Modes and Boundary Behavior of Beta Densities

Reading the Shape of a Beta Density

A beta density can peak in the interior of its support, reach its largest finite value at an endpoint, or become unbounded near one or both endpoints. These possibilities are controlled by whether each shape parameter is below, equal to, or above 1. The mode describes where the density is largest; it is not in general the mean.

Throughout, let $X\sim\operatorname{Beta}(\alpha,\beta)$, with $\alpha>0$ and $\beta>0$. We examine the canonical density on $0<x<1$ and its one-sided limits at 0 and 1. Because a density is defined only up to changes on sets of probability zero, endpoint behavior here means the limiting behavior of this canonical density, not an arbitrary assigned value at an endpoint.

Definition: The beta density is $$ f(x)=\frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)},\qquad 0<x<1, $$ where $B(\alpha,\beta)>0$ is the beta function. An interior mode is an interior point at which this density attains its maximum. A finite endpoint mode refers to a finite endpoint limit that is at least as large as every interior density value. If the density diverges at an endpoint, we call that endpoint a boundary singularity; the density has no finite maximum value there.

The beta function is a positive constant with respect to $x$, so it affects the height of the density but not the location of an interior stationary point. The factors $x^{\alpha-1}$ and $(1-x)^{\beta-1}$ determine what happens as $x$ approaches the two endpoints.

Interior Stationary Points

On $0<x<1$, the density is positive and differentiable. It is therefore convenient to differentiate its logarithm: this gives the same stationary points as differentiating the density, while keeping the algebra simple.

Theorem: If $\alpha+\beta\ne2$, the only possible interior stationary point is $$ x_*=\frac{\alpha-1}{\alpha+\beta-2}, $$ and it is stationary exactly when $x_*\in(0,1)$. If $\alpha+\beta=2$, there is no interior stationary point unless $\alpha=\beta=1$; in that uniform case, every point in $(0,1)$ is stationary.

Proof. Logarithmic differentiation gives $$ \frac{d}{dx}\log f(x)=\frac{\alpha-1}{x}-\frac{\beta-1}{1-x}. $$ Since $f(x)>0$, $f'(x)=0$ exactly when this derivative is zero. Multiplying by $x(1-x)>0$, the stationary-point equation becomes $$ (\alpha-1)(1-x)-(\beta-1)x=0, $$ or, equivalently, $$ (\alpha-1)-(\alpha+\beta-2)x=0. $$ If $\alpha+\beta\ne2$, this linear equation has the unique solution $x_*$ displayed in the theorem; it is an interior stationary point precisely when that solution lies in $(0,1)$. If $\alpha+\beta=2$, the left side of the last equation is the constant $\alpha-1$. It has no solution if $\alpha\ne1$. If $\alpha=1$, then $\beta=1$ as well, and the density is the constant $1$ on $(0,1)$, so every interior point is stationary. $\square$

The formula for $x_*$ is a candidate, not by itself a conclusion that the point is a maximum. In fact, it is an interior maximum when both shape parameters exceed 1, and an interior minimum when both are below 1. When one parameter is on each side of 1, there is no interior stationary point.

Worked Example: An Interior Mode for Beta(3, 4)

Let $X\sim\operatorname{Beta}(3,4)$. Since both parameters exceed 1, an interior mode is expected. The stationary point is $$ x_*=\frac{3-1}{3+4-2}=\frac{2}{5}. $$ To verify that it is a maximum, differentiate the log density twice: $$ \frac{d^2}{dx^2}\log f(x)=-\frac{2}{x^2}-\frac{3}{(1-x)^2}<0,\qquad 0<x<1. $$ Thus the log density is strictly concave, and its stationary point is a strict maximum. The density also tends to 0 at both endpoints because both parameters exceed 1. Hence $2/5$ is the unique global mode.

The mean of this distribution is $$ E[X]=\frac{\alpha}{\alpha+\beta}=\frac{3}{7}. $$ Here $2/5\ne3/7$: the mode locates the peak of the density, while the mean is its probability-weighted average.

Endpoint Limits and Boundary Singularities

Near 0, the factor $(1-x)^{\beta-1}$ tends to 1, so the behavior is governed by $x^{\alpha-1}$. Near 1, the corresponding behavior is governed by $(1-x)^{\beta-1}$. Because $B(\alpha,\beta)$ is finite and positive for positive parameters, it does not alter whether either limit is zero, finite, or infinite.

Theorem: The beta density has the following one-sided endpoint limits: $$ \lim_{x\downarrow0}f(x)= \begin{cases} +\infty,&0<\alpha<1,\\ \beta,&\alpha=1,\\ 0,&\alpha>1, \end{cases} \qquad \lim_{x\uparrow1}f(x)= \begin{cases} +\infty,&0<\beta<1,\\ \alpha,&\beta=1,\\ 0,&\beta>1. \end{cases} $$

Proof. As $x\downarrow0$, $(1-x)^{\beta-1}\to1$, so the limit is determined by $x^{\alpha-1}/B(\alpha,\beta)$. A positive power tends to 0, a zero power is 1, and a negative power diverges to $+\infty$. When $\alpha=1$, $B(1,\beta)=1/\beta$, so the finite limit is $\beta$. For the other endpoint, $x^{\alpha-1}\to1$ as $x\uparrow1$. The same power argument applied to $(1-x)^{\beta-1}$ gives zero, a finite limit, or divergence according as $\beta>1$, $\beta=1$, or $\beta<1$. Since $B(\alpha,1)=1/\alpha$, the finite limit when $\beta=1$ is $\alpha$. $\square$

