Multivariate Gaussian Distribution¶

Suppose random vector $\mathbf{X} = (X_1, \ldots, X_p)^T$ has a multivariate Gaussian distribution:

\[\mathbf{X} \sim \mathcal{N} (\mathbf{\mu}, \mathbf{\Sigma})\]

with probability density function:

\[ f (\mathbf{x}) = \frac{1}{\sqrt{(2 \pi)^p \lvert \mathbf{\Sigma} \rvert}} \exp \left( - \frac{1}{2} (\mathbf{x} - \mathbf{\mu})^T \mathbf{\Sigma}^{-1} (\mathbf{x} - \mathbf{\mu}) \right) \]

where:

$\mathbf{\mu} = (\mu_1, \ldots, \mu_p)^T$ is the mean vector
$\mathbf{\Sigma}$ is the covariance matrix
$\lvert \mathbf{\Sigma} \rvert$ is the determinant of $\mathbf{\Sigma}$.

Parameter Estimation¶

Suppose we have a dataset:

\[\mathcal{D} = \{ \mathbf{x}_i \}_{i = 1}^n\]

where each sample $\mathbf{x}_i$ is a $p$-dimensional vector.

The mean vector is calculated by averaging the samples:

\[ \mathbf{\mu} = \frac{1}{n} \sum_{i = 1}^n \mathbf{x}_i \]

where $n$ is the number of samples.

The covariance matrix is estimated by summing the outer products of the centered samples:

\[ \mathbf{\Sigma} = \frac{1}{n - 1} \sum_{i = 1}^n (\mathbf{x}_i - \mathbf{\mu}) (\mathbf{x}_i - \mathbf{\mu})^T \]

Matrix Form¶

Let $\mathbf{X} = (\mathbf{x}_1, \ldots, \mathbf{x}_n)^T$ be the design matrix with each row being a sample.

The centered design matrix is computed as:

\[ \tilde{\mathbf{X}} = \left( \mathbf{x}_1 - \mathbf{\mu}, \ldots, \mathbf{x}_n - \mathbf{\mu} \right)^T \]

The covariance matrix can be written as:

\[ \mathbf{\Sigma} = \frac{1}{n - 1} \tilde{\mathbf{X}}^T \tilde{\mathbf{X}} \]

Properties of the Covariance Matrix¶

Theorem

Covariance matrix $\mathbf{\Sigma}$ is always symmetric and positive semi-definite.

Proof:

$\mathbf{\Sigma}$ is symmetric because:

\[ \mathbf{\Sigma}^T = \left( \frac{1}{n - 1} \tilde{\mathbf{X}}^T \tilde{\mathbf{X}} \right)^T = \frac{1}{n - 1} \tilde{\mathbf{X}}^T \tilde{\mathbf{X}} = \mathbf{\Sigma} \]

$\mathbf{\Sigma}$ is positive semi-definite, because for any vector $\mathbf{v} \in \mathbb{R}^p$:

\[ \mathbf{v}^T \mathbf{\Sigma} \mathbf{v} = \mathbf{v}^T \left( \frac{1}{n - 1} \tilde{\mathbf{X}}^T \tilde{\mathbf{X}} \right) \mathbf{v} = \frac{1}{n - 1} (\tilde{\mathbf{X}} \mathbf{v})^T (\tilde{\mathbf{X}} \mathbf{v}) = \frac{1}{n - 1} \lVert \tilde{\mathbf{X}} \mathbf{v} \rVert^2 \ge 0 \]

Therefore, $\mathbf{\Sigma}$ is symmetric and positive semi-definite.

\[\tag*{$\blacksquare$}\]

Eigenvalue Interpretation¶

Without giving a formal proof, an equivalent condition to symmetric and positive semi-definite property is that all eigenvalues of the covariance matrix are non-negative [18].

An eigenvalues of the covariance matrix can be interpreted as the amount of variance explained by the corresponding eigenvector, which cannot be negative.
A zero eigenvalue indicates that one or more variables can be expressed as a linear combination of other variables, which effectively makes the covariance matrix singular and not invertible. Also, along the direction represented by the corresponding eigenvector, there is no variability in the data.
In case of a singular covariance matrix:
- Apply Tikhonov regularization by adding a small multiple of the identity matrix to the covariance matrix [23].
- Use the pseudo-inverse of the covariance matrix.
- Perform dimensionality reduction.

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