Converting Factorial Moments to Ordinary Moments

Why Convert Between Moment Types?

For a count \(X\), the ordinary moment \(E[X^k]\) is a familiar summary, but the product \(X(X-1)\cdots(X-r+1)\) can be more convenient in calculations involving counts. These falling factorials vanish whenever the count is less than \(r\), and they arise naturally when counting ordered selections from a set of \(X\) objects. The two kinds of moments describe the same distributional information when the relevant moments exist; Stirling numbers provide the exact conversion.

The conversion is useful in both directions. Factorial moments can make a distribution-specific calculation simpler, after which ordinary moments can be recovered for interpretation. Conversely, ordinary moments already known for a count can be converted into factorial moments. The identities below concern nonnegative integer-valued random variables and integer moment orders.

Falling Factorials and Stirling Numbers

Definition: For an integer \(r\geq 1\), the falling factorial of \(x\) is \((x)_r=x(x-1)\cdots(x-r+1)\), and \((x)_0=1\). If \(X\) is a nonnegative integer-valued random variable, its \(r\)th factorial moment is \(\alpha_r=E[(X)_r]\), when this expectation is finite. Its \(k\)th ordinary raw moment is \(m_k=E[X^k]\), when finite. Set \(\alpha_0=m_0=1\).

For example, \((5)_3=5\cdot4\cdot3=60\), while \(5^3=125\). Factorial moments and ordinary moments are therefore different quantities, even though each is a moment of the same random variable.

Definition: The Stirling number of the second kind \(S(k,r)\) is the number of ways to partition a set of \(k\) labelled elements into \(r\) nonempty, unlabelled groups. Use \(S(0,0)=1\), and set \(S(k,r)=0\) when \(r<0\), \(r>k\), or \(k>0,r=0\). The signed Stirling number of the first kind \(s(k,r)\) is defined by the polynomial identity \((x)_k=\sum_{r=0}^k s(k,r)x^r\); in particular, \(s(0,0)=1\).

The first few Stirling numbers of the second kind give the identities \(x^2=(x)_1+(x)_2\) and \(x^3=(x)_1+3(x)_2+(x)_3\). For the reverse conversion, \((x)_2=x^2-x\) and \((x)_3=x^3-3x^2+2x\). These are polynomial identities, so they hold for every real \(x\), and in particular for every possible value of a count.

The Conversion Identities

Theorem: Let \(X\) be nonnegative integer-valued and let \(k\geq1\). If \(E[X^k]<\infty\), then all factorial moments through order \(k\) are finite and \[ m_k=\sum_{r=0}^k S(k,r)\alpha_r =\sum_{r=1}^k S(k,r)\alpha_r. \] Conversely, if all factorial moments \(\alpha_r\) for \(1\leq r\leq k\) are finite, then \(E[X^k]<\infty\) and the same identity holds. For each \(1\leq k\), whenever these moments are finite, the inverse identity is \[ \alpha_k=\sum_{r=0}^k s(k,r)m_r. \]

Here the \(r=0\) term in the first sum vanishes for \(k\geq1\), because \(S(k,0)=0\). The two finiteness conditions in the theorem are equivalent: for nonnegative integer \(x\), \(0\leq(x)_r\leq x^r\), and the expansion of \(x^k\) in falling factorials has nonnegative coefficients, including coefficient \(1\) on \((x)_k\).

Why the Stirling Identities Hold

The polynomial identity underlying the forward conversion can be obtained by induction. It starts with \(x^0=(x)_0\). The key multiplication rule is \[ x(x)_r=(x)_{r+1}+r(x)_r, \] because \((x)_{r+1}=(x)_r(x-r)\). Multiplying an expansion for \(x^k\) by \(x\) therefore gives the recurrence for the coefficients.

Start with the base case.
\(x^0=(x)_0\), so the coefficient is \(S(0,0)=1\).

Multiply by \(x\).
Use \(x(x)_r=(x)_{r+1}+r(x)_r\) on each term in the expansion.

Collect coefficients.
The coefficient of \((x)_r\) at the next order is \(rS(k,r)+S(k,r-1)\), the recurrence for \(S(k+1,r)\).

