Lower bound on the error of proportion estimation Let $X \sim \operatorname{Bin}(n,p)$. Suppose we estimate $p$ by $\hat{p}=\frac{X}{n}$. By  Hoeffding’s inequality
it holds for all $\delta \in (0,1)$ with probability at least $1-\delta$ that, $$\lvert\hat{p}-p\rvert\le \sqrt{\frac{\log\frac{2}{\delta}}{2n}}.
$$
I am interested in a matching non-asymptotic high probability lower bound.
The expected value $\mathbb{E}[\lvert\hat{p}-p\rvert]$ is lower bounded by $\sqrt{\frac{p(1-p)}{2n}}$ (see reference [1]). I am interested in a high-probability lower bound of the form
$$
\lvert\hat{p}-p\rvert\ge c\sqrt{\frac{\log\frac{1}{\delta}}{n}}
$$ preferably with an explicit constant $c$. Does anyone know of a reference?
Reference
[1] Daniel Berend and Aryeh Kontorovich, "A sharp estimate of the binomial mean absolute deviation
with applications"
Statistics & Probability Letters, Volume 83, Issue 4, April 2013, Pages 1254–1259, MR3041401, Zbl 1268.60021.
 A: $\newcommand{\de}{\delta}$You want a huge deal more than what there is in reality.
Indeed, you want
\begin{equation*}
    p_n:=P\Big(|\hat p-p|\ge c\sqrt{\frac{\ln(1/\de)}n}\,\Big)\ge1-\de \tag{1}\label{1}
\end{equation*}
for some $c\in(0,\infty)$, all large enough $n$, and all small enough $\de>0$.
By the central limit theorem,
\begin{equation*}
    Z_n:=\frac{\hat p-p}{\sqrt{pq/n}}\to Z
\end{equation*}
(as $n\to\infty$), where $q:=1-p$ and $Z\sim N(0,1)$. So, \eqref{1} would imply
\begin{equation*}
    1-\de\le p_n=P\Big(|Z_n|\ge c\sqrt{\frac{\ln(1/\de)}{pq}}\,\Big)
    \underset{n\to\infty}\longrightarrow
    P\Big(|Z|\ge c\sqrt{\frac{\ln(1/\de)}{pq}}\,\Big)
    \underset{\de\downarrow0}\longrightarrow0,
\end{equation*}
which in turn would imply $1\le0$.

On the positive side, by Shevtsova's version of the Berry–Esseen inequality, for any real $t$
\begin{equation}
\begin{aligned}
    P(|\hat p-p|\le t)&=P\Big(|Z_n|\le t\sqrt{\frac n{pq}}\Big) \\ 
    &\le P\Big(|Z|\le t\sqrt{\frac n{pq}}\Big) 
    +\frac{1/2}{\sqrt n}\,\frac{p^3 q+q^3 p}{(pq)^{3/2}} \\ 
    &\le at\sqrt{\frac n{pq}}+\frac{1}{2\sqrt{npq}}, 
\end{aligned}
\end{equation}
where $a:=\sqrt{2/\pi}$.
So, for $n\ge1/(\de^2 pq)$ we have $\frac{1}{2\sqrt{npq}}\le\frac\de2$ and hence
\begin{equation}
\begin{aligned}
    P\Big(|\hat p-p|\le\frac\de{2a}\,\sqrt{\frac{pq}n}\Big)&\le\de,  
\end{aligned}
\end{equation}
that is,
\begin{equation}
\begin{aligned}
    P\Big(|\hat p-p|>\frac\de{2a}\,\sqrt{\frac{pq}n}\Big)&\ge1-\de.   
\end{aligned}
\end{equation}
