I’m guessing that most people are pretty comfortable with the concept of uncorrelated Gaussian noise. It’s the most frequently assumed noise. Even if you don’t realise it, you’re probably assuming Gaussian noise.

Quick check: Are you using a chi-squared test to fit your data? Yes? Well there you go.

Co-variate Gaussian Noise

Here I’m going to talk about multi-variate, or co-variate, Gaussian noise. Co-variate Gaussian noise is the situation where the value of one data point affects the value of another. This kind of co-variance, i.e. variance between things, is usually expressed as a covariance matrix.

If you have N data points, then your covariance matrix will have a size: N x N. The matrix is normally denoted K (or sometimes $latex \mathbf{\Sigma}$).

$latex \mathbf{K(x,x)} = \left( \begin{array}{cccc} k(x_1,x_1) & k(x_1,x_2) & ... & k(x_1,x_n) \\ k(x_2,x_1) & k(x_2,x_2) & ... & k(x_2,x_n) \\ \vdots & \vdots & \vdots & \vdots \\ k(x_n,x_1) & k(x_n,x_2) & ... & k(x_n,x_n)  \end{array} \right) $

Uncorrelated, or independent, Gaussian noise is a special case of the covariance matrix where only the diagonal elements have a non-zero value, i.e.

$latex \mathbf{K(x,x)} = \left( \begin{array}{cccc} k(x_1,x_1) & 0 & ... & 0 \\ 0 & k(x_2,x_2) & ... & 0 \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & ... & k(x_n,x_n)  \end{array} \right) $

The value of each diagonal element corresponds to the variance of a particular data point, e.g. a data point at position $latex x_1$ with a mean value $latex \mu_1 $ would have a variance $latex k(x_1,x_1) = \sigma_1^2 $ ; a data point at position $latex x_2$ with a mean value $latex \mu_2 $ would have a variance $latex k(x_2,x_2) = \sigma_2^2 $ , and so on and so forth.

The actual y-value at each x-position will be drawn from a probability distribution

$latex p(y_i) = N (\mu_i, K_{ij})$

which in the case of a diagonal covariance matrix reduces to

$latex y_i = \mu_i + N(0, \sigma^2_i)$.

However, if we start to add in non-zero values to the other elements of the covariance matrix then this will no longer be the case and the y-value at one position will affect the y-value at another.

Covariate Gaussian Noise in Python

To simulate the effect of co-variate Gaussian noise in Python we can use the numpy library function multivariate_normal(mean,K).

Note: the Normal distribution and the Gaussian distribution are the same thing.

First off, let’s load some libraries:

import numpy as np   # the numpy library
import pylab as pl   # the matplotlib for plotting

then we can call the multivariate_normal(mean,K) function:

# make an array of positions
# these are evenly spaced, but they don’t have to be
x = np.arange(0, 20.,0.01)

# use numpy to draw a sample from the multi-variate
# normal distribution, N(0,K), at all the positions in x
y = np.random.multivariate_normal(np.zeros(len(x)),K)

The multivariate_normal function takes two arguments: (1) an array of noiseless mean values for each of the x-positions, and (2) a covariance matrix for all the x-positions.

In this case, I’ve made all of the  mean values equal to zero.

At the moment we haven’t specified K, so these lines of code won’t work just yet. To create K we need to build a matrix of values that are calculated by the function little k. This function is known as the covariance kernel and it defines how much of an affect one data value has on another.

The Covariance Kernel

To start with we are going to define a squared-exponential covariance kernel. This has the form:

$latex k(x_n,x_m) = h^2 \exp{ \left( \frac{-(x_n - x_m)^2}{\lambda^2} \right)} $

where $latex x_n$ is the x-position of one data point and $latex x_m $ is the x-position of another. The value of the kernel is a function of how far away from each other these data points are, i.e. $latex x_n - x_m $. It also has a couple of hyper-parameters that govern the overall shape of the kernel: $latex h$ and $latex \lambda$. These are referred to as hyper-parameters because they don’t really have any direct physical meaning so they’re not bog-standard normal model parameters. Here the $latex h $ parameter controls the normalisation of the kernel and $latex \lambda $ controls the width of the kernel.

