Tempered Stable Distributions: Understanding Q and R(dx) for Computing R(dx)

[cernlife]

My question is how to compute R(dx). But before I can ask that I have to write down the background to my problem, so bear withme


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A tempered stable distribution is when a stable distribution is tempered by an exponential function of the form e^{-\theta{x}}. In my particular case we are using a tempered stable law defined by Barndorff-Nielsen in the paper "modified stable processes" found here, http://economics.oul...nmsprocnew1.pdf .

In Barndorff's paper, \theta = (1/2)\gamma^{1/\alpha}, hence the tempering function is defined as e^{-(1/2} \gamma^{1/\alpha}{x}.


In Rosinski's paper on "tempering stable processes" (which can be found http://www-m4.ma.tum.de/Papers/Rosinski/tstable.pd or http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.78.7073&rep=rep1&type=pdf) he states that tempering of the stable density f \mapsto f_{\theta} leads to tempering of the corresponding Levy measure M \mapsto M_{\theta}, where M_{\theta}(dx) = e^{-\theta{x}}M(dx).

Rosinski then goes on to say the Levy measure of a stable law in polar coordinates is of the form

M_0(dr, du) = r^{-\alpha-1}dr\sigma(du) \hspace{30mm} (2.1)

and then says the Levy measure of a tempered stable density can be written as

M(dr, du) = r^{-\alpha-1}q(r,u)dr\sigma(du) \hspace{30mm} (2.2)

he then says, the tempering function q in (2.2) can be represented as

q(r,u) = \int_0^{\infty}e^{-rs}Q(ds|u) \hspace{30mm} (2.3)

Rosinski's paper also defines a measure R by

R(A) = \int_{R^d} I_A(x/||x||^2)||x||^{\alpha}Q(dx) \hspace{30mm} (2.5)

and has

Q(A) = \int_{R^d} I_A(x/||x||^2)||x||^{\alpha}R(dx) \hspace{30mm} (2.6)

now I know that for my particular tempered stable density the levy measure M is given by

2^{\alpha}\delta\frac{\alpha}{ \Gamma(1-\alpha)}x^{-1-\alpha}e^{-(1/2)\gamma^{1/\alpha}x}dx


Rosinski then goes on to state Theorem 2.3: The Levy measure M of a tempered stable distribution can be written in the form

M(A)=\int_{R^d}\int_0^{\infty} I_A(tx)t^{-\alpha-1}e^{-t}dtR(dx)

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So the question is, how can I work out what Q is? and what is R(dx)?

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I would greatly appreciate if anyone can help me on this one.

 

[micromass]

Hi cernlife!

Is R just a measure? Then R(dx) most likely represents that we integrate w.r.t. the measure R with variable x. Thus

\int_\Omega{f(x)R(dx)}=\int_\Omega{fdR}.

So R(dx) is just the same as dR, but the dx just means that the variable is x. The reason we need to denote that variable is simply because in a double integral, there are two variables: t and x. And we need to denote the variable by something, because saying

\iint{f(x,y)dRdQ}

wouldn't make clear whether the first integral is w.r.t. x or y...

Does that make sense, or do you think this has nothing to do with it?

 

[cernlife]

Yes R is just a measure.

Yes you make perfect sense, by that I mean in the paper I am talking about they also write

"This yields a change of variable formula"

\int_{R^d} R(dx) = \int_{R^d} F(x/||x||^2)R(dx)

between equations (2.5) and (2.6).

However, I am still as lost as every as how to compute R(dx). Could you explain in more detail how I can compute R(dx) in my specific case?

Thanks for the reply, greatly appreciated.

 

[micromass]

You can't calculate R(dx), since it is just a notation. It just means you got to integrate w.r.t. the measure R and w.r.t. the variable x. There's nothing to calculate here...

 

[I like Serena]

Hi cernlife! Welcome to PF!

Yes R is just a measure.

