Let $X\in\mathbb{R}^{n}$ be a multivariate normal variable with the mean vector $\mu$ and the covariance matrix $\Sigma$. It is well known that if the matrix $\Sigma$ is positive-definite the following cumulative distribution \begin{equation} F\left(a_{1}, a_{2}, \cdots, a_{n}\right) = \mathbb{P}\left(X_{1}<a_{1},X_{2}<a_{2},\cdots,X_{n}<a_{n}\right) \end{equation} is monotonically increasing with respect to $a_{1},a_{2},\cdots,a_{n}$.

Is the above property still valid if the covariance matrix $\Sigma$ is not positive-definite (the degenerate case)?


The definition of a probability (and of a sigma algebra) necessarily implies that the CDF is always monotonically increasing with respect to any of its variables.

So, yes, it is also the case if Σ is degenerate.

Edit : A less theoretical answer consists of stating that if this was not the case, then you could find strictly negative probabilities. If $a_{1} < a_{1}^{'}$ and

\begin{equation} F\left(a_{1}, a_{2}, \cdots, a_{n}\right) > F\left(a_{1}^{'}, a_{2}, \cdots, a_{n}\right) \end{equation}

Then this means that

\begin{equation} \mathbb{P}\left(a_{1}\leq X_{1}<a_{1}^{'},X_{2}<a_{2},\cdots,X_{n}<a_{n}\right) < 0 \end{equation}

Edit : It seems I got tricked by the wording (I am not a native English speaker). The CDF is always nondecreasing, no matter what Σ is. If Σ is degenerate, then the PDF can be "not increasing". The simplest example is $ X=[X_{1}, X_{2}]$ where $X_1$ is a normal distribution and $X_2$ is $0$.

  • $\begingroup$ This answer uses "increasing" where one often uses "nondecreasing". $\endgroup$ – Did Jun 19 '14 at 19:26
  • $\begingroup$ I wanted to use the same notation as the OP. $\endgroup$ – Fezvez Jun 19 '14 at 22:14
  • $\begingroup$ @Did: Could you please give me an example for the case that $F\left(X_{1}<a_{1},X_{2}<a_{2},\cdots,X_{n}<a_{n}\right) = F\left(X_1<a^{'}_{1},X_{2}<a_{2},\cdots,X_{n}<a_{n}\right)$ for $a^{'}_{1}>a_{1}$? (In sigma algebra and $-\infty<a_{j}<\infty$) $\endgroup$ – Van Long DO Jun 20 '14 at 8:46
  • $\begingroup$ You just need to have $0$ probability between $a_{1}$ and $a_{1}^{'}$. For example because F is the uniform distribution in the unit cube, and $a_{1} = 2$ (and $a_{1}^{'} = 3$). $\endgroup$ – Fezvez Jun 20 '14 at 16:47
  • $\begingroup$ I missed that $X$ is a multivariate normal distribution in this case. Then, it is not possible $\endgroup$ – Fezvez Jun 20 '14 at 16:50

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