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 $\mu=1$.



Relating maximum/minimum value and variance of data points
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.


