Uniform, Triangular, and Arcsine Beta Special Cases

Three Familiar Shapes Within the Beta Family

The beta family describes distributions supported on the interval from zero to one, with two shape parameters controlling how probability is distributed across that interval. A few parameter choices produce especially recognizable shapes: a constant density, a symmetric central bulge, and a density that rises sharply near both ends. Comparing these cases shows why the same support and symmetry do not imply the same concentration of probability.

The three cases here are Uniform(0,1), Beta(2,2), and Beta(1/2,1/2), the last of which is also called the arcsine distribution on (0,1). All three are symmetric around one-half, but their densities and probabilities near zero and one differ substantially.

Definitions and Normalizing Constants

Definition: For shape parameters $a>0$ and $b>0$, a beta random variable $X\sim\operatorname{Beta}(a,b)$ has density $$ f(x)=\frac{x^{a-1}(1-x)^{b-1}}{B(a,b)},\qquad 0<x<1, $$ and density zero outside that interval. Here $$ B(a,b)=\int_0^1 x^{a-1}(1-x)^{b-1}\,dx $$ is the beta function. Values assigned to a density at individual endpoints do not affect the distribution.

For Beta(1,1), the integrand in $B(1,1)$ is one, so $B(1,1)=1$. Its density is constant on (0,1), giving the uniform law. For Beta(2,2), direct integration gives $$ B(2,2)=\int_0^1 x(1-x)\,dx =\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 =\frac{1}{6}. $$ Thus its density is $6x(1-x)$.

For Beta(1/2,1/2), the beta integral is $$ B\left(\frac12,\frac12\right) =\int_0^1 \frac{1}{\sqrt{x(1-x)}}\,dx. $$ Use $x=\sin^2\theta$, where $0<\theta<\pi/2$. Then $dx=2\sin\theta\cos\theta\,d\theta$, while $\sqrt{x(1-x)}=\sin\theta\cos\theta$. Therefore the integral equals $\int_0^{\pi/2}2\,d\theta=\pi$. The resulting density has normalizing constant $1/\pi$.

Result: The three densities, understood to be zero outside (0,1), are $$ \begin{aligned} \operatorname{Uniform}(0,1):\quad & f(x)=1,\\ \operatorname{Beta}(2,2):\quad & f(x)=6x(1-x),\\ \operatorname{Beta}\left(\frac12,\frac12\right):\quad & f(x)=\frac{1}{\pi\sqrt{x(1-x)}}. \end{aligned} $$ The last density is unbounded as $x$ approaches either endpoint, but its integral over (0,1) is one.

Worked Example: Identifying the Density From the Parameters

Suppose $X\sim\operatorname{Beta}(2,2)$. Write its density and verify that it integrates to one.

The beta integral for these parameters is $$ B(2,2)=\int_0^1 x(1-x)\,dx =\frac12-\frac13=\frac16. $$ Substituting into the beta density gives $$ f(x)=\frac{x(1-x)}{1/6}=6x(1-x),\qquad 0<x<1. $$ Its total integral is $$ \int_0^1 6x(1-x)\,dx =6\left(\frac12-\frac13\right) =6\left(\frac16\right)=1. $$ The constant is essential: $x(1-x)$ alone has integral $1/6$, so it is not a probability density.

Symmetry and How the Shapes Differ

A density is symmetric around one-half when reflecting a point $x$ to $1-x$ leaves its value unchanged. For each of the three densities, this holds on (0,1). The uniform density is constant, and for the other two, $$ 6(1-x)x=6x(1-x),\qquad \frac{1}{\pi\sqrt{(1-x)x}}=\frac{1}{\pi\sqrt{x(1-x)}}. $$ The same symmetry also means that reflected intervals have equal probability. Symmetry describes balance across the midpoint; it does not say whether probability is concentrated around the midpoint or toward the endpoints.

Result: Each of the three distributions is symmetric about one-half. In particular, if $0\leq u\leq v\leq1$, the probability of $X\in[u,v]$ equals the probability of $X\in[1-v,1-u]$. This follows by substituting $y=1-x$ in the probability integral and using $f(1-y)=f(y)$.

The density $6x(1-x)$ is small near zero and one and larger toward the middle of the interval. The arcsine density behaves in the opposite broad way: its values increase without bound as $x$ approaches either endpoint. This does not mean there is positive probability at either endpoint. A continuous distribution assigns probability zero to a single point; the important comparison is how much probability falls in intervals near the endpoints.

Density height at one point is not itself probability. Probability comes from integrating density over a set. In particular, the arcsine density’s unbounded endpoint behavior is compatible with a total integral of one because the singularities are integrable.

Worked Example: Comparing Density Heights at the Same Point

Compare the three density values at $x=1/4$.

For Uniform(0,1), $f(1/4)=1$. For Beta(2,2), $$ f\left(\frac14\right) =6\left(\frac14\right)\left(\frac34\right) =\frac{18}{16} =\frac98 =1.125. $$ For the arcsine law, $$ f\left(\frac14\right) =\frac{1}{\pi\sqrt{(1/4)(3/4)}} =\frac{1}{\pi\sqrt{3/16}} =\frac{4}{\pi\sqrt{3}} \approx0.735. $$ Thus at this particular point the Beta(2,2) density is highest and the arcsine density is lowest. This comparison is about density height at $1/4$, not about which distribution places the most probability near the endpoints overall.

