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The standard arcsine distribution is a continuous probability distribution on $[0,1]$ with probability density functionIt is the beta distribution with both parameters equal to $1/2$. The density grows without bound toward both endpoints, yet neither endpoint has any probability mass. Its name comes from the inverse sine in its cumulative distribution function.^[ref]

A useful way to see the shape is to choose a point uniformly around a circle and project it onto a diameter. Equal angular intervals do not give equal horizontal distances: the projection moves slowly near either end. An ideal harmonic oscillator sampled at a uniformly chosen time in one complete period has the same endpoint concentration after rescaling its position. The distribution translates a simple geometric slowdown into a precise probability calculation.

For an original comparison, take equal-length regions: the middle half $[1/4,3/4]$, and the two outside quarters together. ThenA uniform distribution on $[0,1]$ instead assigns each region probability $1/2$. Both distributions have mean $1/2$; the mean alone hides their different concentration.

Choose $\Theta$ with uniform distribution on $(0,\pi /2)$ and set $X={\sin }^2\Theta$. Since the sine is increasing on this interval,Differentiation gives $f(x)=1/(\pi \sqrt{x(1-x)})$, and $F(0)=0$, $F(1)=1$ prove normalization. This also evaluates the density's improper integral without treating its divergent endpoint values as infinite probability.

For the circle construction, instead choose $\Phi$ with uniform distribution on $[0,2\pi )$ and put $Y=\cos \Phi$. Each $-1<y<1$ has two preimages in one revolution. The change of variables formula adds both contributions:The affine change of variables $X=(1+Y)/2$ gives the standard arcsine distribution. A single monotone-branch calculation without the second preimage would give only half the total probability.

The same geometry explains the ideal harmonic oscillator $y(t)=A\cos (\omega t+\varphi )$, with amplitude $A>0$ and angular frequency $\omega >0$. Choose $T$ uniformly in any complete period of duration $2\pi /\omega$. Its oscillation phase modulo $2\pi$ is uniform, and $Y=y(T)$ has densityNear a turning point, $|dy/dt|=\omega \sqrt{A^2-y^2}$ is small, so a fixed position interval takes more time to cross. For the chosen amplitude $A=2\mathrm{cm}$, exactly one third of the period is spent in $|y|\le 1\mathrm{cm}$, and two thirds outside that central interval. These are calculations for an ideal sinusoid, rather than measured oscillator data.

The sampling assumption matters. Taking one observation every full period returns the same position each time. It does not produce this probability distribution. Uniformly random observation times and a fixed regularly spaced sampling scheme are different models; aliasing can conceal the motion altogether.

Paul Lévy used a related geometric construction in his work on stochastic processes: project a uniformly chosen point on a circle onto its diameter to divide a segment. The projection description appears explicitly in §5 of his paper, following its arcsine density calculation.^[ref]

Specify whether two parameters mean endpoints or location and width. With endpoints $a<b$, define $Y=a+(b-a)X$ for a standard arcsine distribution variable $X$. ThenThis is a location-scale family with location $a$ and width $b-a$. Its mean is $(a+b)/2$ and its variance is $(b-a{)}^2/8$. A convention using parameters $a,w$ for location and width instead has right endpoint $a+w$, so it must not be read as having endpoints $a,w$.^[ref]

For example, a chosen interval $[-2,6]$ gives mean $2$ and variance $8$, while its density still concentrates toward both endpoints. It is a bounded probability distribution; its endpoint singularities do not make its support infinite.

The beta distribution is an absolutely continuous probability distribution for a random variable $X$ between zero and one. It can describe uncertainty about a probability, such as an unknown coin's chance of heads. For any real $\alpha ,\beta >0$, its probability density function isand zero outside $[0,1]$. The normalizer is the Euler beta function, related to the gamma function by the beta-gamma identity:Thus the probability density function integrates to one. ^[ref].

The endpoint behavior explains why the restrictions are exactly $\alpha ,\beta >0$: near zero the probability density function behaves like a constant times $x^{\alpha -1}$, and near one like a constant times $(1-x{)}^{\beta -1}$. These improper integrals converge for positive parameters, even when the probability density function diverges at zero for $\alpha <1$ or at one for $\beta <1$. There is no probability mass at either endpoint: $\mathbb{P}(X=0)=\mathbb{P}(X=1)=0$. A probability density function's height is not a probability; an interval's probability is the area under it.

Some useful examples of the beta distribution are:

The arcsine distribution curve is plotted only for $0.001\le x\le 0.999$; its finite plotted ends do not represent finite endpoint values of the probability density function. Using the inverse sine, the shaded central interval has exact probability $\mathbb{P}(0.1<X<0.9)=1-4\arcsin (\sqrt{0.1})/\pi \approx 0.5903$, calculated analytically rather than by treating the cropped curve as normalized.

To separate location from spread, write $\kappa =\alpha +\beta$ and $m=\alpha /\kappa$. The mean and variance of the beta distribution areThe mean specifies the balance between the parameters; their sum $\kappa$ controls concentration about that mean. Multiplying both parameters by the same positive number preserves the mean and changes the variance. Larger $\kappa$ puts more area near $m$ and makes the probability density function taller there; the total area stays one. Small $\kappa$ allows much greater spread, sometimes favoring the endpoints rather than the mean itself.

For example, keeping mean $m=0.3$ and taking $\kappa =5,20,80$ gives the beta distributions $\mathrm{Beta}(1.5,3.5)$, $\mathrm{Beta}(6,14)$, and $\mathrm{Beta}(24,56)$. Their variances are $0.035$, $0.01$, and $0.21/81\approx 0.00259259$, respectively. Parameters need not be integers.

