I have a set of noisy data that I am fitting using Gaussian process regression (GPR) with Python's sklearn package using the treatment found here. Below is an example where the error bars on the observational data is simply 20% of the actual value, and the kernel used is just a squared exponential (i.e. radial basis function):

enter image description here

It's a fairly straightforward approach, and the computed prediction appears to fit the data well. But the computed error bars (which are plotted at $1.96\sigma$) are nearly invisible on the plot. Here is the output when I instead plot error bars of $50\sigma$:

enter image description here

The mean fit looks all right, but results for error bars just look incorrect. Should not the $\sigma$ computed by GPR be significantly larger than what's displayed based on the spread and error bars of the original data? It appears the error bars are severely underestimated. Also, why are the errors computed by GPR largest at the tails but quite a bit smaller around $X \approx 20$ where there is greater spread in the data?

Basic Code

For reference, the following code uses Python's sklearn, and full datasets are provided here where the text file can simply be downloaded and used with the below code (which calls this text file and should be named 'output.txt'):

import numpy as np
from matplotlib import pyplot as plt
from sklearn.gaussian_process import GaussianProcessRegressor
from sklearn.gaussian_process.kernels import RBF, ConstantKernel as C
import matplotlib 
matplotlib.rcParams['text.usetex'] = True
matplotlib.rcParams['text.latex.unicode'] = True
import json
with open("output.txt", "r") as infile:
    inputs = json.load(infile)


# Instantiate a Gaussian Process model
kernel = C(1., (1e-1, 1e1))* RBF(10, (1e0, 1e2))  
jump_ind = 1000 #skips ever n elements

X = np.array(inputs[0])
y = np.array(inputs[1])
inds = X.argsort()
X = X[inds]
X = np.atleast_2d(X).T  

total_noise = 0.2*((np.abs(y)))# + np.abs(noise)#np.sqrt(y)#np.abs(np.max(y) - np.min(y)) #np.abs(dy),

# Instantiate a Gaussian Process model
gp = GaussianProcessRegressor(kernel=kernel, alpha = total_noise,
# Fit to data using Maximum Likelihood Estimation of the parameters
gp.fit(X, y) 

total_noise = 0.2*((np.abs(y)))# + np.abs(noise)#np.sqrt(y)#np.abs(np.max(y) - np.min(y)) #np.abs(dy),

# Make the prediction on the meshed x-axis (ask for MSE as well)
x_input_range = np.linspace(np.min(X),np.max(X),jump_ind)
x_range = np.atleast_2d(x_input_range).T 

y_pred, sigma = gp.predict(x_range, return_std=True)  
# Plot the function, the prediction and the 95% confidence interval based on
# the MSE

fig, ax1 = plt.subplots() 
ax1.errorbar(X.ravel(), y, total_noise, fmt='g.', markersize=10, label=u'Observations (Average)')
lns1 = ax1.plot(x_range, y_pred, 'r-', label=u'Prediction')
ax1.fill(np.concatenate([x_range, x_range[::-1]]),
         np.concatenate([y_pred - 1.9600 * sigma,
                        (y_pred + 1.9600 * sigma)[::-1]]),
         alpha=.1, fc='r', ec='None', label='95% confidence interval')

lns = lns1#+lns2
labs = [l.get_label() for l in lns] 
ax1.legend(lns, labs, bbox_to_anchor=(1.08, 1), loc=2, borderaxespad=0.) 
ax1.legend(loc='upper left') 

with the following output:

In[1]: gp.kernel

Out[1]: 1**2 * RBF(length_scale=10)


1 Answer 1


I think that the covariance that you're using is improper. When you set alpha = 0.2 * np.abs(y), what you're doing is using a covariance function that looks like:

$$k(x_i, x_j) = RBF(x_i, x_j) + 0.2 *y_i * \mathcal I (x_i * x_j)$$

What's happening here is that the variance of each data point is set to be $0.2 * y_i$ - which is a standard deviation of $0.447 * sqrt(y_i)$. Assuming that the error bars are plotted as $y_i \pm 2\sigma$, if $y_i = 80$, then $2\sigma = 8.0$. So, I believe that after training, this effect is exacerbated if you observe the predicted $\sigma$s.

My suggestion is that you add a WhiteKernel (as that would include a trainable error term into your GP) and post the results of that. If the effect is still there, I'll edit my answer.

By the way, in my opinion, it isn't (statistically) nice to directly feed in the covariances that depend on $y$ because, in the model, $y$ exists on both sides of the sampling statement:

$$y \sim \mathcal N(0, \Sigma(y))$$

You should consider transforming the problem into one which avoids this dependance, although to get initial answers, it's probably reasonable. I can expand on this if you wish.

  • $\begingroup$ I made the suggested change (i.e. alpha = total_noise**2) but this did not appear to improve the fit. Here is the output and you'll notice the fit is actually worse. I've updated the code to be a full example. That is a good point about the left-hand and right-hand side dependence on the dependent variable, $y$. Based on the data acquisition, though, the error should be dependent on the magnitude of $y$. Are there recommended ways to get around this? Is it truly problematic? $\endgroup$
    – Mathews24
    Oct 26, 2018 at 21:22
  • $\begingroup$ I'm throwing guesses in the air, but what if you add a ConstantKernel() * WhiteKernel() to your kernel with the appropriate parameters? (The guess being that adding a diagonal term using alpha is treated as an untrainable set of params) $\endgroup$
    – adityar
    Oct 26, 2018 at 21:32
  • $\begingroup$ Adding ConstantKernel() * WhiteKernel() appears to help considerably. Could you possibly expand upon why the first term of the kernel (i.e. C(1., (1e-1, 1e1))* RBF(10, (1e0, 1e2)) which involves 2 hyperparameters) is not sufficient? Perhaps I misinterpreting the effect of these hyperparameters on the computed error, but an explanation of why the 2 hyperparameters lead to high bias and limited degrees of freedom for the fit would resolve my question in understanding what valid error bars from GPR actually are. $\endgroup$
    – Mathews24
    Oct 26, 2018 at 21:40
  • $\begingroup$ The RBF essentially enforces continuity - it can never describe errors so the constant leading the RBF essentially changes its scale w.r.t. the y-axis. You can interpret a GP with an RBF +White Kernel as an addition of two GPs, one which is the smooth mean function due to RBF and one is an error term due to the White Kernel $\endgroup$
    – adityar
    Oct 26, 2018 at 21:44
  • $\begingroup$ I talk a little bit more about this in my answer to another post here. $\endgroup$
    – adityar
    Oct 26, 2018 at 21:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.