Beta Distribution Moments

Why Beta Moments Matter

A beta random variable takes values between 0 and 1, so it can represent a random fraction or proportion. Its shape parameters determine not only the distribution’s form but also its average and variability. Computing moments gives concise summaries of those features and supplies quantities used in later statistical calculations.

This tutorial focuses on deriving and calculating raw moments, the mean, and the variance. It assumes the beta density and beta function have been introduced. No probability calculations from the beta density are needed here; the key technique is to express an integral for a moment as another beta-function integral.

Definition: If \(X\sim\operatorname{Beta}(\alpha,\beta)\), where \(\alpha>0\) and \(\beta>0\), its density on \(0<x<1\) is \(f_X(x)=x^{\alpha-1}(1-x)^{\beta-1}/B(\alpha,\beta)\). The beta function is \(B(a,b)=\int_0^1 x^{a-1}(1-x)^{b-1}\,dx\), for \(a,b>0\). The raw moment of integer order \(r\geq 0\) is \(E[X^r]\).

Because the beta density is supported on the bounded interval from 0 to 1, every nonnegative integer raw moment exists. In particular, \(E[X^0]=1\). For positive integer orders, the density definition turns the expected value into an integral whose exponents can be matched to the beta function.

The Beta-Function Identity Behind the Moments

The useful recurrence is \(B(a+1,b)=\frac{a}{a+b}B(a,b)\). It relates a beta function with its first argument increased by one to the original beta function. Here is a derivation for \(a,b>0\).

Split the defining integral.
Since \(1=x+(1-x)\), the integrand for \(B(a,b)\) gives \(B(a,b)=B(a+1,b)+B(a,b+1)\).

Relate the two terms by integration by parts.
Integrating \(B(a+1,b)=\int_0^1 x^a(1-x)^{b-1}\,dx\) by parts, with \(u=x^a\) and \(dv=(1-x)^{b-1}\,dx\), gives \(B(a+1,b)=\frac{a}{b}B(a,b+1)\). The boundary term is zero because \(a,b>0\).

Solve the two relations.
Substitute \(B(a,b+1)=\frac{b}{a}B(a+1,b)\) into the split-integral identity. Then \(B(a,b)=B(a+1,b)(1+b/a)=B(a+1,b)(a+b)/a\), which rearranges to the stated recurrence.

Applying the recurrence repeatedly gives a ratio of products. Define the rising product \((a)^{\overline r}=a(a+1)\cdots(a+r-1)\) for positive integer \(r\), with \((a)^{\overline 0}=1\). The moment formula follows directly from the density:

E[X^r] =\int_0^1 x^r f_X(x)\,dx =\frac{B(\alpha+r,\beta)}{B(\alpha,\beta)} =\frac{(\alpha)^{\overline r}}{(\alpha+\beta)^{\overline r}}, \qquad r=0,1,2,\ldots.

For example, the last ratio is obtained by applying the recurrence \(r\) times: each application contributes a numerator factor \(\alpha+j\) and a denominator factor \(\alpha+\beta+j\), for \(j=0,\ldots,r-1\). The expression with beta functions is often the clearest starting point; the rising-product form is usually quicker for numerical calculations.

Mean and Variance

The mean is the first raw moment. The variance is not the second raw moment: it is the second central moment, found by subtracting the square of the mean from the second raw moment. Using the formula above,

E[X]=\frac{\alpha}{\alpha+\beta}, \qquad E[X^2]=\frac{\alpha(\alpha+1)} {(\alpha+\beta)(\alpha+\beta+1)}.

Therefore,

\begin{aligned} \operatorname{Var}(X) &=E[X^2]-\big(E[X]\big)^2\\ &=\frac{\alpha(\alpha+1)} {(\alpha+\beta)(\alpha+\beta+1)} -\frac{\alpha^2}{(\alpha+\beta)^2}\\ &=\frac{\alpha\beta} {(\alpha+\beta)^2(\alpha+\beta+1)}. \end{aligned}

To see the simplification, put the two terms over the common denominator \((\alpha+\beta)^2(\alpha+\beta+1)\). The numerator becomes \(\alpha(\alpha+1)(\alpha+\beta)-\alpha^2(\alpha+\beta+1)\). Expanding and cancelling leaves \(\alpha\beta\). Since both parameters are positive, this variance is positive, as it must be for a non-degenerate beta distribution.

