Suppose, these are two pictures of plots of conditional probability density of a set of samples.

enter image description here

From the pictures, I can tell that,

(1) the left plot has non-zero covariance between the two dimensions(i.e. off diagonal elements are non-zero), and the means for each dimensions are below 0.

(2) On the right, the covariance matrix is diagonal(i.e. covariance=0. i.e. is spherical), mean=0.

Am I correct?

If YES, what mathematical operation can I apply to the left-plot to make it look like the right one?


Without commenting on whether you are correct, I can tell you the following.

Let $\mathbf{X}$ be a $d$-dimensional random vector:


The joint distribution function can be expressed as:

$$F_{X}(x)=F_{X}(x_{1},\ldots,x_{d})=\text{Pr}(\mathbf{X}\leq \mathbf{x})=\text{Pr}(X_{1}\leq x_{1},\ldots,X_{d}\leq x_{d})$$

With marginal distribution of $X_{i}$:

$$F_{i}(x_{i})=\text{Pr}(X_{i}\leq x_{i})=F_{X}(\infty,\ldots,x_{i},\ldots,\infty)$$

We can say that $\mathbf{X}$ has an elliptical distribution if:


where $\mathbf{Y}\sim S_{k}(\psi)$, $\mathbf{A}\in \mathbb{R}^{d\times k}$ and $\mathbf{\mu}\in\mathbb{R}^{d}$. $S_{k}(\psi)$ means a $k$-dimensional spherical distribution with characteristic generator $\psi$.

Essentially, elliptical distributions are obtained via multivariate affine transformations of spherical distributions. This can be illustrated by the characteristic function $\phi_{X}$:

$$\phi_{X}(t)=\mathbb{E}\big(e^{it'\mathbf{X}}\big)=\mathbb{E}\big(e^{it'(\mathbf{\mu}+\mathbf{A}\mathbf{Y})}\big)=e^{it'\mathbf{\mu}}\mathbb{E}\big(e^{i(\mathbf{A}'t)'\mathbf{Y}}\big)=e^{it'\mathbf{\mu}}\psi(t'\mathbf{\Sigma} t)$$

where $\mathbf{\Sigma}=\mathbf{A}\mathbf{A}'$, which is a positive semidefinite matrix. The elliptical distribution is denoted by:

$$\mathbf{X}\sim E_{d}(\mathbf{\mu},\mathbf{\Sigma},\psi)$$

where $\mu$, $\Sigma$ and $\psi$ are referred to as the location vector, dispersion matrix and characteristic generator of the distribution, respectively. $E_{d}$ denotes a $d$-dimensional ellipitical distribution.

Thus, to transform from an elliptical distribution to a spherical one, the following transformation can be applied:

$$\mathbf{X}\sim E_{d}(\mathbf{\mu},\mathbf{\Sigma},\psi)\Longleftrightarrow \mathbf{\Sigma}^{-\tfrac{1}{2}}(\mathbf{X}-\mathbf{\mu}) \sim S_{d}(\psi)$$

This is analogous to the type of transformation you would do to a univariate normal random variable, $N(\mu,\sigma^{2})$, to obtain a standard normal random variable, $N(0,1)$:

$$Z\sim N(\mu,\sigma^{2})\Longleftrightarrow \sigma^{-1}(Z-\mu) \sim N(0,1)$$

  • $\begingroup$ Please be kind to give explanation to you symbols. What is d,k,j,i,t,Ed etc. $\endgroup$ – ttnphns Nov 17 '16 at 6:50
  • $\begingroup$ Thank you very much. Is there any specific name for the transformation operation you showed? $\endgroup$ – user366312 Nov 17 '16 at 6:56
  • $\begingroup$ @anonymous I'm not sure if there's any specific name for it. It's just an affine transformation, similar to what you'd do to a normal distribution to get a standard normal distribution: $X\sim N(\mu,\sigma^{2})\Longleftrightarrow \frac{X-\mu}{\sigma}\sim N(0,1)$ $\endgroup$ – Ed P Nov 17 '16 at 7:07
  • $\begingroup$ @ttnphns I was pressed for time when you asked. Notation has been added, hopefully I covered everything. $\endgroup$ – Ed P Nov 17 '16 at 23:34
  • $\begingroup$ @StatsPlease a nice simplified answer. $\endgroup$ – Subhash C. Davar Nov 21 '16 at 13:51

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