1

Let $X=\left[x_{1},x_{2},\cdots,x_{n}\right]^{T}\in\mathbb{R}^{n}$ be a multidimensional normal variable with the mean vector $\mu_{X}\in\mathbb{R}^{n}$ and the covariance $\Sigma\in\mathbb{R}^{n\times n}$, where $\Sigma$ is not full rank (the degenerate case), and $b$ be a real number. Let $F\left(b\right)$ be a function defined as follows:

$F\left(b\right)=\mathbb{P}\left(x_{1}<b,x_{2}<b,\cdots,x_{n}<b\right).$

I think that the function $F\left(b\right)$ is always continuous, monotonically non-decreasing with respect to the variable $b$ in some region $\left[b_{1},b_{2}\right]$ and its co-domain is $\left[0,1\right]$. I mean that there always exists $b$ such that $F\left(b\right)=\alpha$, for a given value $\alpha\in\left(0,1\right)$ . However, I am not able give the proof. Could anyone help me with it?

Van Long DO
  • 23
  • 3

1 Answers1

1

Do you mean for any $\alpha$? If so, not sure about that. Consider the "degenerate" Normal distribution that is a constant function. It has mean $\mu \in \mathbb{R}$ and zero variance (so non-invertible covariance matrix), and its distribution satisfies, for $b \geq \mu$, $$ P\left(X \leq b \right) {}={} 1 $$

and for $b < \mu$,

$$ P\left(X \leq b \right) {}={} 0 $$

Clearly, no "b" exists such that $F(b)=\alpha$ for infinitely many "$\alpha$" in $F$'s co-domain. And $F$, though monotonic, is discontinuous for any domain containing $\mu$.

ki3i
  • 5,092