0

I'm trying to draw an ellipsoid of the $3 \times 3$ covariance matrix. Usually, I see the sentence

an ellipsoid corresponding to the eigenvectors and eigenvalues of covariance matrix.

But from the equation

$$(\mathbf{x}-\boldsymbol\mu)^T\boldsymbol\Sigma^{-1}(\mathbf{x}-\boldsymbol\mu)\le \chi_{p,\alpha}^2 \tag{1}$$

in this answer, it seems to me that we do the eigen-decomposition of the inverse of covariance matrix.

Could you please elaborate on this point?

IMHO,

Let $\bar x$ and $\Sigma$ be the mean and covariance matrix of the data. Then the ellipsoid is $\{x \in \mathbb R^d | (x-\bar x)^T \Sigma(x-\bar x) = a\}$ for some $a>0$. The value of $a$ determines the confidence region. We consider the eigen-decomposition $\Sigma = P\Lambda P^{-1} =P\Lambda P^T$. Here $\Lambda$ is the diagonal matrix generated by eigenvalues $\lambda_1, \ldots, \lambda_d$, while each column of $P$ is an eigenvector. Let $y = P^T(x - \bar x)$. Then $(x-\bar x)^T \Sigma(x-\bar x) = y^T \Lambda y = \lambda_1y_1^2+...+\lambda_d y_d^2$. Then $y = P^T(x - \bar x)$ denotes the rotation of $x - \bar x$ by $P^T$.

Akira
  • 381
  • 1
  • 9
  • The eigendecomposition of $\Sigma^{-1}$ is extremely simply related to that of $\Sigma;$ using one is equivalent to using the other. – whuber Nov 03 '20 at 17:12
  • @whuber Could you please confirm if $(x-\bar x)^T \Sigma(x-\bar x) = a$ or $(x-\bar x)^T \Sigma^{-1} (x-\bar x) = a$ is the correct formula? – Akira Nov 03 '20 at 17:14
  • By considering the case of one dimension the answer will be immediately clear. – whuber Nov 03 '20 at 17:16
  • 1
    @whuber Thank you so much for your verification. – Akira Nov 03 '20 at 17:17
  • For the record, I'm not trying to be coy -- I'm only worried about being misunderstood or even mistakenly giving you bad information. It really is better for you to convince yourself concerning how to proceed. – whuber Nov 03 '20 at 17:20
  • @whuber I understand that. That's really deep help. You could just write a short answer "The latter is correct", but you did much more than that. – Akira Nov 03 '20 at 17:22

0 Answers0