**Question summary:**


If I have a two-sided bound, can I immediately get a one-sided bound with tighter constants?


**Question details:**


Let $\mathbf X = X\_1,...,X\_n$ be $n$ i.i.d. real-valued random variables where $X\_i \in [a,b]$ and $\mu = \mathbf E[X\_1]$. For $\delta\in(0,1)$, let $f(\mathbf X, \delta)$ be a function such that:


$$
\Pr\left (\left | \mu -\frac{1}{n} \sum\_{i=1}^n X\_i \right | \leq f(\mathbf X, \delta) \right ) \geq 1-\delta \tag{\*}
$$


For example, if using Hoeffding's inequality, then $f(\mathbf X, \delta) := (b-a)\sqrt{\frac{\ln(1/\delta)}{2n}}$. However, we do not assume that $f$ uses Hoeffding's inequality - it is any $f$ that makes the above equation hold given our assumptions.


The question: Can we conclude that therefore the one-sided bound also holds with $f/2$? That is:


$$
\Pr\left ( \mu -\frac{1}{n} \sum\_{i=1}^n X\_i \leq \frac{1}{2}f(\mathbf X, \delta) \right ) \geq 1-\delta \tag{\*\*}
$$


This holds for Hoeffding's inequality, but does it hold in general?



Symmetry of concentration bounds on mean
The answer is negative. Indeed, for simplicity, let $a=-1$ and $b=1$. 


In the case when $\delta=1/10$, $n=1$, and $X\_1$ is uniformly distributed on $[-1,1]$ (so that $\mu=0$), let $f(\mathbf X,\delta):=1-1/10$. 
Then 
$
\Pr\left (\left | \mu -\frac{1}{n} \sum\_{i=1}^n X\_i \right | \leq f(\mathbf X, \delta) \right ) =
\Pr(|X\_1|\le1-1/10) = 1-1/10=1-\delta
$,
so that $(\*)$ holds -- 
whereas 
$$
\Pr\left ( \mu -\frac{1}{n} \sum\_{i=1}^n X\_i \leq \frac{1}{2}f(\mathbf X, \delta) \right )
=\Pr\left(X\_1\ge-\frac12\,(1-1/10)\right)=29/40 
$$
$$\not\geq 1-1/10=1-\delta
,$$ 
so that $(\*\*)$ fails to hold. 


In all cases other than the one just considered, let e.g. $f(\mathbf X, \delta) := b-a$, so that $(\*)$ hold.


