If I have two signals that are corrupted with Gaussian random noise and I multiply them, what is the noise distribution called? x1 = s+n1 x2 = s+n2 x1*x2 = s*s+(n1+n2)*s+n1*n2 Is noise times noise (n1*n2) the focus of any study? I would be interested to read about it. Thanks.

# Distrubution of noise times noise

Started by ●November 30, 2006

Reply by ●November 30, 20062006-11-30

Hi, On 30 nov, 05:40, "spasmous" <spasm...@gmail.com> wrote:> If I have two signals that are corrupted with Gaussian random noise and > I multiply them, what is the noise distribution called? > > x1 = s+n1 > x2 = s+n2 > x1*x2 = s*s+(n1+n2)*s+n1*n2 > > Is noise times noise (n1*n2) the focus of any study? I would be > interested to read about it.As n1 and n2 are independent random gaussian noise, their product 's distribution is a Bessel distribution. I only found an old french article about it : http://archive.numdam.org/ARCHIVE/RSA/RSA_1968__16_4/RSA_1968__16_4_65_0/RSA_1968__16_4_65_0.pdf> > Thanks.Hope this helps you, Matthieu

Reply by ●November 30, 20062006-11-30

Matthieu Puigt skrev:> Hi, > > On 30 nov, 05:40, "spasmous" <spasm...@gmail.com> wrote: > > If I have two signals that are corrupted with Gaussian random noise and > > I multiply them, what is the noise distribution called? > > > > x1 = s+n1 > > x2 = s+n2 > > x1*x2 = s*s+(n1+n2)*s+n1*n2 > > > > Is noise times noise (n1*n2) the focus of any study? I would be > > interested to read about it.I don't know. n1*n1 has been studied, though. Look for the Rayleigh distribution.> As n1 and n2 are independent random gaussian noise, their product 's > distribution is a Bessel distribution.Is that the same as the Rayleigh distribution? I know French nomenclature can differ from English nomenclature. What is termed "Snell's law" in English books is alledgedly termed "Descarte's law" in the French literaure. Rune

Reply by ●November 30, 20062006-11-30

> > As n1 and n2 are independent random gaussian noise, their product 's > > distribution is a Bessel distribution. > > Is that the same as the Rayleigh distribution? I know French > nomenclature can differ from English nomenclature. What is > termed "Snell's law" in English books is alledgedly termed > "Descarte's law" in the French literaure.They are different distributions. If n1 and n2 are independent and Gaussian, then their product's PDF is: f(z) = (1/(pi*sigma1*sigma2)) * Ko(abs(z)/(sigma1*sigma2)), where Ko(z) is a modified Bessel function of the second kind. Pretty messy. And, n1^2 does not have a Rayleigh distribution; a Rayleigh PDF arises if you have two independent Gaussian random variables n1 and n2, and form a new variable n3 = sqrt(n1^2+n2^2). In this case, n3 will have a Rayleigh distribution. The n1^2 case can be thought of as giving a random variable with a chi-square distribution with one degree of freedom, or a Rayleigh random variable. For the one-degree-of-freedom case, they only differ by a factor of a sqrt(2). Some links: http://en.wikipedia.org/wiki/Gaussian_distribution (where the above function is specified, under "Properties") http://en.wikipedia.org/wiki/Rayleigh_distribution Jason

Reply by ●November 30, 20062006-11-30

cincydsp@gmail.com wrote:> ...If n1 and n2 are independent and > Gaussian, then their product's PDF is: > > f(z) = (1/(pi*sigma1*sigma2)) * Ko(abs(z)/(sigma1*sigma2)), where Ko(z) > is a modified Bessel function of the second kind. Pretty messy. >Ouch. That's not nice. But thank you for sharing your knowledge. Do you know the "rules" for combining two signals with such a distribution? Eg. would summing n1*n2 + n3*n4 (where all are Gaussian, independent noise with the same variance) have the same distrubution as n1*n2 but reduced... variance? I want to know the best way to combine multiple signals that have this distribution.

