Higher-Order Time Correlation Functions of White Noise?

Click For Summary
SUMMARY

The discussion centers on calculating the fourth-order correlation function of Gaussian white noise, specifically represented as . The initial assumption of using 3s4 to account for higher moments was corrected, emphasizing that only Dirac deltas are necessary for the autocorrelation function, while the Kronecker delta is not required. The correct formulation utilizes a standard result from probability theory for zero-mean jointly-Gaussian random variables, which can be derived using characteristic functions.

PREREQUISITES
  • Understanding of Gaussian white noise and its properties
  • Familiarity with Dirac delta and Kronecker delta functions
  • Knowledge of fourth-order correlation functions in statistics
  • Basic principles of probability theory related to jointly-Gaussian variables
NEXT STEPS
  • Study the derivation of correlation functions using characteristic functions
  • Explore the properties of Gaussian distributions in higher-order moments
  • Learn about the application of Dirac and Kronecker delta functions in statistical mechanics
  • Investigate the implications of uncorrelated forces in multi-particle systems
USEFUL FOR

Researchers in statistical mechanics, physicists working with stochastic processes, and anyone studying the properties of Gaussian random variables and their correlations.

Opus_723
Messages
175
Reaction score
3
Suppose I have Gaussian white noise, with the usual dirac-delta autocorrelation function,

<F1(t1)F2(t2)> = s2*d(t1-t2)*D12

Where s is the standard deviation of the Gaussian, little d is the delta function, and big D is the kronecker delta. For concreteness and to keep track of units, say F represents a force.

What is the 4th-order correlation of this same white noise?

<F(t1)F(t2)F(t3)F(t4)> = ?

My first guess, would be to simply let s2 become 3s4 to capture the higher moment in the Gaussian distribution, and add a couple of dirac-deltas and kroneckers to make sure it's only nonzero when all t1,t2,t3,t4 are the same:

<F(t1)F(t2)F(t3)F(t4)> = 3s4*d(t1-t2)d(t2-t3)d(t3-t4)*D12*D23*D34

But this is, of course, wrong. We would need three dirac deltas to ensure that the result is nonzero only when all times are the same, but this introduces too many units of 1/time. It no longer makes any sense as a 4th-order correlation of forces. Somehow this must be accomplished with a different structure.

So what is the right way to do this? Do we need to specify additional properties of the white noise in order to calculate this?
 
Physics news on Phys.org
First, the Kronecker delta is not needed for your autocorrelation function. The Dirac delta is all that you need.

Anyway, since your white noise is Gaussian, I would use a standard result from probability theory that if you have four zero-mean jointly-Gaussian random variables ##x_1, x_2, x_3, x_4##, then
$$ \left\langle x_1 x_2 x_3 x_4 \right\rangle = \left\langle x_1 x_2\right\rangle \left\langle x_3 x_4\right\rangle + \left\langle x_1 x_3\right\rangle \left\langle x_2 x_4\right\rangle + \left\langle x_1 x_4\right\rangle \left\langle x_2 x_3\right\rangle $$

This can be derived using characteristic functions - it isn't a difficult derivation but it is messy.

Jason
 
  • Like
Likes   Reactions: Opus_723
jasonRF said:
First, the Kronecker delta is not needed for your autocorrelation function. The Dirac delta is all that you need.

Anyway, since your white noise is Gaussian, I would use a standard result from probability theory that if you have four zero-mean jointly-Gaussian random variables ##x_1, x_2, x_3, x_4##, then
$$ \left\langle x_1 x_2 x_3 x_4 \right\rangle = \left\langle x_1 x_2\right\rangle \left\langle x_3 x_4\right\rangle + \left\langle x_1 x_3\right\rangle \left\langle x_2 x_4\right\rangle + \left\langle x_1 x_4\right\rangle \left\langle x_2 x_3\right\rangle $$

This can be derived using characteristic functions - it isn't a difficult derivation but it is messy.

Jason

Thank you. This is exactly what I needed, and in retrospect makes a lot of sense. I will note that the Kronecker delta *is* necessary in my particular problem as I have uncorrelated forces acting on different particles, but I probably didn't make that very clear.

Thanks again!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 1 ·
Replies
1
Views
6K
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K