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I know based on the answers to this question Derivative of a Gaussian Process that the derivative of a Gaussian process is another Gaussian process, but I was wondering if someone could tell (or show) me how to explicitly calculate the expression for the derivative of a Gaussian process. For example, assuming that I have the following

$$f\sim GP(X\beta,\sigma^2R)$$ where $R$ is the Gaussian correlation function

$$R=\exp\left\{-\sum_{i=1}^n\frac{|x_{ij}-x_{ik}|^2}{\phi_i}\right\}$$

What would the distribution of $f'$ be (i.e., the derivative of $f$)?

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  • $\begingroup$ @DilipSarwate Fine don't think of it as a GP. I have a random variable $f$ that follows a Normal distribution with mean $X\beta$ and covariance $\sigma^2 R$ and I want to take the derivative of that random variable. What is the solution of taking that derivative? $\endgroup$ – RustyStatistician Dec 23 '15 at 1:50
  • $\begingroup$ @DilipSarwate look at section 9.4 gaussianprocess.org/gpml/chapters/RW9.pdf $\endgroup$ – RustyStatistician Dec 23 '15 at 1:53
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For simplicity I assume x has one dimension.

$\frac{\partial f}{\partial x}$ is normally distributed with expectation:

$E[\frac{\partial f}{\partial x}] = \frac{\partial}{\partial x}E[f]$

And Covariance:

$\text{Cov}(\frac{\partial f_1}{\partial x_1},\frac{\partial f_2}{\partial x_2})$ = $\frac{\partial^2 }{\partial x_2\partial x_1}\text{Cov}(f_1,f_2))$

In case you are using a gaussian correlation function: $\text{Cov}(f_1,f_2)) = \sigma^2\exp(-\frac{1}{2}\frac{(x_1-x_2)^2}{a^2})$, then:

$\text{Cov}(\frac{\partial f_1}{\partial x_1},\frac{\partial f_2}{\partial x_2})$ = $\frac{\sigma^2}{a^2}(1.0-\frac{(x_1-x_2)^2}{a^2})\exp(-\frac{1}{2}\frac{(x_1-x_2)^2}{a^2})$.

If x has more than one dimensions, each of the partial derivatives are normally distributed, and the covariance and expectation for each dimension can be calculated in the same way as above.

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Since your domain for $f$ is $n$-dimensional you will actually have $n$ derivative processes $\frac{\partial f}{\partial x_j}$ with $j=1,\ldots n$. You need to calculate the mean and correlation function of $\frac{\partial f}{\partial x_j}$. From the linked answer you know that those are the derivative of $f$'s mean function and the derivative with respect to both arguments of the correlation $R$. So there is nothing more to do than to calculate those derivatives:

I assume that $\beta$ are constant, hence the mean function is $\frac{\partial X\beta}{\partial x_j}=\beta_j$.

To keep confusion to a minimum let's write $R$ as $R(x,y)=exp\{-\sum_i\frac{(x_i - y_i)^2}{\phi_i}\}$. Then the derivative with respect to the first argument (assuming $\sigma$ is constant) is $$ \frac{\partial R}{\partial x_j}= \left(- 2 \frac{x_j-y_j}{\phi_j}\right) R(x,y)$$ and with respect to both $$ \frac{\partial }{\partial y_j}\frac{\partial R}{\partial x_j}=\frac{\partial }{\partial y_j}\left( R(x,y)\left(- 2 \frac{x_j-y_j}{\phi_j}\right) \right)=\frac{2}{\phi_j}R(x,y)\left( 1 - \frac{2}{\phi_j}(x_j - y_j)^2\right).$$ Which means in your notation that $$ \frac{\partial f}{\partial x_j} \sim \text{GP}\left( \beta_j, \sigma^2 \frac{2}{\phi_j} R(x,y)\left( 1 - \frac{2}{\phi_j}(x_j - y_j)^2\right)\right). $$

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You can use sympy in Python, it will calculate any derivatives including integral defined one.

diffn(ff,x0,kk) :     
 dffk= Derivative(ff(x),x,kk)
 dffk1= simplify( dffk.doit())
 dffx0=  simplify(Subs(dffk1, (x), (x0)).doit())
 return dffx0
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  • $\begingroup$ I dont understand why there is a downvote since this is very useful code (I spent 4 hours to do it properly). $\endgroup$ – quantCode Jan 8 '16 at 7:37
  • $\begingroup$ The code is naturally useful, yet it totally misses the point. $\endgroup$ – mpiktas Apr 17 '16 at 8:35

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