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For a fixed positive sample size $n$ of independent and identically distributed Bernoulli random variables, letThe Hoeffding inequality makes $C_n$ a conservative confidence interval for $p$: ${\mathbb{P}}_p(p\in C_n)\ge 1-\delta$ for every $p\in [0,1]$. Indeed, exclusion from the closed interval implies $|{\hat{p}}_n-p|>r_n$, an event contained in the deviation event bounded by $\delta$. Clipping to $[0,1]$ retains coverage because $p$ already lies there. For $n=738$, $\delta =0.05$, and an observed frequency $369/738=1/2$, $r_n\approx 0.0499924$, giving approximately $[0.450008,0.549992]$.

Coverage concerns repeated independent samples: the unknown $p$ is fixed and the confidence interval is random. This fixed-$n$ guarantee does not justify repeatedly checking the same bound and stopping as soon as it looks favorable. The Hoeffding sample size for Bernoulli frequencies chooses $n$ in advance; guarantees across repeated inspection times require separate control of the additional events, for example through a union bound with an allocated failure probability at each time.

The Hoeffding lemma controls the entire centered moment-generating function using only an interval width. For an essentially bounded real random variable $X$ with finite deterministic bounds $a\le X\le b$ almost surely, every real $\lambda$ satisfiesEquivalently, $\mathbb{E}{e}^{\lambda (X-\mathbb{E}X)}\le e^{{\lambda }^2(b-a{)}^2/8}$. This single-random variable statement needs no independence assumption; independence enters when applying it to the factors for a sum in the Hoeffding inequality. Both signs of $\lambda$ are allowed, and equality holds at $\lambda =0$ or when $a=b$.^[ref]

In the convention $\log \mathbb{E}{e}^{\lambda Y}\le {\lambda }^2{\sigma }^2/2$ for a centered sub-Gaussian random variable $Y$, the Hoeffding lemma supplies ${\sigma }^2=(b-a{)}^2/4$. This upper parameter need not equal the actual variance. For a Bernoulli random variable $B$ of success probability $p$,The actual variance is $p(1-p)$, whereas the supplied parameter is $1/4$ for every $p$. This explains part of the conservatism visible in Hoeffding sample size for Bernoulli frequencies.

A short proof exposes why the constant is $1/8$. Put $Y=X-\mathbb{E}X$ and define exponential tilting byThe tilted probability measure has the same interval of possible values, and differentiating the cumulant-generating function gives ${\psi }^{\prime \prime }(\lambda )={\mathrm{Var}}_{{\mathbb{P}}_{\lambda }}(Y)$. Boundedness justifies these derivatives for every finite $\lambda$. Under any such tilted law, Popoviciu's inequality on variances gives ${\psi }^{\prime \prime }(\lambda )\le (b-a{)}^2/4$: the variance is no larger than the expected value of the squared distance from the interval midpoint, itself at most $(b-a{)}^2/4$. Since $\psi (0)={\psi }^{\prime }(0)=0$, the integral form of Taylor's theorem yieldsThis expression handles either sign of $\lambda$ without changing the argument.

The constant $1/8$ cannot be reduced uniformly over all essentially bounded real random variables. Equal endpoint probabilities at $a$ and $b$ give $Y=\pm (b-a)/2$, a scaled Rademacher random variable, andThus the sharpness is already forced near zero by the variance; the bound need not be equality at nonzero parameters for a particular probability distribution.