Theorem: For \(X\sim\operatorname{Beta}(\alpha,\beta)\) with \(\alpha,\beta>0\), \(E[X^r]=\frac{(\alpha)^{\overline r}}{(\alpha+\beta)^{\overline r}}\) for every nonnegative integer \(r\). In particular, \(E[X]=\frac{\alpha}{\alpha+\beta}\) and \(\operatorname{Var}(X)=\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}\).

Worked Calculations

Worked Example: First Three Moments for a Beta Model

A hypothetical model represents the fraction of a process completed by a randomly selected review point as \(X\sim\operatorname{Beta}(2,3)\). Calculate the first three raw moments, the mean, and the variance. Name the method and check that the variance is consistent with the raw moments.

Method. Apply the beta-function ratio, or equivalently the rising-product formula, separately at each order. Here \(\alpha=2\), \(\beta=3\), and \(\alpha+\beta=5\).

For order one, \(E[X]=2/5=0.4\). For order two,

E[X^2]=\frac{2\cdot3}{5\cdot6}=\frac{6}{30}=\frac{1}{5}=0.2.

For order three,

E[X^3]=\frac{2\cdot3\cdot4}{5\cdot6\cdot7} =\frac{24}{210}=\frac{4}{35}\approx0.1143.

Now calculate the variance from the first two raw moments:

\operatorname{Var}(X)=E[X^2]-\big(E[X]\big)^2 =\frac{1}{5}-\left(\frac{2}{5}\right)^2 =\frac{5}{25}-\frac{4}{25} =\frac{1}{25}=0.04.

The parameter formula checks this result: \(\frac{2\cdot3}{5^2\cdot6}=\frac{6}{150}=\frac{1}{25}\). Both methods agree. The common mistake to avoid is reporting \(E[X^2]=0.2\) as the variance; the correct variance is \(0.2-0.4^2=0.04\).

Worked Example: Higher Moments With Different Shape Parameters

A hypothetical calibration model has a random fraction \(X\sim\operatorname{Beta}(4,2)\). Calculate \(E[X]\), \(E[X^2]\), \(E[X^3]\), and \(\operatorname{Var}(X)\). Also verify the variance using the parameter formula.

Method. The shape parameters are \(\alpha=4\) and \(\beta=2\), so their sum is \(6\). Use the rising-product formula for each raw moment:

E[X]=\frac{4}{6}=\frac{2}{3}, \qquad E[X^2]=\frac{4\cdot5}{6\cdot7} =\frac{20}{42}=\frac{10}{21}.

The third raw moment is

E[X^3]=\frac{4\cdot5\cdot6}{6\cdot7\cdot8} =\frac{120}{336}=\frac{5}{14}.

Calculate the variance using the raw moments:

\begin{aligned} \operatorname{Var}(X) &=\frac{10}{21}-\left(\frac{2}{3}\right)^2\\ &=\frac{30}{63}-\frac{28}{63} =\frac{2}{63}\approx0.03175. \end{aligned}

The parameter formula gives \(\frac{4\cdot2}{6^2\cdot7}=\frac{8}{252}=\frac{2}{63}\), confirming the result. The third raw moment is \(5/14\approx0.3571\); it is larger than the variance, which is possible because a raw moment and a central moment measure different quantities. The common mistake here is to use \(E[X^2]-(E[X])^2\) but then simplify the fractions with different denominators incorrectly.

