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If there are any links for free e-books that can I downloaded to study the autocorrelation function and its properties in details.

Thanks!

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- #1

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If there are any links for free e-books that can I downloaded to study the autocorrelation function and its properties in details.

Thanks!

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EnumaElish

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I meant another links that will provide any free download E-books if you know.

Thanks alot!

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EnumaElish

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http://www.google.com/search?hl=en&q=autocorrelation+finance&btnG=Search

www.springerlink.com/index/M106653187216806.pdf

www.ier.hit-u.ac.jp/Common/publication/DP/DP448.pdf

http://www.stat.tamu.edu/~ljin/Finance/stat689-R.htm [Broken]

www.springerlink.com/index/M106653187216806.pdf

www.ier.hit-u.ac.jp/Common/publication/DP/DP448.pdf

http://www.stat.tamu.edu/~ljin/Finance/stat689-R.htm [Broken]

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- #5

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I am going to find the autocorrelation function for the following equation

Hn(t)= (-1)^n * e^(t^2)* d^n/dt^n * e^(-t^2)

So, I didnt find the previous links can help me to solve this problem.

Hn(t)= (-1)^n * e^(t^2)* d^n/dt^n * e^(-t^2)

So, I didnt find the previous links can help me to solve this problem.

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- #6

EnumaElish

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You need to solve

Lim_{s-->0}∫_{-s}^{s} Hn(t)Hn(t+s)ds / (2s).

Note,

∫_{-s}^{s} Hn(t)Hn(t+s)ds = ∫_{t-z}^{z-t} Hn(t)Hn(z)dz

Lim

Note,

∫

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- #7

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Lims-->0∫-ss Hn(t)Hn(t+s)ds / (2s).

Note,

∫-ss Hn(t)Hn(t+s)ds = ∫t-zz-t Hn(t)Hn(z)dz

but I want to understand why you suggest this. I want to understad exactly by steps how can I find a solution.

Thanks alot.

- #8

EnumaElish

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As for step-by-step solution, I'll suggest to think about how the fundamental theorems of calculus might apply to this AC function. See http://mathworld.wolfram.com/FundamentalTheoremsofCalculus.html

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Thanks alot EnumaElish!

I will try to read these and I will reply you.

Thanks alot.

I will try to read these and I will reply you.

Thanks alot.

- #10

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I couldnt solve the equationYou need to solve

Lim_{s-->0}∫_{-s}^{s}Hn(t)Hn(t+s)ds / (2s).

Note,

∫_{-s}^{s}Hn(t)Hn(t+s)ds = ∫_{t-z}^{z-t}Hn(t)Hn(z)dz

∫

may be I should to follow such courses that explain Autocorrelation function or may be I must make reveiw for the integrals.

I want to advise me, please.

- #11

EnumaElish

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For starters,

∫Hn(t)Hn(z)dz = Hn(t)∫Hn(z)dz.

Then, insert Hn(z) = (-1)^n * e^(z^2)* d^n/dz^n * e^(-z^2). Now you have:

Hn(t)∫(-1)^n * e^(z^2)* d^n/dz^n * e^(-z^2)dz = Hn(t)(-1)^n ∫e^(z^2)* fn(z) dz

where fn(z) = d^n/dz^n * e^(-z^2).

You need to work on solving ∫e^(z^2)* fn(z) dz.

___________________

*CORRECTION:* Lim_{s-->0}∫_{-s}^{s} Hn(t)Hn(t+s)ds *should have been* Lim_{T-->0}∫_{-T}^{T} Hn(t)Hn(t+s)ds.

*Then, defining* z = t+s, *we have* ∫_{-T}^{T} Hn(t)Hn(t+s)ds = ∫_{-T}^{T} Hn(t)Hn(z)dz.

*This does not change the substance of the advice I have given until now.*

∫Hn(t)Hn(z)dz = Hn(t)∫Hn(z)dz.

Then, insert Hn(z) = (-1)^n * e^(z^2)* d^n/dz^n * e^(-z^2). Now you have:

Hn(t)∫(-1)^n * e^(z^2)* d^n/dz^n * e^(-z^2)dz = Hn(t)(-1)^n ∫e^(z^2)* fn(z) dz

where fn(z) = d^n/dz^n * e^(-z^2).

You need to work on solving ∫e^(z^2)* fn(z) dz.

___________________

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- #12

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For starters,

∫Hn(t)Hn(z)dz = Hn(t)∫Hn(z)dz.

Then, insert Hn(z) = (-1)^n * e^(z^2)* d^n/dz^n * e^(-z^2). Now you have:

Hn(t)∫(-1)^n * e^(z^2)* d^n/dz^n * e^(-z^2)dz = Hn(t)(-1)^n ∫e^(z^2)* fn(z) dz

where fn(z) = d^n/dz^n * e^(-z^2).