def cov_kernel(x1,x2,h,lam):

    Squared-Exponential covariance kernel

    k12 = h**2*np.exp(-1.*(x1 - x2)**2/lam**2)
    return k12

We can then use this kernel function to fill our covariance matrix:

def make_K(x, h, lam):

    Make covariance matrix from covariance kernel

    # for a data array of length x, make a covariance matrix x*x:
    K = np.zeros((len(x),len(x)))

    for i in range(0,len(x)):
        for j in range(0,len(x)):

            # calculate value of K for each separation:
            K[i,j] = cov_kernel(x[i],x[j],h,lam)

     return K

and we can then update our earlier call to the numpy multivariate_normal function:

# make an array of 200 evenly spaced positions between 0 and 20:
x = np.arange(0, 20.,0.01) 

# make a covariance matrix:
K = make_K(x,h,lam)

# draw samples from a co-variate Gaussian
# distribution, N(0,K), at positions x1:
y = np.random.multivariate_normal(np.zeros(len(x)),K)

Putting It All Together

We can expand this to look at what happens when we vary the hyper-parameters of the covariance kernel, in particular the width, $latex \lambda $:

# make an array of 200 evenly spaced positions between 0 and 20:
x = np.arange(0, 20.,0.01)

for i in range(0,3):

    h = 1.0

    if (i==0): lam = 0.1
    if (i==1): lam = 1.0
    if (i==2): lam = 5.0

    # make a covariance matrix:
    K = make_K(x,h,lam)

    # five realisations:
    for j in range(0,5):

        # draw samples from a co-variate Gaussian distribution, N(0,K):
        y = np.random.multivariate_normal(np.zeros(len(x)),K)

        tmp2 = '23'+str(i+3+1)

    tmp1 = '23'+str(i+1)
    pl.title(r"$\lambda = $"+str(lam))


The top row of plots is showing the covariance matrix for each of the three different kernel widths. The bottom row is showing five realisations of data y-values that correspond to each of the covariance matrices. Remember that all of these realisations have zero mean so the only contribution to the y-values is (covariate) Gaussian noise.

The narrower the covariance kernel, the more diagonal the covariance matrix becomes and the more like independent noise the y-values appear.

Gaussian Process Modelling

We can invert the probability statement

$latex p(y_i) = N (\mu_i, K_{ij})$

to calculate the value of a hypothetical measurement at a test position, $latex x_{\ast}$, based on existing measurements, y, at positions, x.

Key to this are two things:

  1. Knowing the form of the covariance a priori; or, having sufficient example data points to calculate it;
  2. Having a sufficient number of example data points to provide fixed points in the function a priori. These are our training data.

Normally if you can satisfy the second of these requirements, the first follows naturally.

Inverting the above probability statement allows us to write the following equations. These give us the posterior mean ($latex m_{\ast}$) and the posterior variance ($latex C_{\ast}$) at that point, i.e. the value and the uncertainty on that value:

$latex \mathbf{m_{\ast}} = \mathbf{ K(x_{\ast},x)^T K(x,x)^{-1} y } \\ \mathbf{C_{\ast}} = \mathbf{ K(x_{\ast},x_{\ast}) - K(x_{\ast},x)^T K(x,x)^{-1} K(x_{\ast},x) } $

This is more simply written as:

$latex \mathbf{m_{\ast}} = \mathbf{ k^T_{\ast} K^{-1} y } \\ \mathbf{C_{\ast}} = \mathbf{ k(x_{\ast},x_{\ast}) - k^T_{\ast} K^{-1} k_{\ast} } $

assuming that the original measurements have zero mean. If they don’t it becomes:

$latex \mathbf{m_{\ast}} = \mu_{\ast} + \mathbf{ k^T_{\ast} K^{-1} (y - \mu) } \\ \mathbf{C_{\ast}} = \mathbf{ k(x_{\ast},x_{\ast}) - k^T_{\ast} K^{-1} k_{\ast} }. $

To understand where this comes from in more detail you can read Chapter 2 of Rasmussen & Williams.

So, how do we actually practically do this in Python?

