I saw the following statement in this paper and wanted to understand its meaning:

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1 Answer 1


$\DeclareMathOperator{\diag}{diag}$ $\DeclareMathOperator{\tr}{tr}$

A great question. Most standard multivariate analysis texts only treated the $N > m$ case to give the MLE of $\mu$ and $X$ (or more frequently, $\Sigma = X^{-1}$). The $N \leq m$ case seems always overlooked (or merely qualitatively mentioned).

In short, "the MLE problem is unbounded" means that the objective function \begin{align} l(\mu, X) := -\log\det(X) + \frac{1}{N}\sum_{i = 1}^N(\hat{\xi}_i - \mu)^TX(\hat{\xi}_i - \mu), \quad \mu \in \mathbb{R}^m, X \in S_+^m \end{align} does not have a fixed lower bound (which depends on $\hat{\xi}_1, \ldots, \hat{\xi}_N$ only) when $m \geq N$ (by contrast, when $N > m$, almost every multivariate analysis textbook will show you $l(\mu, X)$ is bounded below). To show that $l(\mu, X)$ can be arbitrarily small, first introduce the following notations:
\begin{align} & Z = \begin{bmatrix} \hat{\xi}_1^T \\ \vdots \\ \hat{\xi}_N^T \end{bmatrix} \in \mathbb{R}^{N \times m}, \; e = \begin{bmatrix} 1 \\ \vdots \\ 1 \end{bmatrix} \in \mathbb{R}^{N \times 1}, \; \bar{\xi} = N^{-1}\sum_{i = 1}^N\hat{\xi}_i, \\ & A = \sum_{i = 1}^N(\hat{\xi}_i - \bar{\xi})(\hat{\xi}_i - \bar{\xi})^T = Z^T(I_{(N)} - N^{-1}ee^T)Z := Z^TPZ \in \mathbb{R}^{m \times m}. \end{align} It can be seen that $P$ is idempotent, whence $A = (PZ)^TPZ$ is positive semi-definite, and \begin{align} r:= \operatorname{rank}(A) = \operatorname{rank}(PZ) \leq \operatorname{rank}(P) = N - 1 < m. \end{align}
Therefore, the spectral decomposition of $A$ can be written as \begin{align} A = O\diag(\Lambda, 0_{(m - r)})O^T, \end{align} where $O$ is an order $m$ orthogonal matrix, $\Lambda = \diag(\lambda_1, \ldots, \lambda_r)$, $\lambda_i (1 \leq i \leq r)$ are positive eigenvalues of $A$.

Using these notations, $l(\mu, X)$ can be rewritten as: \begin{align} l(\mu, X) = -\log\det(X)+\frac{1}{N}\tr(XA) + (\bar{\xi} - \mu)^TX(\bar{\xi} - \mu). \end{align}

Let $X^* = O\diag(\Lambda^{-1}, kI_{(m - r)})O^T$ (where $k > 0$ is to be determined), $\mu^* = \bar{\xi}$, it then follows that \begin{align} & l(\mu^*, X^*) = \sum_{i = 1}^r\log\lambda_i - (m - r)\log k+ \frac{r}{N}, \end{align} whence $l(\mu^*, X^*) \to -\infty$ as $k \to \infty$, i.e., $l$ cannot be bounded from below.

If the above matrix operations are too complicated to understand, you can get some sense by considering the case $m = N = 1$ (univariate normal distribution with only one observation), for which the objective function is \begin{align} l(\mu, X) = -\log X + (\hat{\xi}_1 - \mu)^2X. \end{align} Therefore $l(\hat{\xi}_1, X) = -\log X \to -\infty$ as $X \to +\infty$, i.e., the objective function is unbounded.

  • $\begingroup$ Overwhelming? It's rather precise. +1. $\endgroup$ Commented Dec 23, 2022 at 5:32
  • 2
    $\begingroup$ @User1865345 Thank you. It takes me a while to achieve this precision. :) $\endgroup$
    – Zhanxiong
    Commented Dec 23, 2022 at 5:39
  • $\begingroup$ Changing to $X^\ast, \mu^\ast$ was witty. Keep answering. Appreciate your posts. $\endgroup$ Commented Dec 23, 2022 at 5:41

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