# Relationship between average and characteristic function of a Gaussian process

I'm having trouble understanding an equality given in a book ("Speckle Phenomena in Optics" by Joseph Goodman p.145) for a zero mean, stationary Gaussian process: $\overline{\exp(i [\phi(x_1)-\phi(x_2)])}=\exp(-\sigma^2[1-\mu(x_1-x_2)])$ where $\sigma^2$ is the variance of the process $\phi$ and $\mu$ is the normalized autocorrelation of the process. Could anyone derive this equality for me or give me a reference that explains it? The author says: "from the relationship of the average to characteristic functions we have ..." Thanks!

• Please add the reference and the page number or give a short description of the notation used. Jan 4, 2014 at 13:59
• That's part of the problem - he does not define the average. Maybe it's still possible to solve. Jan 4, 2014 at 14:04
• @Philipp The author is using average to refer to the fact that the characteristic function of a random variable $X$ can be expressed as the expectation of $\exp \left( i X \right)$, i.e. mathematically, $\varphi \left( X \right) = \int_{\Omega_X} \exp \left( i X \right) dF_x$ You also did not answer fg nu's question. Jan 4, 2014 at 14:39

The expression of the characteristic function as stated is a proof that the sum of two autocorrelated 0 mean Gaussian process realizations make a 0 mean Gaussian random variable with variance $2\sigma^2(1 - \mu(x_1 - x_2))$ for that particular autocorrelation process.

It may help to recall that the MGFs (and consequently characteristic functions) of the sum of two independent random variables is just the product of their respective MGFs (CFs). When two Gaussian realizations from the process as described have an autocorrelation of 0, then we would see:

$\int_{\Omega_{x_1, x_2}} \exp \left( i t\left( \phi(x_1) - \phi(x_2) \right) \right) dF_{x_1, x_2}= \exp \left( - \sigma^2 t^2 \right)$

For the sum of dependent realizations, you will have to go through the steps of expressing and integrating the joint densities. In particular, it helps to assume that $\mu(\phi(x_1), \phi(x_2))$ is fixed. Additionally, you may want to review the proof of normality of sums of random normal variables using convolution since the double integral allows you to express $\phi(x_1), \phi(x_2)$'s joint density as a product of marginal and conditional densities respectively.

• Hi AdamO, I don't see where the t^2 is coming from in your formula, it doesn't appear on the LHS of the equation. Without it, it's a special case of the formula above, but still I'm lacking the way to get there. Jan 4, 2014 at 15:06
• If you're not familiar with characteristic functions and their applications then I recommend consulting one of the graduate level probability theory texts like Casella Burger or Hogg and Craig. Jan 4, 2014 at 15:12
• Ok after reading a bit more I am still stuck. Especially I dont get how we get from an expectation value around the exp (on the left side) to an expectation value in the exponent (on the right side), which is in the definition of the autocorrelation. Some more help in computing this would be very helpful. Jan 13, 2014 at 13:49
• There are only two things you need to know. First off, what's the characteristic function of a random variable? A random variable is a probabilistic function defined over a sample space, it's characteristic function is a frequency decomposition of that function (xref any real analysis text book) and the characteristic function uniquely determines any measurable RV. For a normal RV, the characteristic function is $\exp \left( it\mu - \sigma^2 t^2 / 2 \right)$ ($t$ is defined over the frequency domain, e.g. all reals). (ctd) Jan 13, 2014 at 16:33
• (ctd) second, is that the distribution of the difference of two correlated normal RVs ($X_1, X_2$ ) with means $\mu_1, \mu_2$, variance $\sigma^2_1, \sigma_2^2$ and covariance $\mbox{Cov}(X_1, X_2)$, is a normal RV with mean $\mu_1 - \mu_2$ and variance $\sigma^2_1 + \sigma^2_2 - 2 \mbox{Cov}(X_1, X_2)$ Jan 13, 2014 at 16:35