Prediction in Python

Let’s start with the covariate data we made that looked like this:

If we take the final realization from these data, which has $latex \lambda = 5$, and select 5 points from it as our training data then we can calculate the posterior mean and variance at any other point based on those five training points.

First off, let’s randomly select our training points and allocate all the data positions in our realisation as either training or test:

# set number of training points
nx_training = 5

# randomly select the training points:
tmp = np.random.uniform(low=0.0, high=2000.0, size=nx_training)
tmp = tmp.astype(int)

condition = np.zeros_like(x)
for i in tmp: condition[i] = 1.0

y_train = y[np.where(condition==1.0)]
x_train = x[np.where(condition==1.0)]
y_test = y[np.where(condition==0.0)]
x_test = x[np.where(condition==0.0)]

Next we can use those 5 training data points to make a covariance matrix using the function we defined above:

# define the covariance matrix:
K = make_K(x_train,h,lam)

To calculate the posterior mean and variance we’re going to need to calculate the inverse of our covariance matrix. For a small matrix like we have here, we can do this using the numpy library linear algebra functionality:

# take the inverse:
iK = np.linalg.inv(K)

However, be careful inverting larger matrices in Python. The numerical stability of the numpy linear algebra inversion is pretty poor. The scipy linear algebra inversion is better because it always uses BLAS/LAPACK, but whenever possible don’t invert a matrix at all.

And now we’re ready to calculate the posterior mean and variance (or standard deviation, which is the square-root of the variance):

for xx in x_test:

    # find the 1d covariance matrix:
    K_x = cov_kernel(xx, x_train, h, lam)

    # find the kernel for (xx,xx):
    k_xx = cov_kernel(xx, xx, h, lam)

    # calculate the posterior mean and variance:
    m_xx = np.dot(K_x.T,np.dot(iK,y_train))
    sig_xx = k_xx - np.dot(K_x.T,np.dot(iK,K_x))

    sig.append(np.sqrt(np.abs(sig_xx))) # note sqrt to get stdev from variance

Let’s see how we did. Here I’m going to plot the training data points, as well as the original realisation (dashed line) that we drew them from, on the left. On the right I’m going to plot the same data plus the predicted posterior mean (solid line) and the standard deviation (shaded area).

# m and sig are currently lists - turn them into numpy arrays:

# make some plots:

# left-hand plot
ax = pl.subplot(121)
pl.scatter(x_train,y_train)  # plot the training points
pl.plot(x,y,ls=':')        # plot the original data they were drawn from

# right-hand plot
ax = pl.subplot(122)
pl.plot(x_test,m,ls='-')     # plot the predicted values
pl.plot(x_test,y_test,ls=':') # plot the original values

# shade in the area inside a one standard deviation bound:
ax.fill_between(x_test,m-sig,m+sig,facecolor='lightgrey', lw=0, interpolate=True)

pl.scatter(x_train,y_train)  # plot the training points

# display the plot:

We can see that the prediction is pretty good. It’s not exactly the same as the original realisation, but then again it would be pretty surprising if we could predict things perfectly based on incomplete data.

The shaded regions of uncertainty are also useful because they tell us how much accuracy we should expect in any given range of positions. Note that the further away from our training data we go, the larger the uncertainty becomes. On the right hand side it explodes because we’ve run out of training data points to constrain the posterior mean.

Statistically we should not expect our original realisation to lie within the shaded region all the time: it’s a one sigma limit, i.e. we should expect the values of the original realisation to lie within one sigma of the posterior mean ~68% of the time.

One of the most well known examples of using Gaussian Process Modelling for forward prediction is the application described in Rasmussen & Williams, which shows the prediction for the future of atmospheric CO2 levels.

Figure 5.6 from Rasmussen & Williams

Here I’m going to step through how to repeat R&W’s CO2 prediction in Python.

GPM with George

Some basic libraries…

import numpy as np
import pylab as pl
import scipy.optimize as op

In the section above we coded up covariance kernels and a covariance matrix from scratch. This time we’re going to need more than one kernel. We could write a little library that defines a whole load of different covariance kernels… but in fact somebody has already done it for us :-)

The george Gaussian Process Modelling library is pip installable:

pip install george

We need to import the library and the covariance kernels:

import george
from george import kernels

We’re also going to need some data. I downloaded these data from the NOAA website.

