Beta Densities and Shape Parameters

Introduction

A probability density on $(0,1)$ can describe quantities such as a proportion, a probability, or a fraction of a whole. The Beta family is especially useful because its two positive shape parameters can describe a wide range of patterns on this interval: density can be relatively central or spread out, and it can be weighted more toward one side than the other.

The essential construction is simple. Start with the nonnegative function $x^{a-1}(1-x)^{b-1}$ on $(0,1)$, where $a$ and $b$ are positive. The beta integral supplies the constant that makes its total integral equal to one. The parameters then control the two factors separately: $a$ controls the power of $x$, while $b$ controls the power of $1-x$.

Definition and Normalization

Definition: For $a>0$ and $b>0$, the Beta distribution with shape parameters $a$ and $b$, written $X\sim\operatorname{Beta}(a,b)$, has density $$ f_{a,b}(x)=\frac{1}{B(a,b)}x^{a-1}(1-x)^{b-1},\qquad 0<x<1, $$ and density zero outside $(0,1)$. Here $$ B(a,b)=\int_0^1 x^{a-1}(1-x)^{b-1}\,dx =\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}. $$

The integral $B(a,b)$ is the beta function. For positive $a,b$, it is finite and strictly positive: the beta integral converges at both endpoints, and its integrand is positive throughout $(0,1)$. The beta-gamma identity, established with the beta integral, gives its expression in terms of the gamma function. Consequently $1/B(a,b)$ is a well-defined positive constant.

To verify normalization directly, integrate the proposed density over its support:

\int_{-\infty}^{\infty} f_{a,b}(x)\,dx =\frac{1}{B(a,b)}\int_0^1x^{a-1}(1-x)^{b-1}\,dx =\frac{B(a,b)}{B(a,b)} =1.

Thus the function is a probability density. Values assigned at the endpoints $0$ and $1$ do not affect the distribution, since changing a density at finitely many points does not change any integral. The formula is stated on the open interval to avoid needing separate endpoint conventions.

Worked Example: The Uniform Case

Find the density for $X\sim\operatorname{Beta}(1,1)$.

For these parameters, $$ B(1,1)=\int_0^1 x^0(1-x)^0\,dx=\int_0^1 1\,dx=1. $$ Therefore the density is $$ f_{1,1}(x)=1,\qquad 0<x<1. $$ It integrates to one because the interval has length one. This example shows that the Beta family includes the uniform density on $(0,1)$.

Worked Example: A Density Weighted Toward Smaller Values

Normalize the function $x(1-x)^4$ on $(0,1)$, and evaluate its density at $x=1/5$.

The powers identify the parameters: $a-1=1$, so $a=2$, and $b-1=4$, so $b=5$. The beta-gamma identity gives $$ B(2,5)=\frac{\Gamma(2)\Gamma(5)}{\Gamma(7)} =\frac{1!\,4!}{6!} =\frac{24}{720} =\frac{1}{30}. $$ Hence $$ f_{2,5}(x)=30x(1-x)^4,\qquad 0<x<1. $$ Its integral is $30B(2,5)=30(1/30)=1$, confirming the normalizing constant. At $x=1/5$, $$ f_{2,5}(1/5) =30\left(\frac15\right)\left(\frac45\right)^4 =6\cdot\frac{256}{625} =\frac{1536}{625} =2.4576. $$ This density is larger at $1/5$ than at $4/5$: the latter value is $30(4/5)(1/5)^4=24/625=0.0384$. The unequal powers create a strong left-right imbalance in density values.

What the Shape Parameters Control

Each parameter appears in a different factor. If $a$ is larger while $b$ is held fixed, the factor $x^{a-1}$ gives relatively more weight to larger $x$ values than to smaller ones. If $b$ is larger while $a$ is held fixed, $(1-x)^{b-1}$ gives relatively more weight to smaller $x$ values. These comparisons can be made precisely by taking ratios of densities.

