|improve this question

edited Jan 11 at 22:00

Did

248k23223460

asked Mar 7 '13 at 3:24

bouncebounce

6314

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean by "why is it needed"? The Lévy-Khintchine decomposition provides a way of characterizing any Lévy process in terms of three components (the Lévy triplet) - one of which is a measure called the Lévy measure.
  $endgroup$
  – Stefan Hansen
  Mar 7 '13 at 8:20
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Levy measure describes the distribution of the jumps of the process. For example, a Poisson process of parameter $c>0$ has the Levy measure $cdelta(x-1)$, implying that the jump of size 1 occurs with intensity $c$. (In general, the jump part of a Levy process is a compensated sum of the Poisson point process with characteristic measure as the Levy measure.)
  $endgroup$
  – Sangchul Lee
  Mar 7 '13 at 8:30
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SangchulLee You mentioned that the levy measure of poisson process with rate c has levy measure $cdelta(x-1)$, could you please show the calculation explicitly?
  $endgroup$
  – math101
  Feb 25 '16 at 11:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @math101, Following the definition of Lévy measure as in the Lévy-Kintchine formula, you can easily check that $nu(x) = lambda delta(x-1)$ gives $$ phi_{X_1}(xi) = mathrm{e}^{lambda (e^{i xi} - 1)}, $$ which is the characteristic function of the Poisson process.
  $endgroup$
  – Sangchul Lee
  Feb 26 '16 at 0:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes. To match $c(e^{ixi}-1)$ with $int(e^{ixi x}-1)nu(dx)$. The first part, $e^{ixi}=int e^{ixi x}delta_1(dx)=int e^{ixi x} ddelta_1=e^{ixi 1}=e^{ixi}$. The 2nd part, we have $1=intdelta_1(dx).$
  $endgroup$
  – math101
  Feb 26 '16 at 8:26
  
  
  

  
  
  
  

add a comment | 

3

3

3

6

$begingroup$

A Borel measure $nu$ on $mathbb{R}$ is called a Lévy measure if
$nu({0})=0$ and $int_mathbb{R}(1wedge|x^2|) , nu(dx) < infty .$ (https://en.wikipedia.org/wiki/Financial_models_with_long-tailed_distributions_and_volatility_clustering#Infinitely_divisible_distributions)

So, what exactly is $(1wedge|x^2|)$? (Or rather correctly, what is the definition of levy measure saying?)

Edit: OK, but then why is levy measure needed?

real-analysis measure-theory economics

share|cite|improve this question

edited Jan 11 at 22:00

Did

248k23223460

asked Mar 7 '13 at 3:24

bouncebounce

6314

$endgroup$

A Borel measure $nu$ on $mathbb{R}$ is called a Lévy measure if
$nu({0})=0$ and $int_mathbb{R}(1wedge|x^2|) , nu(dx) < infty .$ (https://en.wikipedia.org/wiki/Financial_models_with_long-tailed_distributions_and_volatility_clustering#Infinitely_divisible_distributions)

So, what exactly is $(1wedge|x^2|)$? (Or rather correctly, what is the definition of levy measure saying?)

Edit: OK, but then why is levy measure needed?