Yes you make perfect sense, by that I mean in the paper I am talking about they also write

"This yields a change of variable formula"

\int_{R^d} R(dx) = \int_{R^d} F(x/||x||^2)R(dx)

between equations (2.5) and (2.6).

However, I am still as lost as every as how to compute R(dx). Could you explain in more detail how I can compute R(dx) in my specific case?

Thanks for the reply, greatly appreciated.

I'm not sure if I can help, but we'll see if I can add something useful to what MM already wrote.


First, your formula is a little off. It should be:

\int_{R^d} R(dx) = \int_{R^d} F(x/||x||^2)Q(dx)


I haven't puzzled everything out yet, and this is a little while ago for me, but I think this is mostly a long winded story about converting spherical coordinates to cartesian coordinates in d-dimensional space.

Note that spherical coordinates in d dimensions are identified by coordinates on the unit sphere S^d-1 and the distance to the origin r.
For instance, in 3 dimensions this would be (theta, phi) combined with r.

If you want to calculate anything, you'd simply integrate in whatever coordinate system you like.


Let me give it as a specific example that fits your formulas in 2 dimensions.

We're talking about an entity A here that is described by a function I_A in d-dimensional coordinates.
We can integrate it using spherical coordinates (Q) or we can integrate it using cartesian coordinates (R).

If we integrate I_A in polar coordinates, we have

Q(A)=\int_0^{2\pi} \int_0^{\infty} I_A(r, \phi) \cdot dr \cdot r d\phi

If we do it in cartesian coordinates we have:

R(A)=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} I_A(x, y) \cdot dx \cdot dy

To do it properly we need to make the proper conversions of the function I_A of course.


That's all! Cheers!

(Hope this helps. )

 

[cernlife]

are you sure?

In theorem 2.3 (as written above) they say

M(A) = \int_{R^d} \int_0^{\infty} I_A(tx)t^{-\alpha-1}e^{-t}dtR(dx)

now I know the levy measure M in my case, i.e when the tempering function is

e^{-(1/2)\gamma^{1/\alpha}x}

is

M(dx) = 2^{\alpha}\delta\frac{\alpha}{ \Gamma(1-\alpha)}x^{-1-\alpha}e^{-(1/2)\gamma^{1/alpha}x}

so in order to get the above Levy measure from Theorem 2.3, I need to put something in as R(dx), otherwise how can I evaluate the expression in Theorem 2.3 to get the Levy Measure?

Also in another paper "A new tempered stable distribution and its applications" they deduce an expression for R(dx). although this is of no good to me as they are working with a different tempered stable distribution, (tempering a symmetric stable distribution, where as I am doing what Bardorff-Nielsen did and work with tempering a positive stable distribution). Anyway in there paper "A new tempered stable distribution and its applications" they say "The measure R corresponding to the Levy measure M can be deduced as

R(dx) = (\kappa_+r_+^{-p}|x|^{p-1}+\kappa_-r_-^{-p}|x|^{p-1})dx

there Tempered stable distribution is diffenent so I can just use their R(dx), I need to figure out what my R(dx) is.

So do you have any thoughts on how to work out what R(dx) = ... in my case?

thanks for your reply's it really helps to talk to someone else about all of this as its so important I can work this out and I really have no idea!

 

[cernlife]

Thanks "I like Serena" for the reply, much appreciated.

However I still don't have the slightest clue how to get my

R(dx)=...some kind of equation...

when I say "some kind of equation" I mean like

R(dx) = (2^{\alpha}\delta\frac{\alpha}{ \Gamma(1-\alpha)})dx

or something like that - I am not saying that's correct, its just a guess to show what kind of answer I am looking for - I'm finding all of this calculus, measure stuff and polar things all very abstract and confusing.

 

[I like Serena]

So do you have any thoughts on how to work out what R(dx) = ... in my case?

thanks for your reply's it really helps to talk to someone else about all of this as its so important I can work this out and I really have no idea!