For completeness, the recurrence also follows from the partition definition: in a partition of \(k+1\) elements into \(r\) groups, the last element either joins one of the \(r\) existing groups, or forms a singleton while the other \(k\) elements form \(r-1\) groups. Thus \(S(k+1,r)=rS(k,r)+S(k,r-1)\). The base case and this recurrence establish the polynomial identity for every \(k\).

For the inverse identity, start from \((x)_0=1\) and use \((x)_{k+1}=(x-k)(x)_k\). If \((x)_k=\sum_{r=0}^k s(k,r)x^r\), multiplication by \(x-k\) gives \(s(k+1,r)=s(k,r-1)-k\,s(k,r)\). This is the recurrence for the signed Stirling numbers of the first kind and proves the inverse polynomial expansion by induction.

It remains to pass from polynomial identities to expectations. If \(E[X^k]<\infty\), then \(0\leq(X)_r\leq X^r\leq 1+X^k\) for \(1\leq r\leq k\), so the factorial moments in the forward identity are finite. Taking expectations of the finite polynomial expansion gives the stated formula. Conversely, \(X^k=\sum_{r=1}^k S(k,r)(X)_r\) for \(X\geq0\); every summand is nonnegative, so finite factorial moments through order \(k\) imply \(E[X^k]<\infty\). Taking expectations proves the forward identity in this direction as well. The inverse identity follows in the same way from its finite polynomial expansion. \(\square\)

Worked Conversions

Worked Example: Ordinary Moments of a Poisson Count

Let \(X\sim\operatorname{Poisson}(\lambda)\), where \(\lambda>0\). Its probability mass function is \(P(X=x)=e^{-\lambda}\lambda^x/x!\) for \(x=0,1,\ldots\). For \(r\geq1\), terms with \(x<r\) contribute zero to the factorial moment. Re-indexing the remaining sum gives \[ \alpha_r=\sum_{x=r}^{\infty}\frac{x!}{(x-r)!}\frac{e^{-\lambda}\lambda^x}{x!} =\lambda^r e^{-\lambda}\sum_{y=0}^{\infty}\frac{\lambda^y}{y!} =\lambda^r. \] The last sum is \(e^\lambda\), so the cancellation is exact. Also \(\alpha_0=1=\lambda^0\).

For \(k=2\), the polynomial identity is \(x^2=(x)_1+(x)_2\). Hence \(m_2=\lambda+\lambda^2\). For \(k=3\), use \(x^3=(x)_1+3(x)_2+(x)_3\), obtaining \(m_3=\lambda+3\lambda^2+\lambda^3\).

As a numerical check, take \(\lambda=2\). Then \(\alpha_1=2\), \(\alpha_2=4\), and \(\alpha_3=8\). Thus \(E[X^2]=2+4=6\), and \(E[X^3]=2+3(4)+8=22\). Both ordinary moments are finite, as required for the conversion.

Worked Example: Factorial Moments of a Binomial Count

Let \(X\sim\operatorname{Binomial}(n,p)\), where \(n\) is a positive integer and \(0\leq p\leq1\). Its mass function is \(P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}\). For \(1\leq r\leq n\), use \((x)_r\binom{n}{x}=(n)_r\binom{n-r}{x-r}\) and set \(y=x-r\). Then \[ \alpha_r =\sum_{x=r}^{n}(x)_r\binom{n}{x}p^x(1-p)^{n-x} =(n)_r p^r\sum_{y=0}^{n-r}\binom{n-r}{y}p^y(1-p)^{n-r-y} =(n)_r p^r. \] The remaining sum is \(1\), by the binomial theorem. If \(r>n\), then \((X)_r=0\) for every possible \(X\), so \(\alpha_r=0\).

For example, take \(n=5\) and \(p=0.3\). The first three factorial moments are \(\alpha_1=5(0.3)=1.5\), \(\alpha_2=5\cdot4(0.3)^2=1.8\), and \(\alpha_3=5\cdot4\cdot3(0.3)^3=1.62\). Consequently, \(E[X^2]=\alpha_1+\alpha_2=1.5+1.8=3.3\), and \(E[X^3]=\alpha_1+3\alpha_2+\alpha_3=1.5+5.4+1.62=8.52\).