There is also a reduced version (up to 2001) of this dataset available directly through the Python Statsmodels Datasets package.

# this is a function to read the Mauna Loa data from file
def read_co2(filename):

    co2file = open(filename,'r')

    while True:
        line = co2file.readline()
        if not line: break

        items = line.split()

        if (items[0]!='#') and (items[3]!='-99.99'):



    return time,co2

t,y = read_co2("mauna_loa.dat")

I’m going to pick out the data up to 2003 to use as training data points. Then I’m going to use the training data to predict the future at test data points from 2003 to 2025.

First off, here are the training data:

pl.ylabel(r"CO$_2$ concentration, ppm", fontsize=20)
pl.xlabel("year", fontsize=20)
pl.title(r"Mauna Loa CO$_2$ data: 1958 to 2003")

I’m going to split out my training data as a separate array:

y_to_2003 = y[np.where(t<2003.)]
t_to_2003 = t[np.where(t<2003.)]

To select appropriate kernels to describe the behaviour of the data in the covariance matrix we need to look at the trends that are present.

Firstly, there is a long term increase:

We could include a mean function to model this long term rise, but we can also just use a covariance kernel with a large width:

$latex k(x_i,x_j) = h^2 \exp{ \left( \frac{-(x_i - x_j)^2}{\lambda^2} \right)} $

# Squared exponential kernel
# h = 66; lambda = 67
k1 = 66.0**2 * kernels.ExpSquaredKernel(67.0**2)

Note that I’m including the values of the hyper-parameters for each kernel in the code as they are defined in R&W.

Secondly, there is that periodic behaviour. So we need a periodic kernel.

However, we don’t know if the behaviour is exactly periodic, so we should allow for some decay away from exact periodicity.

$latex k(x_i,x_j) = h^2 \exp{ \left( - \frac{(x_i - x_j)^2}{\lambda^2} - \frac{2}{\gamma^2}\sin^2(\frac{\pi (x_i - x_j)}{P}) \right)} $

# periodic covariance kernel with exponential component to
# allow decay away from periodicity:
# h = 2.4; lambda = 90; gamma = 1.3; P = 1
k2 = 2.4**2 * kernels.ExpSquaredKernel(90**2) * kernels.ExpSine2Kernel(2.0 / 1.3**2, 1.0)

Thirdly, there are the medium term irregularities in the long term behaviour. Technically I believe these are known as a wibble.

$latex k(x_i,x_j) = h^2 \left( 1 + \frac{(x_i - x_j)^2}{2\alpha\beta^2} \right)^{-\alpha} $

# rational quadratic kernel for medium term irregularities.
# h = 0.66; alpha = 0.78; beta = 1.2
k3 = 0.66**2 * kernels.RationalQuadraticKernel(0.78, 1.2**2)

And… finally, there is the noise. These are empirical data and so they will always have some noise component due to the measurement equipment which can introduce both uncorrelated and short range correlated noise contributions:

$latex k(x_i,x_j) = h^2 \exp{ \left( \frac{-(x_i - x_j)^2}{\lambda^2} \right)} + \sigma^2 \delta_{\rm ij} $

# noise kernel: includes correlated noise 
# h = 0.18; lambda = 1.6; sigma = 0.19
k4 = 0.18**2 * kernels.ExpSquaredKernel(1.6**2)

Let’s put all these components together to make our final combined kernel:

kernel = k1 + k2 + k3 + k4

Now we need to feed our combined kernel to the george library. Here I’m including uncorrelated (i.e. white) noise. I’ve given it an initial standard deviation of 0.19 ppm, but by setting fit_white_noise=True I’m making it an optimizable hyper-parameter.

# Create an instance of the Gaussian process:
gp = george.GP(kernel, mean=0.0, fit_mean=False, white_noise=np.log(0.19**2), fit_white_noise=True)

Then we compute the covariance matrix:


and… that’s it.