real-analysis measure-theory economics

real-analysis measure-theory economics

share|cite|improve this question

edited Jan 11 at 22:00

Did

248k23223460

asked Mar 7 '13 at 3:24

bouncebounce

6314

share|cite|improve this question

edited Jan 11 at 22:00

Did

248k23223460

asked Mar 7 '13 at 3:24

bouncebounce

6314

share|cite|improve this question

share|cite|improve this question

edited Jan 11 at 22:00

Did

248k23223460

edited Jan 11 at 22:00

Did

248k23223460

edited Jan 11 at 22:00

Did

248k23223460

248k23223460

asked Mar 7 '13 at 3:24

bouncebounce

6314

asked Mar 7 '13 at 3:24

bouncebounce

6314

asked Mar 7 '13 at 3:24

bouncebounce

6314

6314

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean by "why is it needed"? The Lévy-Khintchine decomposition provides a way of characterizing any Lévy process in terms of three components (the Lévy triplet) - one of which is a measure called the Lévy measure.
  $endgroup$
  – Stefan Hansen
  Mar 7 '13 at 8:20
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Levy measure describes the distribution of the jumps of the process. For example, a Poisson process of parameter $c>0$ has the Levy measure $cdelta(x-1)$, implying that the jump of size 1 occurs with intensity $c$. (In general, the jump part of a Levy process is a compensated sum of the Poisson point process with characteristic measure as the Levy measure.)
  $endgroup$
  – Sangchul Lee
  Mar 7 '13 at 8:30
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SangchulLee You mentioned that the levy measure of poisson process with rate c has levy measure $cdelta(x-1)$, could you please show the calculation explicitly?
  $endgroup$
  – math101
  Feb 25 '16 at 11:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @math101, Following the definition of Lévy measure as in the Lévy-Kintchine formula, you can easily check that $nu(x) = lambda delta(x-1)$ gives $$ phi_{X_1}(xi) = mathrm{e}^{lambda (e^{i xi} - 1)}, $$ which is the characteristic function of the Poisson process.
  $endgroup$
  – Sangchul Lee
  Feb 26 '16 at 0:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes. To match $c(e^{ixi}-1)$ with $int(e^{ixi x}-1)nu(dx)$. The first part, $e^{ixi}=int e^{ixi x}delta_1(dx)=int e^{ixi x} ddelta_1=e^{ixi 1}=e^{ixi}$. The 2nd part, we have $1=intdelta_1(dx).$
  $endgroup$
  – math101
  Feb 26 '16 at 8:26
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean by "why is it needed"? The Lévy-Khintchine decomposition provides a way of characterizing any Lévy process in terms of three components (the Lévy triplet) - one of which is a measure called the Lévy measure.
  $endgroup$
  – Stefan Hansen
  Mar 7 '13 at 8:20
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Levy measure describes the distribution of the jumps of the process. For example, a Poisson process of parameter $c>0$ has the Levy measure $cdelta(x-1)$, implying that the jump of size 1 occurs with intensity $c$. (In general, the jump part of a Levy process is a compensated sum of the Poisson point process with characteristic measure as the Levy measure.)
  $endgroup$
  – Sangchul Lee
  Mar 7 '13 at 8:30
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SangchulLee You mentioned that the levy measure of poisson process with rate c has levy measure $cdelta(x-1)$, could you please show the calculation explicitly?
  $endgroup$
  – math101
  Feb 25 '16 at 11:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @math101, Following the definition of Lévy measure as in the Lévy-Kintchine formula, you can easily check that $nu(x) = lambda delta(x-1)$ gives $$ phi_{X_1}(xi) = mathrm{e}^{lambda (e^{i xi} - 1)}, $$ which is the characteristic function of the Poisson process.
  $endgroup$
  – Sangchul Lee
  Feb 26 '16 at 0:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes. To match $c(e^{ixi}-1)$ with $int(e^{ixi x}-1)nu(dx)$. The first part, $e^{ixi}=int e^{ixi x}delta_1(dx)=int e^{ixi x} ddelta_1=e^{ixi 1}=e^{ixi}$. The 2nd part, we have $1=intdelta_1(dx).$
  $endgroup$
  – math101
  Feb 26 '16 at 8:26
  
  
  

  
  
  
  

2

2

$begingroup$
What do you mean by "why is it needed"? The Lévy-Khintchine decomposition provides a way of characterizing any Lévy process in terms of three components (the Lévy triplet) - one of which is a measure called the Lévy measure.
$endgroup$
– Stefan Hansen
Mar 7 '13 at 8:20

$begingroup$
What do you mean by "why is it needed"? The Lévy-Khintchine decomposition provides a way of characterizing any Lévy process in terms of three components (the Lévy triplet) - one of which is a measure called the Lévy measure.
$endgroup$
– Stefan Hansen
Mar 7 '13 at 8:20

2

2

$begingroup$
Levy measure describes the distribution of the jumps of the process. For example, a Poisson process of parameter $c>0$ has the Levy measure $cdelta(x-1)$, implying that the jump of size 1 occurs with intensity $c$. (In general, the jump part of a Levy process is a compensated sum of the Poisson point process with characteristic measure as the Levy measure.)
$endgroup$
– Sangchul Lee
Mar 7 '13 at 8:30