In 3 dimensions we would have

R(A) = integral something R(dx) = integral something dxdydzPuzzling some more, I guess there's also a kind of fourier-transform involved.

So for a fourier-transform in 1 dimension, let's call it F(A), it seems we would have:

F(A) = integral something R(dx) = integral something e^{-2πi k x} dxI believe all of these formulas are intended to use as short hand notation.
But they do obscure things quite a lot!

 

[cernlife]

yes very obscure to myself

do you have any notion how to get the expression

R(dx) = ...dx

That I so badly need?

 

[I like Serena]

yes very obscure to myself

do you have any notion how to get the expression

R(dx) = ...dx

That I so badly need?

After looking some more I draw back my previous comment about fourier.

I believe R(dx) simply represents an infinitesimal element dxdydz.

What you would have is that M(dx) = something R(dx)


Can you verify that the formulas you gave before are correct?


For instance, I suspect that:
M(dx) = 2^{\alpha}\delta\frac{\alpha}{ \Gamma(1-\alpha)}x^{-1-\alpha}e^{-(1/2)\gamma^{1/alpha}x}
should be:
M(dx) = 2^{\alpha}\delta\frac{\alpha}{ \Gamma(1-\alpha)}x^{-1-\alpha}e^{-(1/2)\gamma^{1/alpha}x} R(dx)

and for:
R(dx) = (\kappa_+r_+^{-p}|x|^{p-1}+\kappa_-r_-^{-p}|x|^{p-1})dx
I would need some context, from which I suspect it will become clear that R(dx) is what we already concluded.

 

[cernlife]

Here is the context, its a bit long, so please bear with me...

M is the Levy measure of the tempered stable distribution, which we can check as follows

We define a stable distribution as (in paramitization)

s_{\alpha,\delta} = s(y; \alpha, (\delta{2^{\alpha}}cos(\pi\alpha/2)^{1/\alpha}, 1, 0)

The characteristic function of a stable distribution

Y {\sim} s(y;\alpha, c, 1, 0)

is

E[e^{-iuY}] = exp(-\frac{c^\alpha}{cos(\pi\alpha/2)}iu^{\alpha})

A tempered stable distribution r.v, say X, is obtained by tilting a stable one, defined

X=d\cdot{e}^{-\theta{y}}\cdot{s(y;\alpha, c, 1, 0)}

To find the normalizing constant d, note that the characteristic function of the tempered stable random variable is

\Psi(u)=E[e^{iux}]=d\cdot\int_{\infty}^{\infty}e^{iux}e^{-(1/2)\gamma^{1/\alpha}x}s(y;\alpha, c, 1, 0)

=d\cdot{e}^{-\delta(\gamma^{1/\alpha}-2iu)^\alpha}

since

\Psi(0)=1

by the definition of characteristic functions, we must have 'd' as

d=e^{\delta\gamma}

so we have shown that the characteristic function of the tempered stable distribution is given by

\Psi(u)=E[e^{iux}]=e^{\delta\gamma-\delta(\gamma^{1/\alpha}-2iu)^{\alpha}}

Now the Levy-Khinchin representation of an infinity divisible law is defined here as

E[e^{iux}]=\int_0^{\infty}(e^{iux}-1)M(dx)

Where M(dx) is the Levy measure of the infinity divisible law. We can see that if the Levy measure is given by

M(dx) = (2^{\alpha}\delta\frac{\alpha}{ \Gamma(1-\alpha)}x^{-\alpha-1}e^{-(1/2)\gamma^{1/\alpha}x})dx

as this would give the required characteristic function.