Predicting the Future

We can now predict the posterior mean and variance at all our test data points:

# range of times for prediction:
x = np.linspace(max(t_to_2003), 2025, 2000)

# calculate expectation and variance at each point:
mu, cov = gp.predict(y_to_2003, x, return_var=True)
std = np.sqrt(cov)

So how does it look?

ax = pl.subplot(111)

# plot the original values

# shade in the area inside a one standard deviation bound:
ax.fill_between(x,mu-std,mu+std,facecolor='lightgrey', lw=0, interpolate=True)

pl.ylabel(r"CO$_2$ concentration, ppm")
pl.title(r"Mauna Loa CO$_2$ data - Initial Prediction")

# display the figure:

Optimizing the Hyper-parameters

But… what if the values of the hyper-parameters from R&W weren’t exactly right? We should probably optimise them for the data. We can do that in simple cases using the optimization function in the scipy library.

To use the scipy library optimization function we need to provide

(1) some objective function to optimize, and (2) the gradient of that function.

I’m going to define the objective function for the optimization as a negative log-likelihood. We could write this function ourselves, but in fact george has a built in log-likelihood that we can simply call directly.

The log-likelihood is computed as:

$latex \log \mathcal{L} \propto (\mathbf{y} - X^T \mathbf{x})^T K^{-1}(\mathbf{y} - X^T \mathbf{x}) $

where y is the variable and x are the points at which it is measured; K is the covariance matrix and X is the operator that maps x onto y.

def nll(p):
    # Update the kernel parameters and compute the likelihood.
    gp.kernel[:] = p
    ll = gp.lnlikelihood(y_to_2003, quiet=True)

    # The scipy optimizer doesn't play well with infinities.
    return -ll if np.isfinite(ll) else 1e25
def grad_nll(p):
    # Update the kernel parameters and compute the likelihood gradient.
    gp.kernel[:] = p
    gll = gp.grad_lnlikelihood(y_to_2003, quiet=True)
    return -gll

We can then run the optimization routine:

# initial guess at parameters (see above):
p0 = gp.kernel.vector

# if you want to view your initial guess values uncomment the line below
# print p0

# run optimization:
results = op.minimize(nll, p0, jac=grad_nll)

To use the results of the optimization, we then need to update the kernel with the results of the optimization:

gp.kernel[:] = results.x

Re-run the prediction with the updated parameters:

# range of times for prediction:
x = np.linspace(max(t_to_2003), 2025, 2000)

# calculate expectation and variance at each point:
mu, cov = gp.predict(y_to_2003, x, return_var=True)
std = np.sqrt(cov)

And see what we get:

ax = pl.subplot(111)

# plot the original values

# shade in the area inside a one standard deviation bound:
ax.fill_between(x,mu-std,mu+std,facecolor='lightgrey', lw=0, interpolate=True)

pl.ylabel(r"CO$_2$ concentration, ppm")
pl.title(r"Mauna Loa CO$_2$ data - Optimised Prediction")

# display the figure:

Well, the original parameters were basically optimal so things haven’t changed much.

Let’s now compare the prediction with the actual measured data from Mauna Kea after 2003:

ax = pl.subplot(111)

# plot the training values

# shade in the area inside a one standard deviation bound:
ax.fill_between(x,mu-std,mu+std,facecolor='lightgrey', lw=0, interpolate=True)

# plot the full dataset

pl.ylabel(r"CO$_2$ concentration, ppm")
pl.title(r"Mauna Loa CO$_2$ data - Comparison")

# display the figure:

It’s not quite as much of a difference from the prediction as the news seems to think:

But it does tell us that our model is not perfect and that perhaps:

(1) Accelerated behaviour has appeared in the data after 2003; or

(2) We need another kernel to account for acceleration in the increase; or

(3) Perhaps we should introduce a non-zero mean [my personal guess].

Reading Material

This is a very brief explanation of covariate Gaussian noise. For a better and more detailed description, I like these references:

The second of these has a particularly nice figure showing the effect of covariate Gaussian noise. It looks like this:

Figure 5 from Roberts et al. 2012.

and you can see that it is the same figure we created above.