Index: The Book of Statistical Proofs ▷ Probability Distributions ▷ Univariate continuous distributions ▷ Arcsine distribution ▷ Cumulative distribution function

Theorem: Let $X$ have the standard arcsine distribution, that is, the beta distribution with both shape parameters equal to $1/2$:

\[\label{eq:arcsine-beta} X \sim \mathrm{Bet}\!\left(\frac{1}{2},\frac{1}{2}\right) \; .\]

Its cumulative distribution function, defined for every real $x$, is

\[\label{eq:arcsine-cdf} F_X(x) = \begin{cases} 0, & x \leq 0, \\ \dfrac{2}{\pi}\arcsin\!\left(\sqrt{x}\right), & 0 < x < 1, \\ 1, & x \geq 1 \; . \end{cases}\]

Proof: By the beta density formula, the density is proportional to $[t(1-t)]^{-1/2}$ on $(0,1)$ and vanishes elsewhere. To determine its normalizing constant, substitute $t=\sin^2\theta$ in the integral defining the beta function. For $0<\theta<\pi/2$, both sine and cosine are positive, so

\[\label{eq:arcsine-normalization} \mathrm{B}\!\left(\frac{1}{2},\frac{1}{2}\right) = \int_0^1 \frac{\mathrm{d}t}{\sqrt{t(1-t)}} = \int_0^{\pi/2} 2 \, \mathrm{d}\theta = \pi \; .\]

The substitution applies first on compact subintervals; taking the endpoint limits gives the displayed improper integral. Consequently, the probability density function is $f_X(t)=1/[\pi\sqrt{t(1-t)}]$ for $0<t<1$.

For $0<x<1$, the same substitution has upper limit $\theta=\arcsin(\sqrt{x})$. Therefore,

\[\label{eq:arcsine-cdf-interior} F_X(x) = \frac{1}{\pi}\int_0^x \frac{\mathrm{d}t}{\sqrt{t(1-t)}} = \frac{2}{\pi}\int_0^{\arcsin(\sqrt{x})}\mathrm{d}\theta = \frac{2}{\pi}\arcsin(\sqrt{x}) \; .\]

Finally, $X$ is absolutely continuous and supported on $[0,1]$. Thus, it has no atoms at $0$ or $1$, giving $F_X(x)=0$ for $x\leq0$ and $F_X(x)=1$ for $x\geq1$. This proves \eqref{eq:arcsine-cdf}.

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Metadata: ID: P565 | shortcut: arcsin-cdf | author: syndexco | date: 2026-10-19, 15:50.