Let $\mu \in \mathbb R^n$ and let $\Sigma$ be a positive-definite matrix of order $n \ge 2$. Fix $t \ge 0$ and define $\alpha(\mu,\Sigma,t) > 0$ by

$$ \alpha(\mu,\Sigma,t) := \sup_{\|w\| = 1}\frac{1}{\|w\|_\Sigma}\varphi\left(\frac{w^\top \mu - t}{\|w\|_\Sigma}\right), $$

where $\|w\|_\Sigma := \sqrt{w^\top \Sigma w}$ and $\varphi$ is the Gaussian pdf. Also define $\beta(\mu,\Sigma,t) \ge 0$ by $$ \beta(\mu,\Sigma,t) := \inf_{\|z\| \le t}\|z-\mu\|_{\Sigma^{-1}}. $$

Question. Is there any functional relationship between $\alpha(\mu,\Sigma,t)$ and $\beta(\mu,\Sigma,t)$ ?

Example: isotropic case

Suppose $\Sigma = I_n$, the identity matrix. Then

$\alpha(\mu,\Sigma,t) = \varphi(r_\star)$, where $$ \begin{split} -r_\star := \inf_{\|w\| = 1}|w^\top \mu - t| &= \inf_{\|w\| = 1}\sup_{s \in \{\pm 1\}}s(w^\top \mu-t) = \sup_{s \in \{\pm 1\}}-st+\inf_{\|w\| = 1}sw^\top \mu\\ & = \sup_{s \in \{\pm 1\}}-st-\|\mu\|=t-\|\mu\|, \end{split} $$ if $\|\mu\| \ge t$, and $r_\star = 0$ otherwise. That is, $r_\star = (\|\mu\|-t)_+$.

On the other hand, one computes $$ \begin{split} \inf_{\|z\| \le 1}\|z-\mu\| &= \begin{cases} 0,&\mbox{ if }\|\mu\| \le t,\\ \|t\mu/\|\mu\|-\mu\| = |t-\mu| = \|\mu\|-t,&\mbox{ else} \end{cases}\\ &=(\|\mu\|-t)_+ = r_\star. \end{split} $$

We conclude

$\alpha(\mu,I_n,t) = \varphi(\beta(\mu,I_n,t))$.

  • $\begingroup$ What is $z$ in the definition of $\beta$? Also, what do you mean by a "functional relationship"? That one of them is a function of the other? $\endgroup$ May 16, 2022 at 13:21
  • $\begingroup$ @IosifPinelis Typo, fixed. Yes that's what I meant by functional relationship. I've also included a worked example when $\Sigma=I_n$. Also, I posted the question here by mistake (as this is not really research-level math); I thought I was posting on SE. Please let me know if I should move this to SE. $\endgroup$
    – dohmatob
    May 16, 2022 at 13:30
  • $\begingroup$ Even your own calculation for $\Sigma=I$ shows that there is no functional relationship or, you may say, only a restricted functional relationship. But then you need to somehow qualify your general question. $\endgroup$ May 16, 2022 at 14:36
  • $\begingroup$ @IosifPinelis Maybe I'm missing something in your comment, but my worked example actually shows that $\alpha = \varphi(\beta)$ in the special case when $\Sigma=I_n$. This is an explicit functional relationship. $\endgroup$
    – dohmatob
    May 16, 2022 at 14:51

1 Answer 1


In fact, even when $\Sigma=I$, we have $$I_1(t,\mu):=\inf_{\|w\|=1}|w^\top\mu-t|=(t-\|\mu\|)_+ \tag{1}\label{1}$$ (think of large $t$), whereas $$I_2(t,\mu):=\inf_{\|z\|\le t}\|z-\mu\|=(\|\mu\|-t)_+ \tag{2}\label{2}$$ (think of large $\|\mu\|$).

So, there is no functional relationship between your $\alpha$ and $\beta$.


Details on \eqref{1}: We have $\{w^\top\mu\colon\|w\|=1\}=[-\|\mu\|,\|\mu\|]$. So, $I_1(t,\mu)$ is the distance from $t\ge0$ to the interval $[-\|\mu\|,\|\mu\|]$. So, \eqref{1} follows.

Details on \eqref{2}: Clearly, $\|z-\mu\|\ge0$. Also, $\|z\|\le t$ implies $\|z-\mu\|\ge\|\mu\|-\|z\|\ge\|\mu\|-t$, so that $\|z-\mu\|\ge(\|\mu\|-t)_+$ and hence $$I_2(t,\mu)\ge(\|\mu\|-t)_+. \tag{3}\label{3}$$ If now $\|\mu\|\le t$, let $z:=\mu$. Then $\|z\|\le t$ and $\|z-\mu\|=0=(\|\mu\|-t)_+$, so that $I_2(t,\mu)\le(\|\mu\|-t)_+$. If $\|\mu\|>t$, let $z:=t\mu/\|\mu\|$. Then $\|z\|=t\le t$ and $\|z-\mu\|=\|\mu\|-t=(\|\mu\|-t)_+$, so that again $I_2(t,\mu)\le(\|\mu\|-t)_+$. So, in view of \eqref{3}, \eqref{2} follows.

Details on the absence of a functional relationship between $\alpha$ and $\beta$: If $\|\mu_j\|=j$ for $j=1,2$, then $I_1(0,\mu_j)=0$ but $I_2(0,\mu_j)=j$. So, $I_2$ is not a function of $I_1$.

If $t_j=j$ for $j=1,2$, then $I_2(t_j,0)=0$ but $I_1(t_j,0)=j$. So, $I_1$ is not a function of $I_2$.

So, there is no functional relationship between $I_1$ and $I_2$. Since $\alpha$ and $\beta$ are functions of $I_1$ and $I_2$, respectively, we conclude that there is no functional relationship between $\alpha$ and $\beta$.

  • $\begingroup$ Indeed, there was an error in my calculations. Thanks! $\endgroup$
    – dohmatob
    May 16, 2022 at 16:20

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