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This problem concerns fitting a curve to a set of data under a number of constraints.

Let $f: \mathbb{R}^+ \to \mathbb{R}$ be a strictly convex function with $f(0) = 0$. Suppose that, as data, one has $N$ triplets of the form $(x_i, y_i, y_i^\prime)$ where $y_i = f(x_i)$ and $y_i^\prime = \frac{d}{dx_i} f(x_i)$, i.e., for a few points we observe the function values and the derivatives. This observation is noisy, but the observation error is really small.

My objective is to "predict" the values of $f$ at a fine grid of values contained in the interval $[\min(x), \max(x)]$.

Things I've tried: (i) fit a generalised additive model (GAM) to the data $\boldsymbol z = (\boldsymbol x, \boldsymbol y)$ and ignore the derivative information; (ii) fit a convex Gaussian process by trying to enforce positivity of the second derivative (which is also a GP).

This second approach could be better done as I was not able to incorporate the derivative data into the modelling process.

What I am looking for is some hint on how to incorporate all of the information I have in order to obtain the best possible interpolator for my goal. Notice I don't really care about overfitting or other similar important concerns when doing modelling in general.

Further details

Here are two plots of the data, differing only in how zoomed in/out they are enter image description here

enter image description here

I'm happy to provide more details and edit my question accordingly.

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    $\begingroup$ One approach is to spline the data, as described at stats.stackexchange.com/questions/428558 $\endgroup$
    – whuber
    Feb 28, 2020 at 15:16
  • $\begingroup$ Thanks for the suggestion, @whuber. I copied their code over and ran on my own data, but things didn't look so rosy as the resulting interpolation did not look any better than GAM [in fact it was much worse]. To be clear, GAM works nicely, but I need even more precision, so I'm looking for something that can allow me to use all of the information available. Let me know if I can provide any more useful information. $\endgroup$ Feb 28, 2020 at 17:06
  • $\begingroup$ The approach in my answer at that thread should work very well, because your data are more extensive and far smoother than the data I used to demonstrate its efficacy. I haven't tested the Kalman filter solution posted there but I like it because it ought to apply directly to your data (which have no repeated measurements). $\endgroup$
    – whuber
    Feb 28, 2020 at 17:20
  • $\begingroup$ It's not clear to me how to produce the plots you show in your post in that thread. $\endgroup$ Feb 28, 2020 at 17:51

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