You need to work on solving ∫e^(z^2)* fn(z) dz.

___________________

CORRECTION:Lim_{s-->0}∫_{-s}^{s}Hn(t)Hn(t+s)dsshould have beenLim_{T-->0}∫_{-T}^{T}Hn(t)Hn(t+s)ds.

Then, definingz = t+s,we have∫_{-T}^{T}Hn(t)Hn(t+s)ds = ∫_{-T}^{T}Hn(t)Hn(z)dz.

This does not change the substance of the advice I have given until now.

let g(z)=∫e^(z^2)* fn(z) dz

for n = 1

fn(z)= d/dz e^-(z^2)= -2 z e-(z^2)

So, g(z) = ∫-2 z * e^(z^2)* e-(z^2) dz

= ∫-2 z *e^[(z^2)-(z^2)] dz

= ∫-2 z dz

=-z^2

is not that true?

- #13

EnumaElish

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Now you need to put in the integration limits (-T, T), multiply with Hn(t)(-1)^n, then take the limit T --> 0.

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Now you need to put in the integration limits (-T, T), multiply with Hn(t)(-1)^n, then take the limit T --> 0.[/QUOTE]

we have fn(z) = -Z^2 for n =1

by multiplying it by (-1)^n

we will have Z^2

by multiplying it by Hn(t),

where Hn(t)= (-1)^n * e^(t^2)* d^n/dt^n [e^-(t^2)]

Then, we will have:

Z^2 * (-1)^n * e^(t^2)* d^n/dt^n [e^-(t^2)], where z = t+s

So, how I will find the integral and I have two variables (t and Z)??

Can you help me, Please.

we have fn(z) = -Z^2 for n =1

by multiplying it by (-1)^n

we will have Z^2

by multiplying it by Hn(t),

where Hn(t)= (-1)^n * e^(t^2)* d^n/dt^n [e^-(t^2)]

Then, we will have:

Z^2 * (-1)^n * e^(t^2)* d^n/dt^n [e^-(t^2)], where z = t+s

So, how I will find the integral and I have two variables (t and Z)??

Can you help me, Please.

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- #15

EnumaElish

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Before you multiply the integral with Hn(t), you need to apply the integ. limits to -z^2. This will make z "disappear." The z^2 will be replaced with a quadratic function of T, say q(T).

Do you understand what q(T) is? Can you derive it explicitly?

- #16

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OK, I will have:

Before you multiply the integral with Hn(t), you need to apply the integ. limits to -z^2. This will make z "disappear." The z^2 will be replaced with a quadratic function of T, say q(T).

Do you understand what q(T) is? Can you derive it explicitly?

-z^2[tex]\left|^{T}_{-T}[/tex]

then I will get

-(T)^2 - [- (-T)^2]= 0

So, the autocorelation for

Hn(t)Hn(t+s)ds = 0

is that true!!!

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- #17

EnumaElish

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It seems to be true for n=1.

Can you solve the general case for arbitrary n?

Can you solve the general case for arbitrary n?

- #18

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Actually, this is exactly what I am going to ask you about.It seems to be true for n=1.

Can you solve the general case for arbitrary n?

So, can you just help me a little bit?

And I have another question:

What about the autocorrelation of

Hn(t)Hm(t)dt , where m not equal to n

- #19

EnumaElish

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I cannot see the LaTex equation for m. There seems to be a typesetting error.

EDIT: I can see it now.

For the Hn(t)Hn(t+s) case, you need to work on solving ∫e^(z^2)* fn(z) dz for arbitrary n.

The Hn(t)Hm(t)dt case is not a regular AC function. Go back to the MathWorld link (http://mathworld.wolfram.com/Autocorrelation.html) and review the definition. The AC function has a single argument (a single function), for example, ac(*t*) = AC(Hn(*t*)) or equivalently *ac* = AC(*Hn*). What you are proposing has two arguments, *Hn* and *Hm*. So it doesn't fit the definition of AC.

EDIT: I can see it now.

For the Hn(t)Hn(t+s) case, you need to work on solving ∫e^(z^2)* fn(z) dz for arbitrary n.

The Hn(t)Hm(t)dt case is not a regular AC function. Go back to the MathWorld link (http://mathworld.wolfram.com/Autocorrelation.html) and review the definition. The AC function has a single argument (a single function), for example, ac(

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- #20

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I meant that what about the autocorrelation function between the first derivative of H(t) and the second derivative of H(t)

Such as :

the autocorrelation function of H1(t)H2(t).

Thanks alot!

- #21

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I cannot see the LaTex equation for m. There seems to be a typesetting error.

EDIT: I can see it now.