Worked Example: A Higher-Order Moment and a Scale Check

A hypothetical index \(X\) follows \(\operatorname{Beta}(2,5)\), and a score is defined as \(W=100X\). Calculate \(E[X^4]\), \(E[X]\), \(\operatorname{Var}(X)\), and the mean and variance of \(W\). Keep exact fractions until the final decimal values.

Method. For \(X\), the parameter sum is \(7\). The fourth raw moment is

E[X^4] =\frac{2\cdot3\cdot4\cdot5}{7\cdot8\cdot9\cdot10} =\frac{120}{5040} =\frac{1}{42}\approx0.02381.

The mean and variance of \(X\) are

E[X]=\frac{2}{7}\approx0.2857, \qquad \operatorname{Var}(X) =\frac{2\cdot5}{7^2\cdot8} =\frac{10}{392} =\frac{5}{196}\approx0.02551.

For a constant multiple \(W=100X\), expectation is multiplied by 100 and variance by \(100^2\). Thus

E[W]=100\cdot\frac{2}{7}=\frac{200}{7}\approx28.57, \qquad \operatorname{Var}(W)=100^2\cdot\frac{5}{196} =\frac{12500}{49}\approx255.1.

The variance scaling uses the square of the multiplier, not the multiplier itself. As a check, \(E[W]\) is 100 times the mean of \(X\), while the variance is 10,000 times the variance of \(X\). The common mistake is to multiply the variance by 100; doing so would give \(2.551\), a factor of 100 too small.

Common Mistakes and Checks

Moment calculations are short, but several errors can change the result substantially. Before finalizing an answer, verify the parameter sum, the moment order, and whether the requested quantity is raw or central.

Confusing raw moment and variance. \(E[X^2]\) is not \(\operatorname{Var}(X)\). Subtract \((E[X])^2\) to obtain the variance.
Using the wrong denominator factors. For \(E[X^r]\), the denominator is \((\alpha+\beta)(\alpha+\beta+1)\cdots(\alpha+\beta+r-1)\). It is not just \((\alpha+\beta)^r\).
Stopping before the requested order. The formula for \(E[X^r]\) has \(r\) factors in both rising products. For example, the third moment has three factors, not four.
Forgetting parameter restrictions. The beta density and the recurrence used here require \(\alpha>0\) and \(\beta>0\). The formulas do not describe a beta distribution with a zero or negative shape parameter.
Scaling variance as if it were a mean. If a variable is multiplied by a constant \(c\), its mean is multiplied by \(c\), but its variance is multiplied by \(c^2\).

Key takeaway: Start with \(E[X^r]=B(\alpha+r,\beta)/B(\alpha,\beta)\), then use the beta-function recurrence to obtain the rising-product formula. Compute variance as \(E[X^2]-(E[X])^2\), or use its equivalent parameter formula, and check that both calculations agree.

Check Your Understanding

For each problem, use the beta raw-moment formula and show enough intermediate work to check the parameter sum and moment order.

Let \(X\sim\operatorname{Beta}(2,4)\). Calculate its mean and variance. (Answer: \(E[X]=1/3\), \(\operatorname{Var}(X)=2/63\).)
Let \(Y\sim\operatorname{Beta}(5,3)\). Calculate \(E[Y^2]\) and \(E[Y^3]\). (Answer: \(E[Y^2]=5/12\), \(E[Y^3]=7/24\).)
For \(Z\sim\operatorname{Beta}(3,2)\), calculate the second raw moment and then use it to find the variance. (Answer: \(E[Z^2]=2/5\), \(\operatorname{Var}(Z)=1/25\).)
Let \(U\sim\operatorname{Beta}(1,3)\). Calculate \(E[U]\), \(E[U^4]\), and \(\operatorname{Var}(U)\). (Answer: \(E[U]=1/4\), \(E[U^4]=1/35\), \(\operatorname{Var}(U)=3/80\).)
Let \(X\sim\operatorname{Beta}(2,3)\) and define \(V=250X\). Calculate the mean and variance of \(V\). (Answer: \(E[V]=100\), \(\operatorname{Var}(V)=2500\).)

What You'll Learn