So "I like Serena" going back to your last post this Levy measure M(dx) is defined as above and is correct. It is not defined as

M(dx) = (2^{\alpha}\delta\frac{\alpha}{ \Gamma(1-\alpha)}x^{-\alpha-1}e^{-(1/2)\gamma^{1/\alpha}x})R(dx)

In Rosinski's paper R(dx) is called a "spectral measure" and is not something I have seen before. This spectral measure is what I am so despartly trying to compute. I need like

R(dx) = (...equation...)dx

which I can use in Rosinski's theorem 2.3, which says; The Levy measure M of a tempered stable distribution can be written in the form.

M(A)=\int_{R^d}\int_0^{\infty} I_A(tx)t^{-\alpha-1}e^{-t}dtR(dx)

and this is my Big problem...

Another issue is that Rosinki is talking in the very general case and a tempered stable, i.e a A tempered stable distribution r.v, say X, is obtained by tilting a stable one, defined by

X=d\cdot{e}^{-\theta{y}}\cdot{s(y;\alpha, c, d, f)}

so he is talking about the general case i.e with tempering exponential

{e}^{-\theta{y}}

however in the construction I am following

\theta = (1/2)\gamma^{1/\alpha}

which makes me think that Rosinki's Theorem correct in the general sense, should be specified in my case as: The Levy measure M of a tempered stable distribution can be written in the form.

M(A)=\int_{R^d}\int_0^{\infty} I_A(tx)t^{-\alpha-1}e^{-(1/2)\gamma^{1/\alpha}t}dtR(dx)

moreover as I know the levy measure M of my specific tempered stable distribution I could write

(2^{\alpha}\delta\frac{\alpha}{ \Gamma(1-\alpha)}x^{-\alpha-1}e^{-(1/2)\gamma^{1/\alpha}x})dx=\int_{R^d}\int_0^{\infty} I_A(tx)t^{-\alpha-1}e^{-(1/2)\gamma^{1/\alpha}t}dtR(dx)

where R(dx) is what Rosinski calls the 'Spectral measure'.

Which comes back to my original question and problem, how can I work out what R(dx) is?

 

[micromass]

OK, I actually took a look at your paper.

The point is, why would you need R(dx)? To calculate integrals, yes? Well, the paper gives you the method how to calculate an integral:

\int_{\mathbb{R}^d}{F(x)R(dx)}=\int_{\mathbb{R}^d}{F\left(\frac{x}{\|x\|^2}\right)\|x\|^\alpha Q(dx)}

This is all you need to know really...

You won't find a nice formula for R(dx), since it doesn't look like R is absolutely continuous w.r.t. Q.

 

[cernlife]

Thank you for the further reply 'micromass'.

The reason why I need the expression for R(dx) is due to another part of my work which relates to the paper "On fractional tempered stable motion" by C.Houdre and R.Kawai. In this paper they define (definition 2.4)

Fractional tempered stable motion L_t^H, is given by

L_t^H=\int_0^t K_{H, \alpha}dX_s^{TS}

where

X_s^{TS}

is a real valued tempered stable Levy process, and

K_{H, \alpha}

is a Volterra Type kernal.

Now I really need to work out the distribution of

L_t^H

however, they note that the one-dimensional marginal distribution as Corollary 2.6, which gives the charateristic function for this fractional tempered stable motion, as

E[e^{iuL_t^H}]=exp[\int_0^t\int_{R\backslash{0}}\phi_{\alpha}(uxK_{H,\alpha}(t,s))R(dx)ds]

and so this is why I need R(dx) as I can't compute the distribution of

L_t^H

without knowing the expression for R(dx).

I would just like to say again a Big thanks for all your reply's.

 

[micromass]

E[e^{iuL_t^H}]=exp[\int_0^t\int_{R\backslash{0}}\phi_{\alpha}(uxK_{H,\alpha}(t,s))R(dx)ds]

and so this is why I need R(dx) as I can't compute the distribution of

L_t^H

without knowing the expression for R(dx).