The calculation uses factorial moments to obtain raw moments without summing \(x^2\) or \(x^3\) separately over the binomial mass function.

Worked Example: Recovering Factorial Moments From Raw Moments

Suppose the first four raw moments of a nonnegative integer-valued \(X\) are known. The inverse identities through order four follow by expanding the falling factorials: \((x)_2=x^2-x\), \((x)_3=x^3-3x^2+2x\), and \((x)_4=x^4-6x^3+11x^2-6x\). Taking expectations gives \[ \alpha_2=m_2-m_1,\qquad \alpha_3=m_3-3m_2+2m_1,\qquad \alpha_4=m_4-6m_3+11m_2-6m_1. \]

Use the binomial count from the previous calculation, for which \(m_1=1.5\), \(m_2=3.3\), \(m_3=8.52\), and \(m_4=\alpha_1+7\alpha_2+6\alpha_3+\alpha_4\). Its fourth raw moment is \(m_4=1.5+7(1.8)+6(1.62)+0.972=24.792\), using the already calculated factorial moments and \(\alpha_4=5\cdot4\cdot3\cdot2(0.3)^4=0.972\).

Now check the inverse conversions: \(\alpha_2=3.3-1.5=1.8\); \(\alpha_3=8.52-3(3.3)+2(1.5)=8.52-9.9+3=1.62\); and \(\alpha_4=24.792-6(8.52)+11(3.3)-6(1.5)=24.792-51.12+36.3-9=0.972\). Each result agrees with the direct binomial factorial-moment formula.

Reading and Using the Conversion

The coefficient \(S(k,r)\) says how many ways to partition \(k\) labelled positions into \(r\) nonempty groups. In the identity for \(x^k\), this coefficient weights the falling factorial of order \(r\). The conversion is triangular: the coefficient on \((x)_k\) is \(S(k,k)=1\), and no factorial moment of order greater than \(k\) is needed. The inverse is triangular for the same reason, since \(s(k,k)=1\).

This triangular structure makes the identities useful even when only a limited number of moments are available. For example, a calculation of a fourth raw moment needs factorial moments only through order four. If those factorial moments are easy to obtain from the count's model, the conversion avoids expanding and summing the raw power directly.

Common Mistake: Treating the Two Moments as Identical

A frequent error is to assume that \(\alpha_k=E[(X)_k]\) equals \(m_k=E[X^k]\), or to substitute one for the other in a formula. Already at order two, \((X)_2=X^2-X\), so \(\alpha_2=m_2-m_1\), not \(m_2\). For a Poisson count with \(\lambda=2\), this gives \(\alpha_2=4\) while \(m_2=6\).

The correction is to write the polynomial identity at the required order before taking expectations. Then apply the matching Stirling-number coefficients and check that the moments involved are finite. This prevents a missing lower-order term from changing the result.

Key takeaway: For nonnegative integer-valued \(X\), ordinary and factorial moments through order \(k\) convert through \(S(k,r)\) and \(s(k,r)\). The moments are not interchangeable; the exact coefficients and the required finiteness conditions matter.

Check Your Understanding

Use the polynomial identities and moment conditions developed above to answer these questions.

Write \(x^4\) as a linear combination of \((x)_1,(x)_2,(x)_3,(x)_4\), and use it to express \(m_4\) in factorial moments.
For a Poisson random variable with parameter \(\lambda\), find \(E[X^2]\) and \(E[X^3]\) from its factorial moments.
For \(X\sim\operatorname{Binomial}(4,0.5)\), calculate \(\alpha_2\) and then \(m_2\).
Given \(m_1=2\), \(m_2=7\), and \(m_3=30\), calculate \(\alpha_3\).
Why does finiteness of \(\alpha_1,\ldots,\alpha_k\) imply finiteness of \(E[X^k]\) for a nonnegative integer-valued \(X\)?