For the Hn(t)Hn(t+s) case, you need to work on solving ∫e^(z^2)* fn(z) dz for arbitrary n.

The Hn(t)Hm(t)dt case is not a regular AC function. Go back to the MathWorld link (http://mathworld.wolfram.com/Autocorrelation.html) and review the definition. The AC function has a single argument (a single function), for example, ac(t) = AC(Hn(t)) or equivalentlyac= AC(Hn). What you are proposing has two arguments,HnandHm. So it doesn't fit the definition of AC.

for n=1

∫e^(z^2)* fn(z) dz = -z^2

for n=2

∫e^(z^2)* fn(z) dz = [tex]\frac{3}{4}[/tex] z^3

So, should I continue for n=3,4,5,...

- #22

EnumaElish

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The AC function has a very specific definition; and it is defined for a single parent function.

I meant that what about the autocorrelation function between the first derivative of H(t) and the second derivative of H(t)

Such as :

the autocorrelation function of H1(t)H2(t).

Thanks alot!

I am going to look at your question that way. You can write the AC function of H1*H2 by following the general definition of AC for function F:

AC

and making the substitution F = H1*H2.

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- #23

EnumaElish

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You can continue and hope that you will "see" the general solution eventually by induction and guesswork. But the more direct approach is to try to solve the general case algebraically.for n=1

∫e^(z^2)* fn(z) dz = -z^2

for n=2

∫e^(z^2)* fn(z) dz = [tex]\frac{3}{4}[/tex] z^3

So, should I continue for n=3,4,5,...

- #24

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The correlation is equal to the average of the product of two random variables and is defined as

cor(X,Y) = E[XY]

= ∫∞ −∞∫∞ −∞ xyf(x,y)dxdy

So, if the above equation seems to be correct, then I can define H1(t) as X and H2(t) as Y.

Can I work like this depending on the above?

And for a given set of variables, we use the correlation coefficient to give us the linear relationship between our variables. The correlation coefficient of two variables is defined in terms of their covariance and standard deviations, as seen below

ρ= cor(X,Y)/σχ σΥ

If there is no relationship between the variables then the correlation coefficient will be zero and if there is a perfect positive match it will be one.

If there is a perfect inverse relationship, where one set of variables increases while the other decreases, then the correlation coefficient will be negative one.

So, my second equastion is how to find the mean and variance for H1(t) and H2(t)

Hint: if Hn(t)= (-1)^n cos(2π fc t)* e^[(t^2)/4] *d^n/dt^n *e^[(t^2)/4]

Thanks alot!!

cor(X,Y) = E[XY]

= ∫∞ −∞∫∞ −∞ xyf(x,y)dxdy

So, if the above equation seems to be correct, then I can define H1(t) as X and H2(t) as Y.

Can I work like this depending on the above?

And for a given set of variables, we use the correlation coefficient to give us the linear relationship between our variables. The correlation coefficient of two variables is defined in terms of their covariance and standard deviations, as seen below

ρ= cor(X,Y)/σχ σΥ

If there is no relationship between the variables then the correlation coefficient will be zero and if there is a perfect positive match it will be one.

If there is a perfect inverse relationship, where one set of variables increases while the other decreases, then the correlation coefficient will be negative one.

So, my second equastion is how to find the mean and variance for H1(t) and H2(t)

Hint: if Hn(t)= (-1)^n cos(2π fc t)* e^[(t^2)/4] *d^n/dt^n *e^[(t^2)/4]

Thanks alot!!

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- #25

EnumaElish

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This is the definition of covariance, not correlation.The correlation is equal to the average of the product of two random variables and is defined as

cor(X,Y) = E[XY]

= ∫∞ −∞∫∞ −∞ xyf(x,y)dxdy

To obtain Cov(X,Y), yes.So, if the above equation seems to be correct, then I can define H1(t) as X and H2(t) as Y.

Can I work like this depending on the above?

The correct formula is ρ = Cov(X,Y)/(σχ σΥ) where ρ is the correlation coefficient between X and Y.And for a given set of variables, we use the correlation coefficient to give us the linear relationship between our variables. The correlation coefficient of two variables is defined in terms of their covariance and standard deviations, as seen below

ρ= cor(X,Y)/σχ σΥ

Correct.If there is no relationship between the variables then the correlation coefficient will be zero and if there is a perfect positive match it will be one.

If there is a perfect inverse relationship, where one set of variables increases while the other decreases, then the correlation coefficient will be negative one.

You should follow the approach in this post: https://www.physicsforums.com/showthread.php?p=1387935#post1387935So, my second equastion is how to find the mean and variance for H1(t) and H2(t)

Hint: if Hn(t)= (-1)^n cos(2π fc t)* e^[(t^2)/4] *d^n/dt^n *e^[(t^2)/4]

Thanks alot!!

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