For instance, for $a_2>a_1>0$ and the same $b>0$, $$ \frac{f_{a_2,b}(x)}{f_{a_1,b}(x)} =\frac{B(a_1,b)}{B(a_2,b)}x^{a_2-a_1}. $$ The factor depending on $x$ increases as $x$ increases. The ratio therefore compares the two densities in a way that favors larger values under the larger first parameter. Similarly, for $b_2>b_1>0$, $$ \frac{f_{a,b_2}(x)}{f_{a,b_1}(x)} =\frac{B(a,b_1)}{B(a,b_2)}(1-x)^{b_2-b_1}, $$ whose $x$-dependent factor decreases as $x$ increases. These are relative-density comparisons; they do not mean the density is larger everywhere, since both densities must still integrate to one.

Equal parameters give a symmetric density about $1/2$. When both equal parameters increase, the density becomes more concentrated around the center. The following result makes this comparison precise for central intervals, without requiring a formula for a mode or moments.

Theorem: Let $0<k_1<k_2$, and let $X_j\sim\operatorname{Beta}(k_j,k_j)$ for $j=1,2$. For every $c$ with $0<c<1/2$, $$ P\left(\left|X_2-\frac12\right|\le c\right) > P\left(\left|X_1-\frac12\right|\le c\right). $$ Thus increasing the common shape parameter increases the probability in every fixed nontrivial central interval.

Proof. Write $g_j$ for the density of $X_j$. Apart from its normalizing constant, $g_j(x)$ is $[x(1-x)]^{k_j-1}$. Set $d=k_2-k_1>0$ and $r(x)=[x(1-x)]^d$. Then $$ g_2(x)=C\,r(x)g_1(x),\qquad C=\frac{B(k_1,k_1)}{B(k_2,k_2)}>0. $$ Let $A=[1/2-c,1/2+c]$, the central interval, and let $A^c$ denote its complement within $(0,1)$. The quantity $x(1-x)=1/4-(x-1/2)^2$ is larger on $A$ than on $A^c$, except at their shared boundary points. Therefore $r(x)$ is strictly larger at every interior point of $A$ than at every point of $A^c$, apart from boundary equality.

Both sets have positive probability under $g_1$. Let $m_A$ and $m_{A^c}$ be the conditional averages of $r(X_1)$ on these sets. The preceding strict comparison gives $m_A>m_{A^c}$. Since $P(X_1\in A)+P(X_1\in A^c)=1$, this also gives $$ m_A>E[r(X_1)] =P(X_1\in A)m_A+P(X_1\in A^c)m_{A^c}. $$ Using the density ratio and cancelling its normalizing factor, $$ P(X_2\in A) =\frac{P(X_1\in A)m_A}{E[r(X_1)]} >P(X_1\in A). $$ This proves the claim. $\square$

This result gives a probability-based meaning to increasing central concentration in the symmetric case. In particular, for a fixed $x\ne1/2$, the ratio of the symmetric density at $x$ to its value at the center is $$ \frac{f_{k,k}(x)}{f_{k,k}(1/2)} =\left[4x(1-x)\right]^{k-1}. $$ For $x\ne1/2$, the bracket lies strictly between zero and one. As $k$ increases, this ratio decreases. The density increasingly favors central values relative to that fixed point away from the center.

Worked Example: Comparing Two Symmetric Shapes

Compare the relative density at $x=1/5$ with the density at $x=1/2$ for $\operatorname{Beta}(2,2)$ and $\operatorname{Beta}(10,10)$.

For any symmetric $\operatorname{Beta}(k,k)$ density, the ratio is $[4x(1-x)]^{k-1}$. At $x=1/5$, $4x(1-x)=4(1/5)(4/5)=16/25=0.64$. Thus for $k=2$, the ratio is $0.64$. For $k=10$, it is $$ (0.64)^9\approx0.018014. $$ So the density at $1/5$, relative to the density at $1/2$, is much smaller for $\operatorname{Beta}(10,10)$.

The normalizing constant also illustrates the density scale. Since $B(10,10)=9!9!/19!$, $$ \frac{1}{B(10,10)} =\frac{19!}{9!9!} =19\binom{18}{9} =19(48620) =923780. $$ It follows that $$ f_{10,10}(1/2) =923780\left(\frac12\right)^{18} =\frac{923780}{262144} \approx3.524. $$ For comparison, $f_{2,2}(1/2)=6(1/2)(1/2)=1.5$. The taller central density for the larger equal parameters is consistent with the increased central concentration; the relative-density calculation above quantifies how much more the density favors the center over $1/5$.