The formulas follow from two short moment calculations. Multiplying the probability density function by $x$ or $x^2$ increases the first parameter of the Euler beta function:Apply the Gamma function recurrence to the beta-gamma identity, then subtract $(\mathbb{E}[X]{)}^2$ from $\mathbb{E}[X^2]$ to obtain the variance.

The beta distribution is especially convenient as a prior distribution for the unknown success probability $P$ of Bernoulli trials. The Beta-binomial conjugacy says that $h$ successes and $t$ failures update a prior distribution $\mathrm{Beta}(\alpha ,\beta )$ to a posterior distribution $\mathrm{Beta}(\alpha +h,\beta +t)$. For example, suppose a coin has prior distribution $\mathrm{Beta}(2,2)$ and we observe eight heads and two tails. Its posterior distribution is then $\mathrm{Beta}(10,4)$, with posterior mean $5/7$. ^[ref].

An unbounded probability density function does not imply an atom. HereThe last equivalence follows from $\arcsin u/u\to 1$. The endpoint interval has much more probability than the same interval under a uniform distribution, whose value is $\epsilon$, but its probability still tends to zero. This is an integrable singularity.

For example, the first 1% of the unit interval contains about $6.377\%$ of the probability; the last 1% contains the same amount because the arcsine distribution is a symmetric probability distribution. Each exact endpoint still has zero probability. Density and interval probability answer different questions.

A numerical integration routine must respect these endpoint integrable singularities. The substitution $x={\sin }^2\theta$ givesIt turns a difficult-looking density integral into a constant measure in angle. Integrating finite bins is also a faithful way to draw the distribution near its endpoints; sampling $f(0)$ to make a bar height is not.

A function has an integrable singularity at a point if it becomes unbounded there but its absolute value has a finite improper integral over a sufficiently small surrounding interval, one-sided at an endpoint. The endpoint behavior of the arcsine distribution's probability density function is an example.

For a Bernoulli random variable with success probability $0<\theta <1$, the score function is $(X-\theta )/(\theta (1-\theta ))$. The expected value of its square is $I(\theta )=1/(\theta (1-\theta ))$, since the variance of $X$ is $\theta (1-\theta )$. The Jeffreys prior is therefore the proper beta distribution $\mathrm{Beta}(1/2,1/2)$, also called the arcsine distribution. The Euler beta function has $B(1/2,1/2)=\pi$, giving the stated normalized prior density.

After $s$ successes in $n$ independent Bernoulli trials, Beta-binomial conjugacy gives posterior distribution $\mathrm{Beta}(s+1/2,n-s+1/2)$ and posterior predictive probability $(s+1/2)/(n+1)$ of one further success. Although its prior density diverges at both boundaries, its integral is finite.

Substitute $x={\sin }^{2}v$, with $0<v<\pi /2$, when integrating the target density. Since $dx=2\sin v\cos vdv$, its cumulative distribution function isInverting gives the arcsine distribution samplerThe inverse transform sampling proof in the root solution shows that it has the required probability density function. Use successive independent uniforms for successive samples. A draw at the endpoint one can be discarded if values strictly inside the support are desired; this event has probability zero under the ideal continuous uniform distribution.

The difference of the two eigenvalues is $2i\sin U$, hence $\Delta ={\sin }^{2}U$. Its expectation isFor $0\le x\le 1$, put $\alpha =\arcsin \sqrt{x}\in [0,\pi /2]$. The set of angles satisfying $|\sin u|\le \sqrt{x}$ consists of $[0,\alpha ]$, $[\pi -\alpha ,\pi +\alpha ]$ and $[2\pi -\alpha ,2\pi ]$, of total length $4\alpha$. Thus the full cumulative distribution function isDifferentiating on $(0,1)$ gives the probability density functionThis is the arcsine distribution, equivalently the beta distribution with both parameters $1/2$. It has no atoms at either endpoint.

Interpret the source's missing variable in the limit arrow as $t\to \infty$. We prove convergence in distribution by showing that the angle of the Brownian motion becomes uniform on the circle group.

Let $Z_t=e^{iB(t)}$. For every integer $m$, its angular Fourier coefficients satisfyThese are precisely the Fourier coefficients of the uniform distribution on the unit circle. Thus expectations of every trigonometric polynomial converge to its uniform angular average. By the Stone-Weierstrass theorem, trigonometric polynomials approximate continuous functions on the unit circle uniformly. Given an approximation error $\epsilon$, the discrepancy of expectations is at most the polynomial discrepancy plus $2\epsilon$. Taking $t\to \infty$ and then $\epsilon \to 0$ proves weak convergence of probability measures to uniform angular measure.

The function $z\mapsto (\mathrm{Im}z{)}^2$ is continuous on the unit circle, so the continuous mapping theorem and part (a)(ii) implyEquivalently, the limiting arcsine distribution has cumulative distribution function $2\arcsin \sqrt{x}/\pi$ for $0\le x\le 1$, and probability density function $1/(\pi \sqrt{x(1-x)})$ for $0<x<1$. This proves the limiting law rather than inferring it merely from the limiting mean.

For one Bernoulli random variable,Its variance is $\theta (1-\theta )$, so the expected Fisher information and Jeffreys prior areThe normalizer is $B(1/2,1/2)=\pi$. ThereforeThis Jeffreys prior for a Bernoulli parameter is proper despite its integrable endpoint singularities. It is the arcsine distribution. It replaces the specified $a,b>1$ prior distribution in part (a) when the Jeffreys prior rule is chosen; it is not one of those restricted priors.