Beta distribution with given mean and variance

It occurred to me recently that a problem I solved numerically years ago could be solved analytically, the problem of determining beta distribution parameters so that the distribution has a specified mean and variance. The calculation turns out to be fairly simple. Maybe someone has done it before.

Problem statement

The beta distribution has two positive parameters, a and b, and has probability density proportional to [1]

x^{a-1} (1-x)^{b-1}

for x between 0 and 1.

The mean of a beta(a, b) distribution is

\mu = \frac{a}{a+b}

and the variance is

\sigma^2 = \frac{ab}{(a+b)^2(a+b+1)}

Given μ and σ² we want to solve for a and b. In order for the problem to be meaningful μ must be between 0 and 1, and σ² must be less than μ(1-μ). [2]

As we will see shortly, these two necessary conditions for a solution are also sufficient.


Graphically, we want to find the intersection of a line of constant mean

with a line of constant variance.

Note that the scales in the two plots differ.


If we set

k = \frac{1-\mu}{\mu}

then b = ka and so we can eliminate b from the equation for variance to get

ka^2 = \sigma^2 (1+k)^2 a^2\left((1+k)a + 1 \right)

Now since a > 0, we can divide by a²

k = \sigma^2 (1+k)^2 \left((1+k)a + 1 \right)

and from there solve for a:

a = \frac{k - \sigma^2(1+k)^2}{\sigma^2(1+k)^3} = \mu\left( \frac{\mu(1-\mu)}{\sigma^2} - 1 \right)

and b = ka.

We require σ² to be less than μ(1-μ), or equivalently we require the ratio of μ(1-μ) to σ² to be greater than 1. It works out that the solution a is the product of the mean and the amount by which the ratio of μ(1-μ) to σ² exceeds 1.


Here is a little code to check for errors in the derivation above. It generates μ and σ² values at random, solves for a and b, then checks that the beta(a, b) distribution has the specified mean and variance.

    from scipy.stats import uniform, beta
    for _ in range(100):
        mu = uniform.rvs()
        sigma2 = uniform.rvs(0, mu*(1-mu))
        a = mu*(mu*(1-mu)/sigma2 - 1)
        b = a*(1-mu)/mu
        x = beta(a, b)
        assert( abs(x.mean() - mu) < 1e-10)
        assert( abs(x.var() - sigma2) < 1e-10)

Related posts

[1] It’s often easiest to think of probability densities ignoring proportionality constants. Densities integrate to 1, so the proportionality constants are determined by the rest of the expression for the density. In the case of the beta distribution, the proportionality constant works out to Γ(a + b) / Γ(a) Γ(b).

[2] The variance of a beta distribution factors into μ(1-μ)/(a + b + 1), so it is less than μ(1-μ).

One thought on “Beta distribution with given mean and variance

  1. This result is on the Wikipedia page about the Beta distribution, so indeed someone has done it before, although I don’t know who.

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