Co-Variance and Signum Function

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SUMMARY

The discussion centers on calculating the covariance between a random variable X and its transformation Y via the signum function, where Y equals 1 for non-negative values of X and -1 for negative values. The covariance is determined using the formula Cov(X, Y) = E(XY) - E(X)E(Y), with specific integrals for E(XY) based on the probability distribution p(x). When X follows a Gaussian distribution with a mean of zero, the covariance simplifies to the difference between the mean values of the positive and negative halves of p(x). MATLAB was utilized to validate the calculations, yielding a covariance of approximately 0.7981.

PREREQUISITES
  • Understanding of covariance and expectation in probability theory
  • Familiarity with the signum function and its properties
  • Knowledge of Gaussian distributions and their characteristics
  • Proficiency in MATLAB for statistical computations
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  • Explore advanced covariance calculations for non-linear transformations
  • Learn about the implications of different probability distributions on covariance
  • Investigate the use of MATLAB for statistical analysis and simulations
  • Study the properties of the signum function in various mathematical contexts
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Microelectronics engineers, statisticians, and researchers in probability theory who are interested in the relationship between random variables and their transformations, particularly in the context of Gaussian distributions.

Hanspi
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I have a problem that turned up in my research. I'm a microelectronics engineer, s I hope this is not a textbook question for you physics specialists :-)

Given a random variable X that produces real numbers x with a distribution p(x).

The random variable Y is generated from X by the signum function; i.e., y=1 for x>=0 and y=-1 for x<0.

How can I calculate the covariance of X and Y in general? And, if there is no general solution, does a solution exist if p(x) is a gaussian distribution with mean zero?

Slainte!
Hanspeter
 
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If we use E(X) to denote the expected value of X, you need to compute
E(XY) - E(X)E(Y). The expectations would be computed as integrals. For example, to compute E(XY)

[tex]E(XY) = \int_0^\infty x(1) p(x) dx + \int_{-\infty}^0 x(-1) p(x) dx[/tex]
[tex]= \int_0^\infty x p(x)dx - \int_{-\infty}^0 x p(x) dx[/tex]
 
Thanks a lot for the reply, this is indeed what I needed. If E(X)=0, then E(Y)=0 and the Covariance is simply the mean value of the positive half of p(x) minus the mean value of the negative half of p(x).

To test my understanding, I did (MATLAB):

c=randn(1,1000000);
d=sign(c);
cov(c,d)

ans =

1.0000 0.7981
0.7981 1.0000

mean(c.*(d>=0))-mean(c.*(d<0))

ans =

0.7981

Now comes the mathematically more difficult part: to calculate p(x) in my system depending on the parameters of the system. But hat I know how to to, so thanks a lot for giving me such a good start.

Hanspeter
 

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