# Functional relationship between two quantities

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

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

• 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? May 16, 2022 at 13:21
• @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. May 16, 2022 at 13:30
• 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. May 16, 2022 at 14:36
• @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. May 16, 2022 at 14:51

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 $$\beta$$.

• Indeed, there was an error in my calculations. Thanks! May 16, 2022 at 16:20