$begingroup$
Levy measure describes the distribution of the jumps of the process. For example, a Poisson process of parameter $c>0$ has the Levy measure $cdelta(x-1)$, implying that the jump of size 1 occurs with intensity $c$. (In general, the jump part of a Levy process is a compensated sum of the Poisson point process with characteristic measure as the Levy measure.)
$endgroup$
– Sangchul Lee
Mar 7 '13 at 8:30

$begingroup$
@SangchulLee You mentioned that the levy measure of poisson process with rate c has levy measure $cdelta(x-1)$, could you please show the calculation explicitly?
$endgroup$
– math101
Feb 25 '16 at 11:04

$begingroup$
@SangchulLee You mentioned that the levy measure of poisson process with rate c has levy measure $cdelta(x-1)$, could you please show the calculation explicitly?
$endgroup$
– math101
Feb 25 '16 at 11:04

$begingroup$
@math101, Following the definition of Lévy measure as in the Lévy-Kintchine formula, you can easily check that $nu(x) = lambda delta(x-1)$ gives $$ phi_{X_1}(xi) = mathrm{e}^{lambda (e^{i xi} - 1)}, $$ which is the characteristic function of the Poisson process.
$endgroup$
– Sangchul Lee
Feb 26 '16 at 0:15

$begingroup$
@math101, Following the definition of Lévy measure as in the Lévy-Kintchine formula, you can easily check that $nu(x) = lambda delta(x-1)$ gives $$ phi_{X_1}(xi) = mathrm{e}^{lambda (e^{i xi} - 1)}, $$ which is the characteristic function of the Poisson process.
$endgroup$
– Sangchul Lee
Feb 26 '16 at 0:15

$begingroup$
yes. To match $c(e^{ixi}-1)$ with $int(e^{ixi x}-1)nu(dx)$. The first part, $e^{ixi}=int e^{ixi x}delta_1(dx)=int e^{ixi x} ddelta_1=e^{ixi 1}=e^{ixi}$. The 2nd part, we have $1=intdelta_1(dx).$
$endgroup$
– math101
Feb 26 '16 at 8:26

$begingroup$
yes. To match $c(e^{ixi}-1)$ with $int(e^{ixi x}-1)nu(dx)$. The first part, $e^{ixi}=int e^{ixi x}delta_1(dx)=int e^{ixi x} ddelta_1=e^{ixi 1}=e^{ixi}$. The 2nd part, we have $1=intdelta_1(dx).$
$endgroup$
– math101
Feb 26 '16 at 8:26

add a comment | 

1 Answer
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$$1wedge|x^2|=begin{cases}1 & text{if} & |x|gt1, \ x^2 & text{if} & |x|leqslant1. end{cases}$$

share|cite|improve this answer

answered Mar 7 '13 at 4:11

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1 Answer
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1 Answer
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5

$begingroup$

$$1wedge|x^2|=begin{cases}1 & text{if} & |x|gt1, \ x^2 & text{if} & |x|leqslant1. end{cases}$$

share|cite|improve this answer

answered Mar 7 '13 at 4:11

DidDid

248k23223460

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add a comment | 

5

$begingroup$

$$1wedge|x^2|=begin{cases}1 & text{if} & |x|gt1, \ x^2 & text{if} & |x|leqslant1. end{cases}$$

share|cite|improve this answer

answered Mar 7 '13 at 4:11

DidDid

248k23223460

$endgroup$

add a comment | 

5

5

5

$begingroup$

$$1wedge|x^2|=begin{cases}1 & text{if} & |x|gt1, \ x^2 & text{if} & |x|leqslant1. end{cases}$$

share|cite|improve this answer

answered Mar 7 '13 at 4:11

DidDid

248k23223460

$endgroup$

$$1wedge|x^2|=begin{cases}1 & text{if} & |x|gt1, \ x^2 & text{if} & |x|leqslant1. end{cases}$$

share|cite|improve this answer

answered Mar 7 '13 at 4:11

DidDid

248k23223460

share|cite|improve this answer

share|cite|improve this answer

answered Mar 7 '13 at 4:11

DidDid

248k23223460

answered Mar 7 '13 at 4:11

DidDid

248k23223460

answered Mar 7 '13 at 4:11

DidDid

248k23223460

248k23223460

add a comment | 

add a comment | 

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