Worked Example: A Boundary Singularity for Beta(1/2, 2)

For $X\sim\operatorname{Beta}(1/2,2)$, the density is $$ f(x)=\frac{x^{-1/2}(1-x)}{B(1/2,2)},\qquad 0<x<1. $$ As $\alpha=1/2<1$, the density diverges at 0; as $\beta=2>1$, it tends to 0 at 1. To see whether the density has an interior peak, its log derivative is $$ \frac{d}{dx}\log f(x)=-\frac{1}{2x}-\frac{1}{1-x}<0,\qquad 0<x<1. $$ The density is strictly decreasing throughout the interior. Thus the boundary at 0 is singular, with no finite maximum density value, and there is no interior mode. Its mean is $$ E[X]=\frac{\alpha}{\alpha+\beta}=\frac{1/2}{5/2}=\frac15. $$ The mean is finite even though the density becomes unbounded near zero.

Classifying Maxima, Minima, and Monotone Shapes

The second derivative of the log density is $$ \frac{d^2}{dx^2}\log f(x)=-\frac{\alpha-1}{x^2}-\frac{\beta-1}{(1-x)^2}. $$ If both parameters exceed 1, this expression is strictly negative. Together with the zero limits at both endpoints, that proves the unique stationary point is the unique global maximum. If both parameters are below 1, it is strictly positive; the stationary point is instead a minimum, while the density diverges at both boundaries. If one parameter is below 1 and the other exceeds 1, the log derivative has one sign throughout the interval, so the density is monotone and has no stationary point.

The cases where a parameter equals 1 are simpler to read from the density itself. If $\alpha=1$, then $f(x)$ is proportional to $(1-x)^{\beta-1}$: it decreases when $\beta>1$, increases when $\beta<1$, and is constant when $\beta=1$. If $\beta=1$, the density is proportional to $x^{\alpha-1}$, with the corresponding increasing, decreasing, or constant behavior.

Worked Example: An Interior Minimum for Beta(1/2, 1/2)

For $X\sim\operatorname{Beta}(1/2,1/2)$, the stationary point is $$ x_*=\frac{1/2-1}{1/2+1/2-2}=\frac{-1/2}{-1}=\frac12. $$ The second derivative of the log density is $$ \frac{1}{2x^2}+\frac{1}{2(1-x)^2}>0. $$ Therefore $x=1/2$ is a strict interior minimum, not a mode. Both shape parameters are below 1, so the density diverges at both endpoints. The mean is $1/2$, but here it happens to equal the location of the minimum. This example makes clear that equality with the mean does not establish that a point is a mode.

Worked Example: A Finite Endpoint Mode for Beta(1, 3)

For $X\sim\operatorname{Beta}(1,3)$, the density simplifies to $$ f(x)=3(1-x)^2,\qquad 0<x<1, $$ because $B(1,3)=1/3$. Its derivative is $f'(x)=-6(1-x)<0$ in the interior. The density decreases from the finite limit $f(0+)=3$ to $f(1-)=0$. Thus the left endpoint is a finite endpoint mode, not a boundary singularity. The mean is $1/(1+3)=1/4$, which differs from the endpoint mode.

Mode and Mean Are Different Summaries

For positive $\alpha,\beta$, the beta mean is $\alpha/(\alpha+\beta)$, as established in the earlier material on beta moments. It is a balance point determined by the distribution’s overall probability mass. A mode, by contrast, is a location where the density reaches its largest value when a finite maximum exists. The two quantities need not coincide, and the beta distribution may have an interior mode, a finite endpoint mode, a flat density, or an unbounded boundary singularity.

For $\alpha,\beta>1$, the interior mode is $(\alpha-1)/(\alpha+\beta-2)$, whereas the mean is $\alpha/(\alpha+\beta)$. For example, with $(\alpha,\beta)=(3,4)$, they are $2/5$ and $3/7$, respectively. For other parameter ranges, the density may not have a finite mode at all, even though its mean remains well-defined.

Common Mistakes

Treating every stationary point as a mode. When both parameters are below 1, the interior stationary point is a minimum, as the positive second derivative of the log density shows.
Using the stationary-point formula without checking its conditions. The formula applies when $\alpha+\beta\ne2$, and its value must lie in $(0,1)$. When $\alpha=\beta=1$, every interior point is stationary instead.
Calling a divergent boundary value a finite mode. If a shape parameter is below 1, the density diverges at its associated endpoint. Describe this as a boundary singularity rather than as a finite maximum.
Assuming the mean is the mode. These are different summaries. Calculate or classify the density’s maximum separately from the mean.

Key takeaway: Compare each shape parameter with 1 to determine its endpoint behavior. An interior mode occurs only when both parameters exceed 1; the uniform case is flat, and other parameter combinations yield an interior minimum, monotonicity, or boundary singularities.

Check Your Understanding

Use the shape parameters and, where needed, the density’s derivative to justify each answer.

For $\operatorname{Beta}(4,2)$, find the interior stationary point and classify it.
What are the endpoint limits for $\operatorname{Beta}(2,1)$, and where is its finite density maximum?
For $\operatorname{Beta}(0.7,0.8)$, is the interior stationary point a maximum or minimum? What happens at each endpoint?
Explain why the stationary-point formula does not identify a unique point for $\operatorname{Beta}(1,1)$.
For $\operatorname{Beta}(1,4)$, compare the mode with the mean and state whether the endpoint behavior is singular.