As I thought, you' just need R(dx) to use it in an integral. Using the formula in my previous post, you can write your integral as

E[e^{iuL_t^H}]=exp[\int_0^t\int_{R\backslash{0}}\phi_{\alpha}(uxK_{H,\alpha}(t,s))R(dx)ds]=exp[\int_0^t\int_{R\backslash{0}}\phi_{\alpha}(u\frac{x}{\|x\|^2}K_{H,\alpha}(t,s))\|x\|^\alpha R(dx)ds]

 

[cernlife]

"I like Serena", in answer to your second comment

In the paper "A new tempered stable distribution and its applications to finance" by Rachev and Kim, they define a tempered stable distribution by tempering a symmetric stable distribution (defined on the whole real line), which is slightly different from my work as I am tempering a 'positively skewed to the right' stable distribution (defined only on the positive real line). In their work, Rachev and Kim define what they call a KR tempered distribution, and say as follows...

Coonsider a tempered stable distribution on the reals whose Levy measure M in polar coordinates is

M(dv,du)=v^{-\alpha-1}q(v,u)dv\sigma(du)

where

\sigma(A)=\frac{k_+r_+^{\alpha}}{\alpha+p_+} I_A(1) + \frac{k_-r_-^{\alpha}}{\alpha+p_-} I_A(-1)

and

q(v,1) = (\alpha+p_+)r_+^{-\alpha-p_+}\int_0^{r_+} e^{-v/s}s^{\alpha+p_+{-1}}ds

q(v,-1) = (\alpha+p_-)r_-^{-\alpha-p_-}\int_0^{r_-} e^{-v/s}s^{\alpha+p_-{-1}}ds

Then the spectral measure R corresponding to the Levy measure M can be deduced as

R(dx)=(k_+r_+^{-p_+} I_{(0,r_+)}(x)|x|^{p_+-1}+k_-r_-^{-p_-} I_{(-r_-,0)}(x)|x|^{p_-{-1}})dx

so here they are able to deduce R(dx), although I am unsure how they have done it and in any case it won't be the same as the R(dx) I am looking for as they are considering a slightly different tempered stable distribution.

 

[cernlife]

"micromass" thanks again, that sounds promising, so if my kernal was given by

K_{\alpha}(t,s)=\frac{1}{ \Gamma(\alpha)A}(t-s)^{\alpha-1}

and

\phi(s)=\frac{ \Gamma(1-\alpha)}{\alpha}((1/2)\gamma^{1/\alpha})^{\alpha}-((1/2)\gamma^{1/\alpha}-is)^{\alpha}

I guess I am trying to evaluate

E[e^{iuL_t}]=exp[(\frac{ \Gamma(1-\alpha)}{\alpha}((1/2)\gamma^{1/\alpha})^{\alpha}-((1/2)\gamma^{1/\alpha}-i(ux\frac{1}{ \Gamma(\alpha)A}(t-s)^{\alpha-1}))^{\alpha})||x||^{\alpha}R(dx)ds]

which I have absolutely no idea how to evaluate!

 

[I like Serena]

Yes, I see now.

R(dx) is an arbitrary measure, which is basically a function of location x and the infinitesimal dimensions dx.

So R(dx) could for instance be: R(dx) = (3x+y) dxdydz.

It seems that a lot of the article is about solving measures to match an alpha stable distribution.

Btw, in your initial post you have hyperlinks to 3 articles.
One does not work, one has a typo, but once corrected yields an article that looks *a lot* like the third article.
What's up with that?

The spectral density measure R(dx) is apparently constructed from a previous defined measure Q(dx) by inverting all the points in the unit sphere.
By inversion I mean that each point with a distance r to the origin is replaced by an one with distance 1/r.

Furthermore, this spectral density measure R(dx) is related to a Levy measure M in a sense that a transform is used, like in fourier, which explains why it's called "spectral".

That's as far as I got.
I haven't quite puzzled out how to deduce R(dx).

 

[I like Serena]

Hey cernlife!

Haven't seen you in a little while.
Did you puzzle it out?
Are you satisfied?