Unequal Parameters and Asymmetry

To describe left-right asymmetry directly, compare the density at $x$ with the density at the reflected point $1-x$. First, the beta integral has the symmetry $B(a,b)=B(b,a)$: substituting $u=1-x$ in its defining integral interchanges $a$ and $b$. Consequently, $$ \frac{f_{a,b}(x)}{f_{a,b}(1-x)} =\left(\frac{x}{1-x}\right)^{a-b},\qquad 0<x<1. $$ This ratio is greater than one for $x>1/2$ when $a>b$, and less than one there when $a<b$. For $x<1/2$, the comparisons reverse. Thus $a>b$ weights reflected density comparisons toward the right side, whereas $b>a$ weights them toward the left. If $a=b$, the ratio is one at every reflected pair, so the density is symmetric.

This ratio describes density asymmetry, not a moment-based numerical measure of skewness. It is enough to show why exchanging $a$ and $b$ reverses the left-right imbalance. It also provides a useful check on calculations: a claimed density for $a>b$ should satisfy the displayed ratio.

Worked Example: Reversing the Imbalance

Compare $\operatorname{Beta}(2,5)$ and $\operatorname{Beta}(5,2)$.

The first density was normalized as $f_{2,5}(x)=30x(1-x)^4$. Since $B(5,2)=B(2,5)=1/30$, the second density is $$ f_{5,2}(x)=30x^4(1-x),\qquad 0<x<1. $$ At $x=1/5$, the first density is $1536/625=2.4576$, while the second is $$ 30\left(\frac15\right)^4\left(\frac45\right) =\frac{24}{625} =0.0384. $$ At the reflected point $x=4/5$, these two values swap. More generally, the density ratio for $\operatorname{Beta}(2,5)$ is $$ \frac{f_{2,5}(x)}{f_{2,5}(1-x)} =\left(\frac{x}{1-x}\right)^{-3}. $$ At $x=4/5$, this is $(4)^{-3}=1/64$, matching $0.0384/2.4576=1/64$. The unequal shape parameters produce opposite imbalances when exchanged.

Common Pitfalls and Why the Beta Family Matters

Forgetting the normalizing constant. The function $x^{a-1}(1-x)^{b-1}$ is not generally a density by itself. Its integral is $B(a,b)$, so it must be divided by this quantity.
Interchanging the parameters without changing the interpretation. The exponent $a-1$ belongs to $x$, and $b-1$ belongs to $1-x$. Swapping $a$ and $b$ reverses the density imbalance.
Assuming that increasing a parameter raises the density everywhere. Normalization forces the total area to remain one. A parameter change redistributes density rather than uniformly increasing it.
Confusing asymmetry with a particular endpoint value. The density ratio at reflected points gives a precise comparison across the interval. One value alone does not establish the overall shape.

The Beta family is useful whenever a random quantity must lie between zero and one and different plausible shapes are needed. The parameters provide a flexible way to describe symmetric concentration or left-right imbalance while retaining a tractable normalizing constant. This tutorial has focused on the density and its shape parameters; probabilities from the density require integrating over a specified interval.

Key takeaway: For $a,b>0$, the Beta density is $x^{a-1}(1-x)^{b-1}/B(a,b)$ on $(0,1)$. Equal parameters give symmetry, larger equal parameters increase central concentration, and unequal parameters create a measurable left-right density imbalance.

Check Your Understanding

Try these questions using the density definition and the beta integral.

For $X\sim\operatorname{Beta}(3,2)$, write the unnormalized density kernel and identify which factor has the larger exponent.
Why does dividing $x^{a-1}(1-x)^{b-1}$ by $B(a,b)$ produce a density for positive $a$ and $b$?
For a symmetric $\operatorname{Beta}(k,k)$ density, what is the ratio $f_{k,k}(x)/f_{k,k}(1/2)$? What happens to this ratio as $k$ increases for a fixed $x\ne1/2$?
Use the reflected-density ratio to compare $f_{3,1}(x)$ and $f_{3,1}(1-x)$ when $x>1/2$.
Explain why a larger shape parameter does not imply a larger density at every point of the interval.