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Integral transforms of random processes

  1. Nov 29, 2011 #1

    I considered a statistically independent continuous random process f(x) such that Cov(f(x),f(y))=0 for x[itex]\neq[/itex]y and Cov(f(x),f(x))=σ2.
    Then I would like to compute the correlation function of the Fourier transform of f, that is [itex]Cov\left( F(u),F(v)\right)[/itex].

    The result I got from my calculations is that [itex]Cov\left( F(u),F(v)\right)=0[/itex] when u[itex]\neq[/itex]v, and [itex]Cov\left( F(u),F(u)\right)=\sigma^2[/itex].
    So again, also the Fourier transform of f is supposed to be statistically independent.

    At first I thought that this makes sense, but then I started to wonder why F(u) and F(-u) are supposed to be uncorrelated? We know that F(u)=F(-u)*, so the negative frequencies theoretically depend fully on the positive frequencies (they are actually redundant as one is the complex conjugate of the other).

    How should I interpret this result? Is it wrong?
  2. jcsd
  3. Nov 30, 2011 #2

    Stephen Tashi

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    Science Advisor

    I don't know if you calculations are wrong, but, as a generality, two random variables can be completely dependent and still have zero correlation. Roughly speaking, correlation has to do with the ability to use one to do a linear prediction of the other. The implication that does hold is: independence implies zero correlation, not the converse.

    Instead of two random variables, you are dealing with stochastic processes, which are indexed collections of random variables. So we would have to examine the definition of covariance in that context to see if the above fact explains your result.
  4. Nov 30, 2011 #3
    Hi Stephen!

    thanks for your help.
    I checked my calculations, and I can't seem to spot any mistake.
    I believe that in my case the problem can be reduced in interpreting the following (simpler) question:

    - given a complex random variable x, what is [itex]Cov\left( x , x^{\ast} \right)[/itex] ? where x* denotes the complex conjugate of x.

    Unless I haven't done any mistake, it turns out that if both real and imaginary parts of x have the same variance, then Cov(x,x*) = 0. If the variances of the real and imaginary part are different then the covariance is not zero.

    I think I must interpret somehow this statement to answer the original question.
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