Reply by ●November 30, 20062006-11-30

> Do you know the "rules" for combining two signals with such a > distribution? Eg. would summing n1*n2 + n3*n4 (where all are Gaussian, > independent noise with the same variance) have the same distrubution as > n1*n2 but reduced... variance? I want to know the best way to combine > multiple signals that have this distribution.That would really come down to the properties of the Bessel distribution that results, as n1*n2 + n3*n4 would be the sum of two such random variables. I don't have a good reference for that particular distribution, so I'm not sure. Jason

Reply by ●November 30, 20062006-11-30

spasmous, What you have written in the equations is the same signal corrupted by two different gaussian noise signals and multiplied, which is different than what is asked in the question ("..two signals that are corrupted .."). Then you ask about what the noise distibution of the product of the noisy signals, which is not the (n1*n2) that you subsequently ask about. The noise distribution of the product of the noisy signals has a 's' signal dependent term not included in (n1*n2). So what is the real question? Dirk spasmous wrote:> If I have two signals that are corrupted with Gaussian random noise and > I multiply them, what is the noise distribution called? > > x1 = s+n1 > x2 = s+n2 > x1*x2 = s*s+(n1+n2)*s+n1*n2 > > Is noise times noise (n1*n2) the focus of any study? I would be > interested to read about it. > > Thanks.

Reply by ●November 30, 20062006-11-30

dbell wrote:> spasmous, > > What you have written in the equations is the same signal corrupted by > two different gaussian noise signals and multiplied, which is different > than what is asked in the question ("..two signals that are corrupted > .."). > > Then you ask about what the noise distibution of the product of the > noisy signals, which is not the (n1*n2) that you subsequently ask > about. The noise distribution of the product of the noisy signals has a > 's' signal dependent term not included in (n1*n2). > > So what is the real question?The real question is "Will you do my homework for me". VLV

Reply by ●December 1, 20062006-12-01

spasmous wrote:> Ouch. That's not nice. But thank you for sharing your knowledge. > > Do you know the "rules" for combining two signals with such a > distribution? Eg. would summing n1*n2 + n3*n4 (where all are Gaussian, > independent noise with the same variance) have the same distrubution as > n1*n2 but reduced... variance? I want to know the best way to combine > multiple signals that have this distribution.I'm not following what you want to get out of this. Are you trying to estimate s? Are you trying to estimate some property of the noise? What are your observations, and what is it that you want? I have a feeling that it will be a lot simpler if you rethink your problem statement. Julius

Reply by ●December 2, 20062006-12-02

dbell wrote:> spasmous, > > What you have written in the equations is the same signal corrupted by > two different gaussian noise signals and multiplied, which is different > than what is asked in the question ("..two signals that are corrupted > .."). > > Then you ask about what the noise distibution of the product of the > noisy signals, which is not the (n1*n2) that you subsequently ask > about. The noise distribution of the product of the noisy signals has a > 's' signal dependent term not included in (n1*n2). > > So what is the real question? > > Dirk >Alright - I didn't want to be accused of asking you to do my "homework" ;) So I just asked a little snippet. Here is the full problem: x1 = A*s + n1 x2 = B*s + n2 x3 = C*s + n3 etc. y1 = A*t + n4 y2 = B*t + n5 y3 = C*t + n6 etc. A,B,C are known and n is Gaussian, independent noise. I want to find the best estimate of s*t. One way is to solve for s and t then form s*t. This is what I'm currently doing. Another way is to form products first, like x1*y1, x1*y2, etc, and add them in some way. I was trying to think what the best way to do this might be. Perhaps it is the minimizer of the least squares difference between s*t and the noise corrupted estimate - I can't figure out what this is. The use of least squares might not be appropriate for this noise distribution either. I'm just looking for hints - hopefully someone is interested in this too or has done it before. Your comments much appreciated.