Relating maximum/minimum value and variance of data points Given the expression
$T = \max(|\beta_{\max}-1|,|\beta_{\min}-1|)$
Can we relate $T$ to variance $\sigma^2$ of values $\beta_i^2$, assuming we know $\sigma^2$, $\beta_{\max}$ and $\beta_{\min}$.
PS: What if values $\beta_i^2$ are distributed with mean $\mu=1$.
 A: Let $b_i:=\beta_i$. Then $1-T\le b_i\le1+T$ for all $i$, and these inequalities provide the best bounds on the $b_i$'s given the information we have. So,
$$A\le b_i^2\le B \tag{1}$$
for all $i$, where
$$A:=\max[0,1-T]^2,\quad B:=(1+T)^2,$$
and inequalities (1) provide the best bounds on the $b_i^2$'s given the information we have.
So, for the variance $\sigma^2$ of the $b_i^2$'s we have
$$0\le\sigma^2\le(B-\mu)(\mu-A), \tag{2}$$
where $\mu$ is the mean of the $b_i^2$'s. Moreover, inequalities (2) provide the best bounds on $\sigma^2$ given the information we have.
The latter two statements follow from

Lemma: Let $X$ be any random variable such that $EX=\mu$ and $A\le X\le B$ for some real $A$ and $B$. Then for the variance $\sigma^2$ of $X$ we have
$$0\le\sigma^2\le(B-\mu)(\mu-A). \tag{3}$$
Moreover, inequalities (3) provide the best bounds on $\sigma^2$ in the conditions of this lemma.

Proof. Let $Y:=X-\mu$. Then $EY=0$ and $-a\le Y\le b$, where $b:=B-\mu$ and $a:=\mu-A$. So,
$$0\le E(b-Y)(Y+a)=ba-EY^2=ba-\sigma^2.$$
So,
$$\sigma^2\le ba=(B-\mu)(\mu-A),$$
so that (3) holds. Moreover, $\sigma^2=ba=(B-\mu)(\mu-A)$ if $Y$ takes values $-a,b$ with probabilities $\frac b{a+b},\frac a{a+b}$, respectively, that is, if $X$ takes values $A,B$ with probabilities $\frac b{a+b},\frac a{a+b}$, respectively.
A: If you know $\beta_{\max}$ and $\beta_{\min}$ then I would have thought you know $T$.
If you do not know  $\beta_{\max}$ and $\beta_{\min}$  then you can say $T \ge \sigma$, with equality only when you have a distribution with half the probability at $1-\sigma$ and half the probability at $1+\sigma$ (and a mean of $1$)
