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Suppose I have a gaussian process which takes 2D inputs x and y and gives a 1D output z. I understand based on Calculating the expression for the derivative of a Gaussian process that each of the partial derivatives will be normally distributed. But are the partial derivatives statistically independent? How could that be proved? Would that be generalizable to a non-gaussian stochastic process?

I tried solving it computationally assuming a kernel that depends only on the total distance $\Delta x^2 + \Delta y^2 $ and they indeed seem to be uncorrelated which is equivalent to being independent for the case of gaussian processes. To perform the computations I generated a realization of the process on a 100x100 mesh and computed the derivatives in each direction. Afterwards I plotted a 2D histogram where you can see that the distribution is not 'tilted'. To check further, I computed the covariance of z_x and z_y vectors by multiplying the z_x and z_y slopes at each point (Mean of z_x and mean of z_y are 0) and plotted a histogram which is symmetric around 0 with mean 0.

MymanPJ
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  • The derivatives are not just "normally distributed". They are again Gaussian processes! The dependence of the derivatives follows from the correlation function of the Gaussian Process. The linked answers provide the formulas, which show that they are not independent in general. The non-Gaussian case is much harder, starting with the definition of differentiability. You probably don't want and don't need to go there. – g g Feb 07 '24 at 11:11
  • Maybe I am missing something, I have to admit the generalization for multiple dimensions is not easy to me. But are not the formulas in the linked answer specific to the 1 dimensional case? Will I not have another covariance matrix for the other dimension (let's call it y)? How do I establish uncorrelation based on two different covariance matrices? – MymanPJ Feb 07 '24 at 15:38

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