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



> 
> $\alpha(\mu,I\_n,t) = \varphi(\beta(\mu,I\_n,t))$.
> 
> 
> 



Functional relationship between two quantities
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:**


*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 $\beta$.


