I was in a seminar today and the lecturer said that the gaussian distribution is isotropic. What does it mean for a distribution to be isotropic? It seems like he is using this property for the pseudo-independence of vectors where each entry is sampled from the normal distribution.

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    $\begingroup$ In general, a multivariate normal distribution can be anisotropic depending on the covariance matrix. There has. clearly been some miscommunication somewhere along the way. $\endgroup$ – Brian Borchers Oct 30 '16 at 18:52

TLDR: An isotropic gaussian is one where the covariance matrix is represented by the simplified matrix $\Sigma = \sigma^{2}I$.

Some motivations:

Consider the traditional gaussian distribution:

$$ \mathcal{N}(\mu,\,\Sigma) $$

where $\mu$ is the mean and $\Sigma$ is the covariance matrix.

Consider how the number of free parameters in this Gaussian grows as the number of dimensions grows.

$\mu$ will have a linear growth. $\Sigma$ will have a quadratic growth!

This quadratic growth can be very computationally expensive, so $\Sigma$ is often restricted as $\Sigma = \sigma^{2}I$ where $\sigma^{2}I$ is a scalar variance multiplied by an identity matrix.

Note that this results in $\Sigma$ where all dimensions are independent and where the variance of each dimension is the same. So the gaussian will be circular/spherical.

Disclaimer: Not a mathematician, and I only just learned about this so may be missing some things :)

Hope that helps!

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    $\begingroup$ The variables of the multivariate Gaussian may not be independent, even if they have zero covariance. Covariance is a measure of only linear association. Independence implies zero covariance but not vice versa. See this example. $\endgroup$ – mloning Nov 29 '19 at 9:10
  • $\begingroup$ @mloning, your comment is a bit misleading here. When a random vector has as a multivariate normal, 0 covariance does indeed imply independence. All of this gets its own wikipedia page: en.wikipedia.org/wiki/… The notation above suggests to me multivariate normality, so, it's fine. Another way in which the normal distribution is magical $\endgroup$ – RMurphy Apr 24 at 19:56
  • $\begingroup$ Yes, wasn't aware of that, thanks for the clarification! $\endgroup$ – mloning Apr 25 at 15:00

I'd just like to add a bit of visuals to the other answers.

When the variables are independent, i.e. the distrubtion is isotropic, it means that the distribution is aligned with the axis.

For example, for $\Sigma = \begin{pmatrix}1 & 0 \\ 0 & 30\end{pmatrix}$, you'd get something like this:

image of 2D gaussian with higher Y variance

So, what happens when it is not isotropic? For example, when $\Sigma = \begin{pmatrix}1 & 15 \\ 15 & 30\end{pmatrix}$, the distribution appears "rotated", no longer aligned with the axes:

image of 2D gaussian with covariance between X and Y

Note that this is just an example, the $\Sigma$ above is invalid since it is not PSD.


import numpy as np
from matplotlib import pyplot as plt

pts = np.random.multivariate_normal([0, 0], [[1,15],[15,31]], size=10000, check_valid='warn')

plt.scatter(pts[:, 0], pts[:, 1], s=1)
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I am not a math major student but I will give a try to describe my understanding: an isotropic gaussian distribution means a multidimensional gaussian distribution with its variance matrix as an identity matrix multiplied by the same number on its diagonal. Each dimension can be seen as an independent one-dimension gaussian distribution (no covariance exists).

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