Quantifying Probability Near the Endpoints

For $0<\varepsilon\leq1/2$, the two endpoint intervals $(0,\varepsilon)$ and $(1-\varepsilon,1)$ do not overlap. By symmetry, their combined probability is twice the probability of the interval near zero. Direct integration gives the following comparison.

Result: For $0<\varepsilon\leq1/2$, the probability within distance $\varepsilon$ of either endpoint is $$ \begin{aligned} \operatorname{Uniform}(0,1):\quad &P(X<\varepsilon\text{ or }X>1-\varepsilon)=2\varepsilon,\\ \operatorname{Beta}(2,2):\quad &P(X<\varepsilon\text{ or }X>1-\varepsilon)=6\varepsilon^2-4\varepsilon^3,\\ \operatorname{Beta}\left(\frac12,\frac12\right):\quad &P(X<\varepsilon\text{ or }X>1-\varepsilon) =\frac{4}{\pi}\arcsin\sqrt{\varepsilon}. \end{aligned} $$

For the uniform distribution, the probability is the total length of the two intervals, $2\varepsilon$. For Beta(2,2), symmetry and integration give $$ 2\int_0^\varepsilon 6x(1-x)\,dx =12\left(\frac{\varepsilon^2}{2}-\frac{\varepsilon^3}{3}\right) =6\varepsilon^2-4\varepsilon^3. $$ For the arcsine case, an antiderivative of its density is $2\arcsin(\sqrt{x})/\pi$. Indeed, differentiating this expression for $0<x<1$ gives $1/(\pi\sqrt{x(1-x)})$. The probability in one endpoint interval is therefore $2\arcsin(\sqrt{\varepsilon})/\pi$, and doubling yields the stated expression.

Worked Example: Probability Within 0.1 of an Endpoint

Let $\varepsilon=0.1$. The endpoint intervals are $(0,0.1)$ and $(0.9,1)$.

For Uniform(0,1), the combined probability is $$ 2(0.1)=0.2. $$ For Beta(2,2), it is $$ 6(0.1)^2-4(0.1)^3 =0.06-0.004 =0.056. $$ For the arcsine distribution, it is $$ \frac{4}{\pi}\arcsin(\sqrt{0.1}) \approx\frac{4}{\pi}(0.32175) \approx0.410. $$ The comparison is $0.056<0.2<0.410$. In this interval, the Beta(2,2) law assigns the least probability near the endpoints, the uniform law assigns an intermediate amount, and the arcsine law assigns the most.

Worked Example: Using the Arcsine CDF for One Endpoint

For $X\sim\operatorname{Beta}(1/2,1/2)$, calculate $P(X<1/4)$.

Using the antiderivative above, the cumulative distribution function for $0<x<1$ is $$ F(x)=\frac{2}{\pi}\arcsin(\sqrt{x}). $$ Consequently, $$ P\left(X<\frac14\right) =F\left(\frac14\right) =\frac{2}{\pi}\arcsin\left(\frac12\right) =\frac{2}{\pi}\cdot\frac{\pi}{6} =\frac13. $$ By symmetry, $P(X>3/4)=1/3$ as well. The two intervals are disjoint, so their combined probability is $2/3$.

A Naming Pitfall and When the Comparison Matters

Beta(2,2) is sometimes described informally as a symmetric triangular-shaped beta density because it is low at the ends and higher in the middle. However, its graph is curved: $6x(1-x)$ is a quadratic function. The conventional symmetric triangular distribution on [0,1] has a different, piecewise-linear density: $$ g(x)= \begin{cases} 4x,&0<x\leq1/2,\\ 4(1-x),&1/2<x<1. \end{cases} $$ These are distinct probability distributions. Calling Beta(2,2) simply “the triangular distribution” can therefore be misleading; specifying the beta parameters avoids ambiguity.

The distinction among these shapes matters when a model represents a proportion, a fractional measurement, or another quantity restricted to [0,1]. A uniform model gives equal density throughout the interval. Beta(2,2) gives relatively more weight to central values, whereas the arcsine law gives relatively more probability to values near zero and one. The appropriate choice depends on the intended distribution, not just on the shared support or symmetry.

Uniform(0,1): constant density and endpoint-interval probability $2\varepsilon$.
Beta(2,2): density $6x(1-x)$, with less probability near the endpoints than the uniform law for the comparison above.
Beta(1/2,1/2): integrable endpoint singularities and substantially more probability close to either endpoint.

Check Your Understanding

Use the definitions and comparisons in this tutorial to answer the following.

What are the density and beta-function normalizing constant for Beta(2,2)?
Why does an unbounded arcsine density near zero not imply positive probability at zero?
Explain why symmetry around one-half does not determine how much probability lies near the endpoints.
For $0<\varepsilon\leq1/2$, derive the combined endpoint probability for Beta(2,2) by integrating its density.
How does the conventional symmetric triangular distribution differ from Beta(2,2)?