= Simultaneous empirical distribution confidence band
{title2=$[L_n(x),U_n(x)]$}
Fix a failure <probability> $0<\alpha<1$ and assume an <independent and identically distributed> <statistical sample>. A finite-sample <confidence band> for its population <cumulative distribution function> is
$$
\varepsilon_{n,\alpha}=\sqrt{\frac{\log(2/\alpha)}{2n}},\qquad
L_n(x)=\max\{0,F_n(x)-\varepsilon_{n,\alpha}\},\qquad
U_n(x)=\min\{1,F_n(x)+\varepsilon_{n,\alpha}\}.
$$
Here $\log$ is the <natural logarithm>. The <DKW inequality> implies
$$
\mathbb P\!\left(L_n(x)\leq F(x)\leq U_n(x)\text{ for every }x\in\mathbb R\right)\geq1-\alpha.
$$
This is simultaneous coverage over the whole <real line>, valid at each fixed sample size.<All of Nonparametric Statistics by Larry Wasserman>[§2.1, printed p. 15, Theorem 2.6 and the derivation preceding it]{ref}
To obtain the band, solve $2e^{-2n\varepsilon^2}=\alpha$ for the positive $\varepsilon$. Outside a failure <event> of <probability> at most $\alpha$, $|F_n(x)-F(x)|\leq\varepsilon_{n,\alpha}$ for every $x$. Since a <cumulative distribution function> already lies in $[0,1]$, clipping the endpoints to this <interval> preserves the coverage <event>. The guarantee is at least $1-\alpha$, and need not equal it.
Choose $\alpha=0.1$, giving at least $90\%$ simultaneous coverage under the stated model. For $n=5,50,500$, respectively, the half-widths before clipping are
$$
\varepsilon_{5,0.1}\approx0.54733,\qquad
\varepsilon_{50,0.1}\approx0.17308,\qquad
\varepsilon_{500,0.1}\approx0.054733.
$$
For the chosen five observations, the band at $x=1$ is $[0,0.94733]$, while at $x=4$ it is approximately $[0.25267,1]$. At $x=3$ the unclipped band would be $[0.05267,1.14733]$, so clipping gives $[0.05267,1]$. The broad uncertainty is informative: five observations give a precise description of themselves, but a coarse bound for an unknown population <cumulative distribution function>.
\Image[/empirical-distribution-simultaneous-uncertainty.png]
{title=Finite-sample simultaneous uncertainty around an <empirical distribution function>}
{description=Left: the chosen five observations give $F_5$ and its band clipped to $[0,1]$, using $\alpha=0.1$. Right: theoretical half-widths for $n=5,50,500$, with no additional samples constructed. The <confidence band> has coverage at least $90\%$ under <independent and identically distributed> sampling; the drawing illustrates the procedure and does not establish its coverage.}
A collection of $90\%$ pointwise <confidence intervals> only promises coverage separately at each preselected threshold; it does not promise $90\%$ coverage of an entire <cumulative distribution function>. Here the one simultaneous <event> permits inspecting any threshold after seeing the data. The <probability> refers to repeated sampling of the entire <statistical sample>, with $F$ fixed. It does not assign $90\%$ <probability> to a fixed $F$ after observing this particular band, and it is not a <prediction interval> for a future observation. The displayed $n=5,50,500$ comparison gives a guarantee for each fixed $n$ separately, not simultaneous validity over all sample sizes or an optional stopping rule. The chosen observations are an illustration; the coverage claim follows from the <DKW inequality>, not from their appearance or a simulation. Dependence or observations with different sampling laws require another argument.
At fixed $\alpha$, multiplying $n$ by $100$ divides the half-width by $10$. Conversely, a desired half-width $\delta>0$ is ensured by $n\geq\log(2/\alpha)/(2\delta^2)$, rounded up to an <integer>. For $\alpha=0.1$ and $\delta=0.1$, this sufficient sample size is $150$. This makes the <confidence band> useful for planning how many observations are needed before estimating threshold <probabilities> to a chosen simultaneous accuracy.
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